Answer:
compressable
Explanation:
as liquid nitrogen came out from container the force exerted on it which changes it into liquid ends making it gas
liquid nitrogen boils at 77 K or -196° C
Name the following compound: Cuzs
O sulfur copperide (ll)
O sulfur copperide (1)
O copper(I) sulfide
copper(ll) sulfide
Answer:
THE ANSWER IS: copper(I) sulfide.
hope this helped <3
Explanation:
What happens to the concentration of hydron What happens to the pH of a buffer when a small amount of acid is added? It will decrease by 10 points. It will increase by 10 points. It will stay about the same.ium ions as the pH of a solution increases? Disabled A. hydronium ion concentration increases Student Selected Incorrect B. hydronium ion concentration stays the same Disabled C. hydronium ion concentration decreases
Answer:
B
Explanation:
the concentration will be the same disabled
Write a formula for the ionic compound that forms from magnesium
and oxygen.
Answer:
MgO
Explanation:
One source of aluminum metal is alumina, Al2O3. a. Determine the percentage composition of Al in alumina. b. How many pounds of aluminum can be extracted from 2.0 tons of alumina.?
Answer:
sorry i cant give the answer but you can gi end check in answer sheet of this becoz i had same question in exam in chemistry so i revised then i checked the answer if you want to check answer go to www.coachscotchemistry.com there you xan find the answer
Balance each of the following equations. Then, drag and drop each equation to match the coefficient of H2O in the balanced chemical equation. A coefficient for water may be used once, more than once, or not at all. Drag and drop your selection from the following list to complete the answer:
C2H5OH + O2 + CO2 + H2O NH3 + O2 + NO2 + H20 C3H2 + O2 + CO2 + H2O H2SO4 + NaOH → Na2SO4 + H20 NO2 + H2O → HNO3 + NO
Answer:
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
3 NO₂ + H₂O → 2 HNO₃ + NO
Explanation:
We will balance the equation using the trial and error method.
C₂H₅OH + O₂ → CO₂ + H₂O
1) We balance C atoms by multiplying CO₂ by 2 and H atoms by multiplying H₂O by 3.
C₂H₅OH + O₂ → 2 CO₂ + 3 H₂O
2) We balance O atoms by multiplying O₂ by 3.
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
NH₃ + O₂ → NO₂ + H₂O
1) We balance H atoms by multiplying NH₃ by 2 and H₂O by 3.
2 NH₃ + O₂ → NO₂ + 3 H₂O
2) We balance N atoms by multiplying NO₂ by 2.
2 NH₃ + O₂ → 2 NO₂ + 3 H₂O
3) We balance O atoms by multiplying O₂ by 3.5
2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O
C₃H₈ + O₂ → CO₂ + H₂O
1) We balance C atoms by multiplying CO₂ by 3 and H atoms by multiplying H₂O by 4.
C₃H₈ + O₂ → 3 CO₂ + 4 H₂O
2) We balance O atoms by multiplying O₂ by 5.
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
H₂SO₄ + NaOH → Na₂SO₄ + H₂O
1) We balance Na atoms by multiplying NaOH by 2.
H₂SO₄ + 2 NaOH → Na₂SO₄ + H₂O
2) We balance H and O atoms by multiplying H₂O by 2.
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
NO₂ + H₂O → HNO₃ + NO
1) We balance H atoms by multiplying HNO₃ by 2.
NO₂ + H₂O → 2 HNO₃ + NO
2) We balance N atoms by multiplying NO₂ by 3.
3 NO₂ + H₂O → 2 HNO₃ + NO
Epinephrine (adrenaline) is a hormone secreted into the bloodstream in times of danger and stress. It is 59.0% carbon, 7.15% hydrogen, 26.20% oxygen, and 7.65% nitrogen by mass and has a molar mass of 183 g/mol. Determine the empirical formula for Epinephrine.
Answer: The empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]
Explanation:
Let the mass of the compound be 100 g
Given values:
% of C = 59.0%
% of H = 7.15%
% of O = 26.20%
% of N = 7.65%
Mass of C = 59.0 g
Mass of H = 7.15 g
Mass of O = 26.20 g
Mass of N = 7.65 g
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol
Molar mass of N = 14 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of C}=\frac{59.0g}{12g/mol}=4.917 mol[/tex]
[tex]\text{Moles of H}=\frac{7.15g}{1g/mol}=7.15 mol[/tex]
[tex]\text{Moles of O}=\frac{26.20g}{16g/mol}=1.6375 mol[/tex]
[tex]\text{Moles of N}=\frac{7.65g}{14g/mol}=0.546 mol[/tex]
Step 2: Calculating the mole ratio of the given elements.Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.546 moles
[tex]\text{Mole fraction of C}=\frac{4.917}{0.546 }=9[/tex]
[tex]\text{Mole fraction of H}=\frac{7.15}{0.546 }=13[/tex]
[tex]\text{Mole fraction of O}=\frac{1.6375}{0.546 }=2.99\approx 3[/tex]
[tex]\text{Mole fraction of N}=\frac{0.546}{0.546 }=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O : N = 9 : 13 : 3 : 1
The empirical formula of the compound becomes [tex]C_9H_{13}O_3N_1=C_9H_{13}O_3N[/tex]
To calculate the molecular formula, the number of atoms of the empirical formula is multiplied by a factor known as valency that is represented by the symbol, 'n'.
[tex]n =\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex] .....(2)
We are given:
Mass of molecular formula = 183 g/mol
Mass of empirical formula = 183 g/mol
Putting values in equation 2, we get:
[tex]n=\frac{183g/mol}{183g/mol}=1[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{1\times 9}H_{1\times 13}O_{1\times 3}N_{1\times 1}=C_9H_{13}O_3N[/tex]
Hence, the empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]
In a titration to find the concentration of 30ml of a H2SO4 solution, a student found that 40ml of 0.2M KOH solution was needed to reach the endpoint. What's the concentration of the H2SO4?
Question 21 options:
A) 0.27M
B) 0.53M
C) 0.4M
D) 1.1M
Answer:
it's B
Explanation:
Write the balanced equation: H2SO4 + 2KOH → K2SO4 +2H2O. So 2(moles KOH) = (moles H2SO4); 2(volume KOH)(concentration KOH) = (volume H2SO4)(concentration H2SO4); 2(40ml)(0.2M) = (30ml)(x); x = 0.53M
The concentration of H₂SO₄ solution is equal to 0.133 M.
What is a neutralization reaction?A neutralization reaction can be described as a chemical reaction in which an acid and base react together to form respective salt and water. When a strong acid such as HCl will react with a strong base such as NaOH the salt can be neither acidic nor basic.
When H₂SO₄ (a strong acid) reacts with KOH, the resulting salt will be K₂SO₄ and water.
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
Given, the concentration of KOH solution = 0.2 M
The volume of the KOH solution = 40 ml = 0.040 ml
The number of moles of KOH, n = M × V = 0.2 × 0.04 = 0.008 mol
The volume of the H₂SO₄ = 30 ml = 0.03 L
The number of moles of H₂SO₄, n = 0.008/2 = 0.004 mol
The concentration of H₂SO₄ solution = 0.004/0.03 = 0.133 M
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A chemistry student needs 15.0 g of methanol for an experiment. She has available 320. g of 44.4% w/w solution of methanol in water. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.
Answer:
33.8 g Solution
Explanation:
A chemistry student needs 15.0 g of methanol for an experiment. The concentration of ethanol in the solution is 44.4% w/w, that is, there are 44.4 g of methanol every 100 g of solution. The mass of solution that would contain 15.0 g of methanol is:
15.0 g Methanol × 100 g Solution/44.4 g Methanol = 33.8 g Solution
Since 33.8 g are required and 320. g are available, there is enough solution for the requirements.
the atom as a whole
doesn’t carry any electric charge. It’s called a ____________ _________.
Answer:
Neutron has no charge while electron has a negative charge and proton has a positive charge
Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or diastereotopic. a) Methyls a & b: _________ b) Ha & Hc: _________
Answer:
Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or diastereotopic. a) Methyls a & b: _________ b) Ha & Hc: ________
Explanation:
Homotopic hydrogens:
Consider two hydrogens in the given molecule and replace one by one with a different atom say for example deuterium, then if the two molecules formed by replacing hydrogens are the same then the two hydrogens are called homotopic hydrogens.
After replacing the two hydrogens with a different atom then, enantiomers are formed then, the two hydrogens are called enantiotopic hydrogens.
After replacing the two hydrogens with a different atom then, diastereomers are formed then, the two hydrogens are called diastereotopic hydrogens.
In the methyl group, select two hydrogens and replace one hydrogen atom with a D-atom name the compound.
Again replace another hydrogen atom with D-atom.
Name it.
If both are the same then, the hydrogens are homotopic and they are shown below:
Hence, they are homotopic protons.
One of the most common causes of inaccurate melting point ranges is rapid heating of the compound. Under these circumstances, how will the observed MP range compare to the true MP range
Answer:
INCREASE in the difference between the melting point measured and the true melting temperature.
Explanation:
Melting point of a compound is defined as the temperature at which the soils compound changes into liquid at the atmospheric pressure. There are different circumstances that can lead to inaccurate melting point. These include:
--> presence of impurities in the compound,
--> Molecular composition,
--> Force of attraction, and
--> Rapid heating of the compound.
Under the circumstances of rapid heating of the compound, there would be an increase in the melting point range when compared with the true melting point range of the compound.
The higher the heating rate, the more rapid the rise in oven temperature, increasing the difference between the melting point measured and the true melting temperature.
Please help me name these organic compounds
Answer:
Aldehydes and Ketones
Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively:

In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms:


As text, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–.
In both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits sp2 hybridization. Two of the sp2 orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining sp2 hybrid orbital forms a σ bond to the oxygen atom. The unhybridized p orbital on the carbon atom in the carbonyl group overlaps a p orbital on the oxygen atom to form the π bond in the double bond.
Like the C=OC=O bond in carbon dioxide, the C=OC=O bond of a carbonyl group is polar (recall that oxygen is significantly more electronegative than carbon, and the shared electrons are pulled toward the oxygen atom and away from the carbon atom). Many of the reactions of aldehydes and ketones start with the reaction between a Lewis base and the carbon atom at the positive end of the polar C=OC=O bond to yield an unstable intermediate that subsequently undergoes one or more structural rearrangements to form the final product (Figure 1).
Figure 1. The carbonyl group is polar, and the geometry of the bonds around the central carbon is trigonal planar.
The importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol—for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbon-oxygen bond. The reverse reaction—replacing a carbon-oxygen bond by a carbon-hydrogen bond—is a reduction of that carbon atom. Recall that oxygen is generally assigned a –2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of C–O and C–H bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms:
A chemist adds 370.0mL of a 1.41/molL potassium iodide KI solution to a reaction flask. Calculate the millimoles of potassium iodide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answer: The millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.
Explanation:
Given: Volume of KI = 370.0 mL (1 mL = 0.001 L) = 0.37 L
Molarity of KI solution = 1.41 mol/L
Now, moles of KI (potassium iodide) is calculated as follows.
[tex]Moles = Volume \times Molarity \\= 0.37 L \times 1.41 M\\= 0.5217 mol[/tex]
Convert moles into millimoles as follows.
1 mol = 1000 millimoles
0.5217 mol = [tex]0.5217 mol \times \frac{1000 millimoles}{1 mol} = 521.7 millimoles[/tex]
This can be rounded off to the value 522 millimoles.
Thus, we can conclude that the millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.
The combustion of ethylene proceeds by the reaction
C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (g)
When the rate of disappearance of C2H4 is 0.13 M s-1, the rate of appearance of CO2 is ________ M s-1.
2.67 Determine the density (g/mL) for each of the following:
a. A 20.0-mL sample of a salt solution has a mass of 24.0 g.
The density (g/mL) for a 20.0-mL sample of a salt solution has a mass of 24.0 g is 1.2 g/ml.
What is density?Density is the mass per unit volume. Density is a scalar quantity. It is denoted by d and the symbol for density is given as rho, a Greek symbol. Density is calculated as mass divided by volume.
Mass is the quantity of matter in a physical body. The product of the molar mass of the compound and the moles of the substance are defined as mass.
Volume is the space occupied by a three-dimensional object. Volume is calculated by dividing mass by density.
Given, the 20.0-mL sample of a salt solution, which is the volume.
The mass of the solution is 24.0 g
To calculate the density
mass/volume
24.0 / 20.0 = 1.2 g/ml
Thus, the density of the given salt solution is 1.2 g/ml.
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what is the chemical fomula for water
Answer:
H2O.....................
Calculate the mass of water produced when 9.57 g of butane reacts with excess oxygen.
Express your answer to three significant figures and include the appropriate units.
Answer:
14.9 g
Explanation:
Step 1: Write the balanced equation
C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O
Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀
The molar mass of C₄H₁₀ is 58.12 g/mol.
9.57 g × 1 mol/58.12 g = 0.165 mol
Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀
0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O
Step 4: Calculate the mass corresponding to 0.825 mol of H₂O
The molar mass of H₂O is 18.02 g/mol.
0.825 mol × 18.02 g/mol = 14.9 g
3. Calculate the answers to the appropriate number of significant figures. e) 43.678 x 64.1 = f) 1.678/0.42 =
Label each formula and name pair as correct or incorrect.
Formula Name Correct/Incorrect
Aluminum tribromide
Sulfur dioxide
Beryllium hydride
Magnesium(II) oxide
Copper(II) oxide
Calcium sulfate
Nitric acid
Answer:
Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.
Sulfur dioxide: SO₂.
Beryllium hydride: BeH₂
Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.
Copper(II) oxide: CuO.
Calcium sulfate: CaSO₄
Nitric acid: HNO₃.
Explanation:
Hello there!
In this case, it seems that the formulas were not given, however, we can write the correct one for each given compound according to the widely used nomenclature rules as shown below:
Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.
Sulfur dioxide: SO₂.
Beryllium hydride: BeH₂
Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.
Copper(II) oxide: CuO.
Calcium sulfate: CaSO₄
Nitric acid: HNO₃.
Regards!
why ionic compound are good conduct of electricity in their molten state ??
Answer:
Ionic compounds conduct electricity when molten (liquid) or in aqueous solution (dissolved in water), because their ions are free to move from place to place. Ionic compounds cannot conduct electricity when solid, as their ions are held in fixed positions and cannot move.
Explanation:
because their ions are free to move from place to place.
Suppose you need to prepare 21.0 mL of formate buffer with a ratio of 4 of [sodium formate]/[formic acid] by mixing 0.10 M formic acid and 0.10 M sodium formate. How many milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid)
Answer: A volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).
Explanation:
Given: Total volume of the buffer = 21.0 mL
[tex]\frac{[HCOONa]}{[HCOOH]} = 4[/tex] ... (1)
It is assumed that the volume of HCOONa is x. Hence, volume of HCOOH is (21.0 - x) mL.
Hence,
[HCOONa] = Molarity [tex]\times[/tex] Volume
= 0.10 [tex]\times[/tex] x
= 0.1x mmol
Similarly, [HCOOH] = Molarity [tex]\times[/tex] Volume
= 0.10 [tex]\times[/tex] (21.0 - x) mmol
Using equation (1),
[tex]\frac{[HCOONa]}{[HCOOH]} = 4\\\frac{0.1x}{(21.0 - x)} = 4\\0.1x = 84.0 - 4x\\4.1x = 84.0\\x = 20.49 mL[/tex]
As x is the volume of sodium formate. Hence, 20.49 mL of sodium formate is required to make the buffer.
Thus, we can conclude that a volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).
f. . A metal cylinder has a mass of 100.00 g is heated to 95.50 celcius and then put in 245.5 g of water whose initial temperature is 22.50 Celsius. The final temperature of the mixture is 24.17 Celsius what is the specific heat of the metal.
[tex]\large\colorbox{orange}{May Be Helpful ✌️ Dear ✌️}[/tex][tex]\large\colorbox{orange}{May Be Helpful ✌️ Dear ✌️}[/tex]
Calculate the concentration of a solution with 0.8g of NaCl in 280mL of water.
Answer: The molarity of NaCl solution is 0.0489 M
Explanation:
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}}[/tex] .....(1)
We are given:
Given mass of NaCl = 0.8 g
Molar mass of NaCl = 58.44 g/mol
Volume of the solution = 280 mL
Putting values in equation 1, we get:
[tex]\text{Molarity of solution}=\frac{0.8\times 1000}{58.44\times 280}\\\\\text{Molarity of solution}=0.0489M[/tex]
Hence, the molarity of NaCl solution is 0.0489 M
How many moles are present in a sample if it consists of 5.61x1022 particles? Report your answer to 3 decimal places. Do not include units.
Answer:
The mole is defined as a collection of 6.022 × 1023 particles.
The atomic mass given on a periodic table that is given in grams is the mass of
one mole (6.022 × 1023 particles) of that element
Explanation:
Why is there a huge diversity of substances?
A. Atoms can combine in many ways to form various compounds
and molecules.
B. Any element can combine with any other element.
C. Atoms break apart to form new types of atoms during chemical
reactions.
D. There is an unlimited number of elements.
A. Atoms can combine in many ways to form various compounds
and molecules.
What is the concentration of a solution in which 15 grams of sugar is dissolved in 0.2 L of water?
Answer:
0.2 M
Explanation:
Step 1: Given data
Mass of sugar (sucrose): 15 gVolume of water: 0.2 L (we will assume it is the volume of the solution)There are different ways to express the concentration of a solution. We will calculate molarity, which is one of the most used.
Step 2: Calculate the moles of sucrose
The molar mass of sucrose is 342.3 g/mol.
15 g × 1 mol/342.3 g = 0.044 mol
Step 3: Calculate the molarity of the solution
Molarity is equal to the moles of solute divided by the liters of solution.
M = 0.044 mol/0.2 L = 0.2 M
PLZZZ HELPPPPPPPPPPPPPPPPPPPPP
Answer:
1 +-0.05cm
Explanation:
because this is more suitable
which of the following molecules would you expect to have a dipole moment of zero? a,CH2 Ch3
bH2C=0
cCH2cl
dNH3
Answer: The molecule [tex]CH_{3}-CH_{3}[/tex] is expected to have a dipole moment of zero.
Explanation:
The product of magnitude of the charge calculated in electrostatic units is called dipole moment.
Formula for dipole moment is as follows.
Dipole moment = Charge (in esu) [tex]\times[/tex] distance (in cm)
Non-polar molecules have zero dipole moment.
For example, [tex]CH_{3}-CH_{3}[/tex] is a non-polar molecule so its dipole moment is zero.
[tex]H_{2}C=O[/tex] is a polar molecule so it will have dipole moment.
[tex]CH_{2}Cl_{2}[/tex] is a polar molecule so it will have dipole moment.
[tex]NH_{3}[/tex] has nitrogen atom as more electronegative than hydrogen atom. So, net dipole moment will be in the direction of nitrogen atom.
Thus, we can conclude that the molecule [tex]CH_{3}-CH_{3}[/tex] is expected to have a dipole moment of zero.
For the neutralization reaction between pyridine and propanoic acid, draw curved arrows to indicate the direction of electron flow. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answer:
For the neutralization reaction between pyridine and propanoic acid, draw curved arrows to indicate the direction of electron flow.
Draw curved arrows to show the movement of electrons in this step of the mechanism.
Explanation:
According to Bronsted acid-base theory, an acid is a substance which is a proton donor.
Base is the proton acceptor.
In the given example, acid is propanoic acid and it loses the proton.
Pyridine is the base and it accepts the proton from propanoic acid.
The entire reaction is shown below:
Select all the correct answers
When two generalizations can be made based on what you know about cycles of matter in a closed system?
New matter is added, and old matter is destroyed.
Matter changes its physical form, allowing it to return to its original state.
The amount of matter within the system remains the same
Matter and energy can cross the boundaries of the system.
The cycle has a well-defined starting and Stopping point
Answer:
A
Explanation:
