The concentration of the potassium permanganate solution is 72 mol/L. The answer has three significant figures because the volume of the flask has only one significant figure.
How The answer was obtainedThe concentration of the potassium permanganate solution can be calculated using the formula:
concentration = moles of solute / volume of solution
where the volume of solution is in liters.
First, we need to convert the volume of the flask from milliliters to liters:
500 mL = 0.500 L
Next, we can calculate the concentration of the solution:
concentration = 36. mol / 0.500 L
concentration = 72 mol/L
The concentration of the potassium permanganate solution is 72 mol/L. The answer has three significant figures because the volume of the flask has only one significant figure.
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Q6. Draw a Lewis dot structure for sulfuric acid, H2SO4, in such a way that the octet rule is obeyed for all atoms except H. What is the formal charge on the sulfur atom?
The Lewis structure of the H2SO4 has been shown in the image attached.
How does the atoms in H2SO4 obey the octet rule?In H2SO4, there are a total of 32 valence electrons available for bonding. The central atom in H2SO4 is sulfur (S), which has 6 valence electrons. To achieve an octet, sulfur needs to form six covalent bonds.
The two hydrogen atoms (H) in H2SO4 each contribute one valence electron to form a single covalent bond with sulfur. This leaves sulfur with 4 valence electrons.
The four oxygen atoms (O) in H2SO4 each contribute 6 valence electrons to form a total of 24 valence electrons in four covalent bonds with sulfur. This brings the total number of valence electrons around sulfur to 28.
To complete the octet, each oxygen atom also has two lone pairs of electrons, bringing the total number of valence electrons around sulfur to 32.
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Each of the properties that follow is a characteristics of the carbon atom. in each case, indicate how the property contributes to the role of the carbon atom as the most important atom in biological molecules.
a. the carbon atom has a valence of four.
b. the carbon-carbon bond has a bond energy that is above the energy of photons of light in the visible range(400-700)
c. carbon is one of the lightest elements to form a covalent bond.
d. carbon can form single, double and triple bonds.
e. the carbon atom is a tetrahedral structure.
Carbon's unique properties such as having a valence of four, the ability to form various types of bonds including double and triple bonds, and its tetrahedral structure.
What are the properties of carbon bonds?
a. The carbon atom's valence of four enables it to form up to four covalent bonds with other atoms, allowing for the formation of diverse organic molecules. This property makes carbon the backbone of many biological molecules, including carbohydrates, lipids, proteins, and nucleic acids.
b. The high bond energy of carbon-carbon bonds makes them stable and resistant to breaking under normal physiological conditions, contributing to the stability of biological molecules. This property allows for the formation of complex macromolecules, such as enzymes and DNA, which are essential to life.
c. Carbon's relatively low atomic weight allows it to form strong covalent bonds without adding significant mass to the molecule. This property is essential for the formation of large and complex biological molecules, which require many carbon atoms to function properly.
d. The ability of carbon to form single, double, and triple bonds allows for the formation of diverse molecular structures, including cyclic structures and branching chains. This property contributes to the diversity of organic molecules found in living organisms, allowing for the creation of molecules with specific functions.
e. The tetrahedral structure of the carbon atom enables it to form strong and stable bonds with other atoms while maintaining a relatively stable geometry. This property is essential for the formation of complex three-dimensional structures in proteins and other biological molecules, allowing them to perform specific functions within cells.
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Menthol is composed of C, H, and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g H2O. What is the empirical and molecular formula for menthol?
The empirical formula, CH2O9(menthol) is multiplied by 5 to get the molecular formula, C10H20O.The empirical and molecular formula for menthol are CH2O and C10H20O, respectively.
The molecular formula for menthol is C5H10O.
This can be determined by dividing the molar mass of the empirical formula (156.26 g/mol) by the molar mass of CO2 (44.01 g/mol). This gives a ratio of 3.55, which is equal to the ratio of C atoms in the empirical formula, C10H20O.
Therefore, the molecular formula is C5H10O.
Given:
Menthol is composed of C, H, and O0.1005g sample of menthol is combusted and produces0.2829g of CO2 0.1159g H2O
1. Find: Empirical and molecular formula for menthol.
Let's first calculate the number of moles of CO2 produced. The balanced equation for combustion of menthol is:
C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)
From the above equation, we can see that for 10 moles of CO2 produced 1 mole of menthol is required.
2. By taking the number of moles of CO2 produced, we can calculate the number of moles of menthol burned.
Moles of CO2 = 0.2829g / 44.01g/mol= 0.00643 mol
C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)
From the balanced equation,1 mole of C10H20O requires 10 moles of CO2
Moles of C10H20O burned = 10 * 0.00643= 0.0643 mol
Next, we can calculate the number of moles of H2O produced.
Moles of H2O = 0.1159g / 18.015g/mol= 0.00643 mol
C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)
From the balanced equation,1 mole of C10H20O requires 10 moles of H2O
Moles of C10H20O burned = 10 * 0.00643= 0.0643 mol
3. Now we can calculate the empirical formula of menthol. The empirical formula can be calculated as follows:
Empirical formula = CH2O (Divide all moles by smallest moles) The molecular weight of CH2O = 30 g/mol
The empirical formula mass of the compound is:
mass = (12.011 + 2*1.008 + 15.999) = 30.026
Empirical formula mass of CH2O is 30.026g/mol, and the given sample weighs 0.1005 g.
The number of empirical formula units in the sample is 0.1005 g / 30.026 g/mol = 0.003348Units.
Empirical formula = CH2OThe empirical formula weight of menthol is CH2O, which is equal to 30.026g/mol.
4. To find the molecular formula, we need to know the molecular weight of the menthol. We can calculate it as follows:
Molecular formula mass = Empirical formula mass x n
Where n = integer Molecular formula mass of menthol is 156 g/mol, and the empirical formula mass is 30.026 g/mol.
So, n = 156 g/mol ÷ 30.026 g/mol = 5.192
Thus the empirical formula, CH2O is multiplied by 5 to get the molecular formula, C10H20O.The empirical and molecular formula for menthol are CH2O and C10H20O, respectively.
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which of the following statements may be true regarding a biochemical oxidation-reduction (redox) reaction?
A few statements may be true regarding a biochemical oxidation-reduction (redox) reaction. The statements are as follows: A redox reaction occurs when there is a transfer of electrons between molecules or atoms.
The electron donor becomes oxidized, and the electron acceptor is reduced, causing a transfer of energy. A redox reaction produces ATP, which is the primary energy currency of the cell. Oxidation and reduction are complementary reactions that occur simultaneously in the same reaction, resulting in the release of energy. Redox reactions are vital in metabolic pathways, and the electron carriers NAD+ and FAD+ are essential in these reactions. Oxygen is frequently used as a final electron acceptor in redox reactions. Redox reactions can also occur in non-cellular environments, such as photosynthesis, respiration, and combustion. The significance of redox reactions is enormous, and they play an essential role in sustaining life on earth. They help in generating energy, breaking down complex molecules, synthesizing molecules, and many other cellular processes.
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The activation energy, Ea, for a particular reaction is 13.6 kJ/mol. If the rate constant at 475 K is 0.0450 1/min, then what is the value of the rate constant at 769 K? (R = 8.314 J/mol • K)
At 769 K, the rate constant equals 2.22 1/min.
When the reaction's EA is zero, what is the reaction's rate constant equal to?The final expression is either k=A or k=A. This implies that the response rate will be equal to the value of the collision frequency rather than the temperature when the activation energy is zero.
The Arrhenius equation, which connects the rate constant (k) to the activation energy (Ea) and temperature (T), can be used to solve this issue:
[tex]A = * exp (-Ea / (R * T))[/tex]
With the rate constant (k) at 475 K, we can utilize this knowledge to calculate the pre-exponential factor (A) as follows:
0.0450 1/min = A * exp(-13.6 kJ/mol / (8.314 J/mol•K * 475 K))
[tex]A = 5.74 x 10^9 min^-1[/tex]
The rate constant (k) at 769 K can now be calculated using the Arrhenius equation once more as follows:
[tex]k = 5.74 x 10^9 min^-1[/tex] * exp(-13.6 kJ/mol / (8.314 J/mol•K * 769 K))
k = 2.22 1/min (rounded to two significant figures)
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the denaturing of proteins occurs when the stabilizing forces are altered. below is the set of processes in which proteins are denatured.
The correct answer is bleaching of hair with hydrogen peroxide, curdling of milk by rennin and straightening frizzy hair by conditioner.
What is protein denaturation?Protein denaturation is the process by which a protein loses its shape and function due to exposure to certain environmental conditions such as heat, pH, pressure, or chemical agents. The denaturation process can cause the protein's structure to unravel, leading to the loss of its biological activity. This happens because the protein's three-dimensional structure is crucial to its function, and when the structure is altered, the protein may no longer be able to perform its intended function.
Protein denaturation can occur reversibly or irreversibly depending on the extent of the damage to the protein's structure. Reversible denaturation occurs when the protein regains its structure and function once the environmental stressor is removed, while irreversible denaturation occurs when the damage is permanent and the protein cannot regain its function.
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The full question is:
The denaturing of proteins occurs when the stabilizing forces are altered. Below is the set of processes in which proteins are denatured. Write the chemicals necessary for the denaturation to occur
1. Bleaching of hair
2. Curdling of milk
3. Straightening frizzy hair
Which of the following would INITIATE a Sea Beeze?
A.
Pressure Differences
B.
Temperature Differences
C.
Differences in Friction
D.
Air Mass Differences
Consider the following silica gel TLC plate of compounds A, B, and C developed in hexanes:
Consider the following silica gel TLC plate of com
a) Determine the R f values of compounds A, B, and C run on a silica gel TLC plate using hexanes as the solvent
b) Which compound, A, B, or C, is the most polar?
c) What would you expect to happen to the R f values if you used acetone instead of hexanes as the eluting solvent? (Think polarity of solvents)
The R f values for compounds A, B, and C on a silica gel TLC plate developed in hexanes would be determined by measuring the distance each compound traveled compared to the distance the solvent traveled.
a) There is a 4 cm gap between the origin and the solvent front. The Rf value for spot A is[tex]\frac{1.5}{4}= 0.375[/tex], because it travelled 1.5 cm. Due to the 3.5 cm movement of Spot B, its Rf is[tex]\frac{3.5}{4} = 0.875[/tex]. Spot C shifted 3 cm, making its Rf [tex]\frac{3}{4} = 0.75[/tex].
b)Due to its shorter travel distance than the other two compounds, compound A is the most polar. Recall that polar substances adhere to the adsorbent more readily, move less, and have a lower Rf value.
c)Hexanes is less polar than acetone as a solvent. Each of the three compounds would move more quickly if the same method were employed to elute them.The chemicals can be removed from the polar adsorbent more effectively with a more polar eluting solvent. Each compound would have a higher Rf value if acetone were used to elute the TLC plate as opposed to hexanes because each compound travels more quickly.
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State whether M=[-1 -4] has an inverse. If the inverse exists, find it.
No, the inverse of the matrix represented as M=[-1 -4] does not exist.
What is an inverse matrix?An inverse matrix is a square matrix that, when multiplied by its original matrix, yields the identity matrix. It allows for solving linear equations involving the original matrix.
To determine if a matrix has an inverse, we can compute its determinant. If the determinant is nonzero, then the matrix has an inverse; if the determinant is zero, then the matrix does not have an inverse.
The given matrix M is a 1x2 matrix, so it's not square and it doesn't have an inverse.
Therefore, we can't find the inverse of matrix M.
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The activation energy Ea for a particular reaction is 50.0 kJ/mol. How much faster is the reaction at 314 K than at 310.0 K? (R = 8.314 J/mol • K)
In comparison to 310.0 K, the reaction happens 1.28 times faster at 314 K.
A higher K value: what does it mean?A high K value (higher than 1) denotes an equilibrium with more products than reactants, whereas a low K value (less than 1) denotes an equilibrium with more reactants than products.
The Arrhenius equation, which connects the rate constant (k) to the activation energy (Ea) and temperature (T), can be used to determine how much faster the reaction happens at 314 K than it does at 310.0 K. k = A * exp(-Ea / (R * T))
where T is the temperature in Kelvin, R is the gas constant, and A is the preexponential factor.
For this reaction, we can assume that the pre-exponential component is fixed and that the sole variable is temperature.
exp[(Ea / R) * (1/T1 - 1/T2)] = k2 / k1
where Ea is the activation energy, R is the gas constant, k1 is the rate constant at 310.0 K, and k2 is the rate constant at 314 K.
k2 / k1 = exp[(50.0 kJ/mol / (8.314 J/mol•K)) * (1/310.0 K - 1/314 K)] is the result of substituting the provided numbers.
If we condense this phrase, we get:
k2 / k1 = 1.28
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Reaction:
N2 + 3H2 ------> 2NH3
Question 1: Calculate the mass of N2 needed to react with 10 g of H2
Question 2: Calculate the mass of N2 needed to produce 15 g of NH3
Explanation:
The reactant contains 2N and 6H
The product contains 2N and 6H
Therefore, the chemical equation is balanced
From the equation, for every 1 mole of N2 that reacts, 3 moles of H2 are required.
We know 28.6 grams of N2 reacted, but we don’t know the mass ratio but just the mole ratio, so we have to convert 28.6 grams of N2 to the corresponding moles of N2.
From the periodic table, the molar mass of N is about 14 g/mol, so the molar mass of nitrogen gas or N2 is two times of that which is 28 g/mol.
With this, we can calculate moles of N2, but we also need to make sure the equation is setted up the right way.
Looking at the units, if we cancel out the grams, we are left with mol. We also know that in multiplication, numerator of one number cancel with the denominator of another number and vice versa
So the equation looks like this 28.6g * mol/28g = 1.021 mol N2
So the number of moles of H2 required is 1.021 mol N2 * 3 mol H2/1 mol N2 = 3.063 mol H2 (notice that mol N2 canceled out, so the equation is set up correctly)
However, the question ask for number of grams of H2 needed, so we need the molar mass of hydrogen gas or H2, which is 1*2 = 2 g/mol
3.063 mol H2 * 2 g H2/ mol H2 = 6.126 g H2
Ans: 6.126 g H2
1. Choose the atom with the smaller atomic size.
Select one:
a. Nitrogen
b. Bismuth
2. Choose the atom with the smaller atomic size.
Select one:
a. Arsenic
b. Bromine
Its atomic radius increases form top to bottom inside a group, then decreases from left and right across a period. As a result, francium is indeed the largest element while helium is the smallest.
Which atomic size has the smaller diameter?Atomic radii inside the periodic table decrease across a row form left to right and increase across a column from top to bottom. Due to these two patterns, the periodic table's lower left and upper right corners, respectively, contain the largest and smallest atoms.
Which atom is the smallest?The atomic radius grows form top to bottom inside a group and decreases form left to right during a period, as seen in the images below. As a result, francium is indeed the largest element while helium is the smallest.
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a. The atom with the smaller atomic size is: Nitrogen
a. The atom with the smaller atomic size is: Arsenic.
How is atomic size of elements calculated?Atomic size, also known as atomic radius, is the distance between the nucleus of an atom and its outermost electrons. It is typically measured in picometers (pm) or angstroms (Å). The atomic size of an element can be calculated by finding the distance between the nucleus and the outermost electron shell of an atom of that element. This distance can be determined using various methods, including X-ray diffraction and spectroscopy. The atomic size of elements generally decreases from left to right across a period and increases from top to bottom down a group in the periodic table.
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1. Use the bond enthalpies to calculate the enthalpy change for this reaction. Is the reaction exothermic or endothermic?
C2H6O + 3 O2 = 2 CO2 + 3 H2O
The enthalpy change of the reaction is 7,227 kJ/mol and the reaction is endothermic.
What is the enthalpy change of the reaction?To calculate the enthalpy change of the reaction using bond enthalpies, we need to find the total energy required to break the bonds in the reactants and the energy released when new bonds are formed in the products.
The enthalpy change of the reaction can be calculated as follows:
Reactants:
1 mole of C2H6O requires breaking 2 C-H bonds and 1 C-O bond.
3 moles of O2 requires breaking 3 O=O bonds.
Products:
2 moles of CO2 releases forming 4 C=O bonds.
3 moles of H2O releases forming 6 O-H bonds.
The bond enthalpies for the relevant bonds are:
C-H = 413 kJ/molC-O = 358 kJ/molO=O = 495 kJ/molC=O = 745 kJ/molO-H = 467 kJ/molThe enthalpy change for the reaction is:
(2 × 745 kJ/mol) + (3 × 6 × 467 kJ/mol) - (2 × 413 kJ/mol) - (1 × 358 kJ/mol) - (3 × 495 kJ/mol) = 7,227 kJ/mol
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How do the number of collisions affect the size of the balloon?
Answer:
As the number of gas particles increases, the frequency of collisions with the walls of the container must increase. This, in turn, leads to an increase in the pressure of the gas. Flexible containers, such as a balloon, will expand until the pressure of the gas inside the balloon once again balances the pressure of the gas outside.
Explanation:
What is the pH of 0.335 M trimethylammonium chloride, (CH3)3NHCI? The Kb of trimethylamine, (CH3)3N, is 6.3 x 10-5. (value = 0.02)
The pH of 0.335 M trimethylammonium chloride, (CH3)3NHCI is approximately 5.676.
What is pH?
To find the pH of the solution, we need to first determine if (CH3)3NHCI acts as an acid or base. Since (CH3)3NHCI is a salt composed of a weak base, trimethylamine, and a strong acid, hydrochloric acid (HCI), it will undergo hydrolysis in water.
The hydrolysis reaction is given by:
(CH3)3NH+ (aq) + H2O (l) ⇌ (CH3)3N (aq) + H3O+ (aq)
The Kb expression for the equilibrium reaction is:
Kb = [ (CH3)3N ] [ H3O+ ] / [ (CH3)3NH+ ]
Since (CH3)3NH+ and HCl dissociate completely in water, the initial concentration of (CH3)3NH+ is equal to the concentration of (CH3)3NHCI, which is 0.335 M.
Using the Kb value given, we can solve for the concentration of H3O+:
Kb = 6.3 x [tex]10^{-5}[/tex] = [ (CH3)3N ] [ H3O+ ] / 0.335
[ H3O+ ] = Kb x (CH3)3NH+ / (CH3)3N
[ H3O+ ] = 6.3 x [tex]10^{-5}[/tex] x 0.335 / 1
[ H3O+ ] = 2.1095 x [tex]10^{-6}[/tex] M
Finally, we can calculate the pH using the expression:
pH = -log [H3O+]
pH = -log (2.1095 x [tex]10^{-6}[/tex])
pH = 5.676
Therefore, the pH of the solution is approximately 5.676.
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Complete question is: The pH of 0.335 M trimethylammonium chloride, (CH3)3NHCI is approximately 5.676.
1) what is the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29% oxygen? worksheet
The empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29% oxygen is CH2O.
To calculate this, you need to first convert the percentage composition into mass composition. This is done by multiplying the percentages by the molecular weight of each element.
Carbon: 65.5% x 12 g/mol = 0.786 g/mol
Hydrogen: 5.5% x 1 g/mol = 0.055 g/mol
Oxygen: 29% x 16 g/mol = 0.464 g/mol
Now that you have the mass composition, you can calculate the moles of each element by dividing the mass of each element by its molar mass.
Carbon: 0.786 g/mol / 12 g/mol = 0.065 mol
Hydrogen: 0.055 g/mol / 1 g/mol = 0.055 mol
Oxygen: 0.464 g/mol / 16 g/mol = 0.029 mol
Finally, divide each element's moles by the smallest moles to get the empirical formula.
Carbon: 0.065 mol / 0.029 mol = 2.24 = 2 mol
Hydrogen: 0.055 mol / 0.029 mol = 1.90 = 1 mol
Oxygen: 0.029 mol / 0.029 mol = 1.00 = 1 mol
Therefore, the empirical formula of the molecule is CH2O.
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How many calories are required to raise the temperature of a 35.0 g sample from 35 °C to 85 °C? The sample has a specific heat of 0.108 cal/g °C.
Answer:
First, we need to calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 85 °C - 35 °C
ΔT = 50 °C
Next, we can use the following formula to calculate the heat energy required:
Q = m·C·ΔT
where Q is the heat energy in calories, m is the mass of the sample in grams, C is the specific heat in cal/g °C, and ΔT is the change in temperature in °C.
Plugging in the given values, we get:
Q = 35.0 g · 0.108 cal/g °C · 50 °C
Q = 189.0 calories
Therefore, 189.0 calories are required to raise the temperature of the sample from 35 °C to 85 °C
!!!50 points!!!
Problem 1. What masses of 15% and 20% solutions are needed to prepare 200 g of 17% solution?
Problem 2. What masses of 18% and 5% solutions are needed to prepare 300 g of 7% solution?
Problem 3. 200 g of 15% and 350 g of 20% solutions were mixed. Calculate mass percentage of final solution.
Problem 4. 300 g of 15% solution and 35 g of solute were mixed. Calculate mass percentage of final solution.
Problem 5. 400 g of 25% solution and 150 g of water were mixed. Calculate mass percentage of final solution.
Answer:
See Below.
Explanation:
Problem 1
Let x be the mass of 15% solution needed and y be the mass of 20% solution needed. Then, we have the following system of equations:
x + y = 200 (total mass of solution)
0.15x + 0.20y = 0.17(200) (total amount of solute)
Solving this system of equations gives:
x = 60 g (mass of 15% solution)
y = 140 g (mass of 20% solution)
Therefore, 60 g of 15% solution and 140 g of 20% solution are needed to prepare 200 g of 17% solution.
Problem 2
Let x be the mass of 18% solution needed and y be the mass of 5% solution needed. Then, we have the following system of equations:
x + y = 300 (total mass of solution)
0.18x + 0.05y = 0.07(300) (total amount of solute)
Solving this system of equations gives:
x = 120 g (mass of 18% solution)
y = 180 g (mass of 5% solution)
Therefore, 120 g of 18% solution and 180 g of 5% solution are needed to prepare 300 g of 7% solution.
Problem 3
The total mass of the final solution is
200 g + 350 g = 550 g
The total amount of solute in the final solution is:
0.15(200 g) + 0.20(350 g) = 95 g + 70 g = 165 g
Therefore, the mass percentage of the final solution is:
(mass of solute / total mass of solution) x 100% = (165 g / 550 g) x 100% = 30%
Therefore, the mass percentage of the final solution is 30%.
Problem 4
The total mass of the final solution is
300 g + 35 g = 335 g
The total amount of solute in the final solution is:
0.15(300 g) + 35 g = 75 g + 35 g = 110 g
Therefore, the mass percentage of the final solution is:
(mass of solute / total mass of solution) x 100% = (110 g / 335 g) x 100% = 32.8%
Therefore, the mass percentage of the final solution is 32.8%.
Problem 5
The total mass of the final solution is
400 g + 150 g = 550 g
The total amount of solute in the final solution is
0.25(400 g) = 100 g
Therefore, the mass percentage of the final solution is
(mass of solute / total mass of solution) x 100% = (100 g / 550 g) x 100% = 18.2%
Therefore, the mass percentage of the final solution is 18.2%.
Can some please help with the picture below
The completed table of maximum moles of water, limiting reactant and excess reactant is as follows:
Q: 6 moles, O₂, 1 mole H₂
R: 6 moles, O₂, 2 moles H₂
S: 5 moles, none, none
T: 5 moles, H₂, 2.5 moles O₂
U: 8 moles, H₂, 2 moles O₂
What is the mole ratio of the reaction of hydrogen and oxygen to form water?The mole ratio of the reaction of hydrogen and oxygen to form water is obtained from the equation of the reaction.
The equation of the reaction is given below:
2 H₂ + O₂ --> 2 H₂O
The mole ratio of hydrogen to oxygen is 2:1 in both the water molecule and the reactants, hydrogen gas (H2) and oxygen gas, as can be seen from the balanced equation (O2).
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For the following reaction, which of the reactants would be the acid?
HNO2 ( aq ) + HS - ( aq ) → NO2 - ( aq ) + H2S ( aq )
Select one:
a.
HS -
b.
H2O
c.
NO2 -
d.
HNO2
(Chem 2 Quiz 3.1)
The acid in the reaction would donate a proton and that would be HNO2.
How do you know an acid in a reaction?An acid in a chemical reaction can be identified by the presence of hydrogen ions (H+): Acids are compounds that produce hydrogen ions when dissolved in water. In a chemical reaction, an acid may donate a hydrogen ion to another compound or accept a pair of electrons from a base.
When we look at the reaction, we can see that the specie that has given out the replaceable hydrogen ion is HNO2 thus it is the acid in the reaction.
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Look at the picture below
The claim was correct . All elements have same number of particles in one mole and have different number of particles in a mole based on atomic number .
What is mole ?In the International System of Units, the mole (symbol mol) is the unit of substance amount (SI). The amount of substance is a measurement of how many elementary entities of a given substance are present in an object or sample. An elementary entity can be an atom, a molecule, an ion, an ion pair, or a subatomic particle such as an electron, depending on the substance. For example, despite having different volumes and masses, 10 moles of water (a chemical compound) and 10 moles of mercury (a chemical element) contain equal amounts of substance, and the mercury contains exactly one atom for each molecule of water.
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2. Hydrogen bromide reacts with propene to form either 1-bromopropane or 2-bromopropane. Explain why
2-bromopropane is the major product.
3. Explain how the reaction with bromine can be used to test for an alkene. Include the mechanism for the reaction between hex-1-ene and bromine in your answer.
a) Describe the process of addition polymerisation.
b) Show the repeating unit of the polymer that is formed from the addition polymerisation of chloroethene monomers. Name and give at least one use for this polymer.
Answer:
Explanation:
2-bromopropane is the major product because the reaction mechanism involves the formation of the most stable carbocation intermediate. When hydrogen bromide reacts with propene, the hydrogen atom from HBr adds to the carbon atom of the double bond that has fewer hydrogen atoms attached, resulting in the formation of a carbocation intermediate. The intermediate can either form 1-bromopropane or 2-bromopropane depending on the position of the carbocation. The 2-bromopropane is the major product because the secondary carbocation formed in this case is more stable than the primary carbocation formed in the case of 1-bromopropane.
To test for an alkene, bromine water can be used. When an alkene reacts with bromine water, the bromine molecule adds across the double bond, forming a colorless dibromoalkane product. The mechanism for the reaction between hex-1-ene and bromine involves the formation of a cyclic bromonium ion intermediate, followed by the attack of water on the intermediate, resulting in the formation of the dibromoalkane product.
a) Addition polymerization is a process in which unsaturated monomers are joined together to form a polymer. The process involves breaking the double bond of the monomer and joining the monomers together to form a long-chain polymer. The process requires a catalyst to initiate the reaction.
b) The repeating unit of the polymer formed from the addition polymerization of chloroethene monomers is -CH2-CHCl-. This polymer is called polyvinyl chloride (PVC), and it has a wide range of uses, including pipes, electrical cables, and vinyl flooring.
Ethanol (C2H5OH) boils at a temperature of 78.3 degrees C. What amount of energy, in joules, is necessary to heat to boiling and then completely vaporize a 13.1 g sample of ethanol initially at a temperature of 11.1 degrees C? The specific heat of ethanol is approximately constant at 2.44 JK−1g−1. The heat of vaporization of ethanol is 38.56 kJ mol−1.
The total amount of energy necessary to heat and vaporize a 13.1 g sample of ethanol initially at a temperature of 11.1 degrees C is 7.15 kJ.
To calculate the amount of energy, in joules, necessary to heat and vaporize a 13.1 g sample of ethanol initially at a temperature of 11.1 degrees C, we must first calculate the heat necessary to heat the sample to the boiling point of ethanol, 78.3 degrees C. The formula to calculate the amount of energy is: Q = mcΔT, where m is the mass of the sample, c is the specific heat of ethanol, and ΔT is the temperature change from 11.1 degrees C to 78.3 degrees C. Thus, the amount of energy necessary to heat the sample is: Q = 13.1 g * 2.44 JK−1g−1 * (78.3-11.1) = 1,623.08 J.
Next, we must calculate the amount of energy necessary to completely vaporize the sample. To do so, we must use the heat of vaporization of ethanol, which is 38.56 kJ mol−1. To convert from moles to grams, we must use the molar mass of ethanol, which is 46 g/mol. Thus, the amount of energy necessary to vaporize the sample is: Q = (13.1 g/46 g/mol) * 38.56 kJ/mol = 7.15 kJ.
Finally, to calculate the total amount of energy necessary to heat and vaporize the sample, we must add the two values together: Q = 1,623.08 J + 7.15 kJ = 7.15 kJ. the total amount of energy necessary to heat and vaporize a 13.1 g sample of ethanol initially at a temperature of 11.1 degrees C is 7.15 kJ.
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Molar Volume of Hydrogen continued volume of hydrogenhydrogen gas at STP by the theoretical number of moles of hydrogen to calculate the molar ume of hydrogen fo 4. Divide the volume of r Trials 1 and 2 Results Table Number of moles of H, gas Vapor pressure of water Partial pressure of H2 gas Calculated volume of H2 gas at STP Molar volume of H2 gas Average molar volume 5. What is the average value of the molar volume of hydrogen? Look up the literature value of the molar volume of a gas and calculate the percent error in your experimental determination of the molar volume of hydrogen. l Experimental value - Literature value I Literature value x 100% Percent error 6. One mole of hydrogen gas has a mass of 2.02 g. Use your value of the molar volume of hydrogen to calculate the mass of one liter of hydrogen gas at STP This is the density of hydrogen in g/L. How does this experimental value of the density compare with the literature value? (Consult a chemistry handbook for the density of hydrogen.) Laboratory Experiments for Geทeral, Organic and Biolo Molar Volume of Hydrogen continued 7. In setti e water bath. What effect would this have on the measured volume of hydrogen gas? Would the c r voltume of hydrogen be too high or too low as a result of this error? Explain. invertenx u) this experiment, a student noticed that a bubble of air leaked into the graduated cylinder when it was d in the te 8. A student noticed that the silver and shiny. Wh magnesium ribbon appeared to be oxidized-the metal surface was black and dull rather at effect would this error have on the measured volume of hydrogen gas? Would the cal than culated molar volume of hydrogen be too high or too low as a result of this error? Explain. 9. (Optional) Your instructor wants to scale up this experiment for demonstration purposes and would like to collect the gas in an inverted 50-mL, buret at room temperature. Use the ideal gas law to calculate the maximum amount or length of magnesium ribbon that may be used. Laboratory Experiments for General, Organic and Biological Cbemistry7
The average value of the molar volume of hydrogen is 24.0 liters per mole (L/mol).
To calculate the percent error in the experimental determination of the molar volume of hydrogen, you must subtract the experimental value from the literature value and divide by the literature value.
Then, multiply this result by 100% to obtain the percent error.One liter of hydrogen gas at STP has a mass of 0.090 grams, which is the experimental value of the density of hydrogen. This value is lower than the literature value, which is 0.089 grams per liter (g/L).
In this experiment, if a bubble of air leaked into the graduated cylinder when it was placed in the water bath, the calculated molar volume of hydrogen would be too high as a result of this error.
This is because the presence of the bubble of air would increase the measured volume of hydrogen gas.If the magnesium ribbon appeared to be oxidized, the calculated molar volume of hydrogen would be too low as a result of this error.
This is because the oxidation of the magnesium ribbon would reduce the amount of hydrogen gas produced, resulting in a lower measured volume of hydrogen gas. For demonstration purposes, the ideal gas law may be used to calculate the maximum amount or length of magnesium ribbon that may be used.
The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the amount of substance, R is the ideal gas constant, and T is the temperature. Knowing the desired volume of the gas, the amount of substance can be calculated.
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How many moles are in 3.19 × 1016 molecules of NOs?
There are approximately 0.005302 moles of NOs in 3.19 × 10^16 molecules.
What is moles ?
Mole is an SI unit used to measure the amount of any substance.
To calculate the number of moles of NOs in 3.19 × 10^16 molecules, we need to use Avogadro's number, which is 6.022 × 10^23 molecules per mole.
First, we need to convert the number of molecules to moles using the formula:
moles = molecules / Avogadro's number
moles of NOs = 3.19 × 10^16 molecules / 6.022 × 10^23 molecules per mole
moles of NOs = 0.005302 moles (rounded to 4 significant figures)
Therefore, there are approximately 0.005302 moles of NOs in 3.19 × 10^16 molecules.
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The number of moles present in 3.19×10¹⁶ molecules of nitrogen dioxide, NO₂ is 5.30×10⁻⁸ mole
How do i determine the number of moles present?The number of moles present in 3.19×10¹⁶ molecules of NO₂ can be obtained by using the Avogadro's hypothesis as illustrated below:
Number of molecules = 3.19×10¹⁶ moleculesNumber of mole of NO₂ =?From Avogadro's hypothesis,
6.022×10²³ molecules = 1 mole of NO₂
Therefore,
3.19×10¹⁶ molecules = 3.19×10¹⁶ / 6.022×10²³
3.19×10¹⁶ molecules = 5.30×10⁻⁸ mole of NO₂
Thus, we can conclude that the number of mole is 5.30×10⁻⁸ mole
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Classify each of the following as an exothermic or endothermic reaction. Drag the appropriate statements to their respective bins. Reset Help The energy level of the reactants is lower than that of the products. The combustion of wood provides energy A reaction releases 101 J. Exothermic Endothermic Submit Request Answer
1. The energy level of the reactants is lower than that of the products. - Endothermic
2. The combustion of wood provides energy - Exothermic
3. A reaction releases 101 J. - Exothermic
What is an Endothermic?
An endothermic process or reaction is a chemical or physical process that absorbs heat from the surroundings. In an endothermic process, energy is required to break the bonds in the reactants and form new bonds in the products. As a result, the temperature of the surroundings decreases, and the process feels cold to the touch. Examples of endothermic processes include melting ice, cooking an egg, and evaporating water.
What is an Exothermic?
An exothermic process or reaction is a chemical or physical process that releases heat into the surroundings. In an exothermic process, energy is released as the bonds in the reactants are broken and new bonds are formed in the products. As a result, the temperature of the surroundings increases, and the process feels warm to the touch. Examples of exothermic processes include burning a candle, combustion reactions, and the reaction between baking soda and vinegar.
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How many electrons can occupy the following sub-shells: (a) 1s, (b) 3p, (c) 3d, and (d) 6g?
The maximum number of electrons that can occupy the 1s sub-shell is 2, the 3p sub-shell is 6, the 3d sub-shell is 10, and the 6g sub-shell is 32.
For the 1s sub-shell, due to the Pauli Exclusion Principle, two electrons of opposite spin can exist in the same orbital. This means that there is a maximum of two electrons that can occupy the 1s sub-shell. For the 3p sub-shell, three orbitals exist with a maximum of two electrons each. Since two electrons of opposite spin can occupy each orbital, there is a maximum of six electrons that can occupy the 3p sub-shell. For the 3d sub-shell, five orbitals exist with a maximum of two electrons each. Since two electrons of opposite spin can occupy each orbital, there is a maximum of 10 electrons that can occupy the 3d sub-shell. Finally, for the 6g sub-shell, seven orbitals exist with a maximum of two electrons each. Since two electrons of opposite spin can occupy each orbital, there is a maximum of 32 electrons that can occupy the 6g sub-shell.
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Which of the following is the major organic product of the condensation of ammonia or a primary amine with the carbonyl group of an aldehyde or ketone?
Imine
The major organic product of the condensation of ammonia or a primary amine with the carbonyl group of an aldehyde or ketone is an imine.
A functional group or organic substance with a carbon-nitrogen double bond (C=N) is known as an imine. A hydrogen atom or an organic group may be joined to the nitrogen atom. (R). The carbon atom is connected to two more single bonds. Imines are present in numerous processes and are frequently found in manufactured and naturally occurring chemicals.
The five core atoms for ketimines and aldimines, C2C=NX and C(H)C=NX, respectively, are coplanar. The sp2-hybridization of the mutually double-bonded nitrogen and carbon atoms yields planarity. For nonconjugated imines, the C=N distance is 1.29-1.31, whereas for conjugated imines, it is 1.35. The C-N distances in amines and nitriles, on the other hand, are 1.47 and 1.16, respectively. Slow rotation occurs around the C=N bond. E- and Z-isomers were detected using NMR spectroscopy of aldimines have been detected. Owing to steric effects, the E isomer is favored.
An imine is formed when a primary amine reacts with a carbonyl group (C=O) of an aldehyde or ketone to form a new C-N bond. This reaction is known as a condensation reaction, as it involves the loss of a small molecule (e.g. water) to form the product.
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The correct questions is :
What is the major organic product of the condensation of ammonia or a primary amine with the carbonyl group of an aldehyde or ketone?
which of the following describes an experimental technology being used to reduce carbon dioxide emissions from coal?
Carbon capture and storage is one experimental method being utilised to lower carbon dioxide emissions from coal (CCS).
One experimental technique being used to reduce carbon dioxide emissions from coal is carbon capture and storage (CCS). With CCS, carbon dioxide emissions from factories or power plants are captured and either stored underground in geological formations or used to improve oil recovery.
Coal and other fossil fuels have the potential to drastically cut their carbon dioxide emissions, but CCS technology currently in the experimental stage. Unfortunately, because of its expensive cost and technical implementation difficulties, the technology is not yet extensively employed. In order to address the current climate problem, efforts to cut CO2 emissions are essential.
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What experimental technology is being used to reduce carbon dioxide emissions from coal?
2.75g NaCL is present
in 650g of Water. Is the solution saturated or unsaturated?
(Solubility of wäter is 0.33%6)
Answer:
To determine if the solution is saturated or unsaturated, we need to compare the amount of NaCl in the solution to its solubility in water at the given temperature.
The solubility of NaCl in water at room temperature (25°C) is approximately 36 grams per 100 grams of water, or 0.36 g/g. This means that at 25°C, water can dissolve up to 0.36 grams of NaCl per gram of water.
In this case, we have 2.75 grams of NaCl dissolved in 650 grams of water. To find the concentration of NaCl in the solution, we divide the mass of NaCl by the total mass of the solution:
concentration of NaCl = mass of NaCl / total mass of solution
concentration of NaCl = 2.75 g / (2.75 g + 650 g)
concentration of NaCl = 0.0042 g/g
Comparing the concentration of NaCl in the solution to its solubility in water at 25°C, we see that:
0.0042 g/g < 0.36 g/g
Since the concentration of NaCl in the solution is less than its solubility in water, the solution is unsaturated.
Explanation: