Answer:
4.46 mol of NH3
Explanation:
The equation of he reaction is given as;
2N + 3H2 --> 2NH3
From the stochiometry of the reaction, 1 mol of Nitrogen produces 2 mol of Ammonia.
Mass of Nitrogen = 31.2g
Molar mass of Nitrogen = 14g/mol
Number of moles = Mass / Molar mass = 31.2 / 14 = 2.23 mol
Since 1 mol of N = 2 mol of NH3;
2.23 mol of N2 would produce x
x = 2.23 * 2 = 4.46 mol of NH3
At standard temperature and pressure conditions, the volume of an ideal gas contained in a jar is 55.3 L. How many molecules are in the jar. This question is to be answered in scientific notation.(eg. 1.5 e5)
Answer:
1.49e24
Explanation:
Standars temperature and pressure are 273.15K and 1atm, respectively.
Using ideal gas law, we can find moles of an ideal gas if we know its pressure, temperature and volume as follows:
PV = nRT
PV / RT = n
Where P is pressure (1atm), V is volume (55.3L), R is gas constant (0.082atmL/molK), T is temperature (273.15K) and n moles of the ideal gas.
Replacing:
PV / RT = n
1atm*55.3L / 0.082atmL/molK*273.15K = n
2.47 moles = n
Now, the question is about the number of molecules in the jar. By definition, 1 mole = 6.022x10²³ molecules.
As we have 2.47 moles:
2.47 mol × (6.022x10²³ molecules / 1 mole) =
1.49x10²⁴ molecules that are in the jar
In scientific notation:
1.49e24The decomposition of ethylene oxide(CH₂)₂O(g) → CH₄(g) + CO(g)is a first order reaction with a half-life of 58.0 min at 652 K. The activation energy of the reaction is 218 kJ/mol. Calculate the half-life at 629 K.
Answer:
Half-life at 629K = 252.4min
Explanation:
Using Arrhenius equation:
[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]
And as Half-life in a first order reaction is:
[tex]t_{1/2}=\frac{ln2}{K}[/tex]
We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:
[tex]58.0min=\frac{ln2}{K}[/tex]
K = 0.01195min⁻¹ = K₁
[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]
[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]
[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]
K₂ = 2.75x10⁻³ min⁻¹
And, replacing again in Half-life expression:
[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]
Half-life at 629K = 252.4minThe half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.
The activation energy of a reaction is related to its rate constant as follows:
[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex] (1)
Where:
k: is the rate constant A: is the pre-exponential factor[tex]E_{a}[/tex]: is the activation energy of the reaction = 218 kJ/mol R: is the gas constant = 8.314 J/(K*mol)T: is the temperature
We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:
[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex] (2)
Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min
Hence, the rate constant at 652 K is:
[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]
Now, from equation (1) we can find the pre-exponential factor (A):
[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]
With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):
[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]
Finally, the half-life at 629 K is (eq 2):
[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]
Therefore, the half-life at 629 K is 251.1 min.
Find more about activation energy here:
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Write the equation for the reaction described: A solid metal oxide, , and hydrogen are the products of the reaction between metal and steam. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answer:
Pb + 2H2O --> PbO2 + 2H2
Explanation:
Products:
Solid metal; PbO2
Hydrogen; H
Reactants:
Metal; Pb
Steam; H2O
Reactants --> Products
Pb + H2O --> PbO2 + H2
Upon balancing we have;
Pb + 2H2O --> PbO2 + 2H2
What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the ions in solution were [Cd2+] = 0.5 M and [Zr4+] = 0.5 M at 298K?
Answer:
1.05 V
Explanation:
Since;
E°cell= E°cathode- E°anode
E°cathode= -0.40 V
E°anode= -1.45 V
E°cell= -0.40-(-1.45) = 1.05 V
Equation of the process;
2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)
n= 8 electrons transferred
From Nernst's equation;
Ecell = E°cell - 0.0592/n log Q
Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]
Since log 1=0
Ecell= E°cell= 1.05 V
Which of these species would you expect to have the lowest standard entropy (S°)?
a. CH4(g)
b. H2O(g)
c. NH3(g)
d. HF(g)
Answer:
d. HF(g)
Explanation:
Hello,
In this case, the standard entropy S° could be predicted by looking at the amount of bonds the compound has, thus, the fewer the number bonds, the lower the standard entropy, it means that d. HF(g) has lowest value as it has one bond only whereas methane has four bonds, water two bonds and ammonia three bonds.
Best regards.
Fill in the blanks with the words given below- [Atoms, homogeneous, metals, true, saturated, homogeneous, colloidal, compounds, lustrous] 1.An element which are sonorous are called................ 2.An element is made up of only one kind of .................... 3.Alloys are ............................. mixtures. 4.Elements chemically combines in fixed proportion to form ........................ 5. Metals are................................... and can be polished. 6. a solution in which no more solute can be dissolved is called a .................... solution. 7. Milk is a .............. solution but vinegar is a .................. solution. 8. A solution is a ................... mixture. pls help, could not get these answers
Answer:
1. metals
2. atom
3. homogeneous
4. compounds
5. lustrous
6. saturated
7. colloidal
8. homogeneous
Explanation:
A student determines the value of the equilibrium constant to be 1.5297 x 107 for the following reaction: HBr(g) + 1/2 Cl2(g) --> HCl(g) +1/2 Br2(g) Based on this value of Keq, calculate the Gibbs free energy change for the reaction of 2.37 moles of HBr(g) at standard conditions at 298 K.
Answer:
[tex]\Delta G=-97.14kJ[/tex]
Explanation:
Hello,
In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:
[tex]\Delta G=-RTln(K)[/tex]
Hence, we compute it as required:
[tex]\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol[/tex]
And for 2.37 moles of hydrogen bromide, we obtain:
[tex]\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ[/tex]
Best regards.
Identify four general properties that make an NSAID unique as compared to the NSAID aspirin. List specific properties that make aspirin, naproxen, and ibuprofen unique from one another
Answer:
NSAIDs are steroidal anti-inflammatories, their action is on the phospholipase A2 enzyme, this enzyme is responsible for breaking down the phospholipids of the membrane to trigger an inflammatory response. This is how steroidal anti-inflammatory drugs inhibit ALL inflammatory pathways (not like NSAIDs that they only inhibit the COX pathway).
These corticosteroid drugs cannot exceed the systemic mineralocorticoid value 1 in the body, since this corticosteroid hormone is also released by the adrenal cortex.
The NSAIDs generate: sporadic peaks in blood glucose, hypertension, fluid retention, increase in body fat mass, possible suppression of the adrenal cortex over time, inhibiting endogenous synthesis of corticosteroids.
On the other hand, naproxen and ibuprofen are NSAIDs, that is, non-steroidal anti-inflammatory drugs that do not influence both routes of inflammation, but only COX, this enzyme is abbreviated as COX but is called cyclooxygenase, and is responsible for a single route of inflammation.
NSAIDs such as naproxen and ibuprofen can cause gastric disorders such as ulcers or gastritis if they are consumed in a very repetitive manner.
In addition, both drugs are anti-inflammatory, analgesic and antipyretic. Although its two main functions are the first two, it was shown to have an effect in lowering body temperature.
That they are anti-inflammatory means that they inhibit the path of inflammation and analgesics the path of pain.
Explanation:
Both types of drugs generate the same effect but by different mechanisms.
Some are steroids and others are not, although steroids are considered to have a greater risk of benefit that is why they are administered against more systematically compromised instances such as anaphylactic shock.
NSAIDs such as naproxen and ibuprofen are the most prescribed today, since they have few risks and very good benefits, meaning that their adverse effects are not lethal or highly relevant and have a good effect on symptoms.
Both must be administered with care when treating a diabetic patient since corticosteroids generate glycemic peaks or increase in blood glucose, and NSAIDs compete for plasma protein with oral hypoglycemic agents, thus generating that these are in higher free concentrations. high diffusing better through the tissues and increases the potency of the adverse effects of these.
An actacide tablet containing Mg(OH)2 (MM = 58.3g / (mol)) is titrated with a 0.100 M solution of HNO3. The end point is determined by using an indicator. Based on 20.00mL HNO3 being used to reach the endpoint, what was the mass of the Mg * (OH) in the antacid tablet? * 0.0583 g 0.583 5.83 g 58.3 g
Answer:
0.0583g
Explanation:
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
From the question, number of moles of HNO3 reacted= concentration × volume
Concentration of HNO3= 0.100 M
Volume of HNO3 = 20.00mL
Number of moles of HNO3= 0.100 × 20/1000
Number of moles of HNO3 = 2×10^-3 moles
From the reaction equation;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2
But
n= m/M
Where;
n= number of moles of Mg(OH)2
m= mass of Mg(OH)2
M= molar mass of Mg(OH)2
m= n×M
m= 1×10^-3 moles × 58.3 gmol-1
m = 0.0583g
Atomic mass is calculated by _____. subtracting protons from neutrons averaging the mass of isotopes adding protons and neutrons subtracting neutrons from protons
Answer:
Atomic mass is calculated by adding protons and neutrons.
Explanation:
Atomic mass is the sum of protons and neutrons in an atomic nucleus. For example, the element Oxygen has 8 protons (derived from the atomic number) and 8 neutrons (derived from subtracting the amount of protons from the atomic mass).
We can craft an equation to show the relationship between these variables.
M - N = P, where M = Mass, N = Neutrons, and P = Protons
This equation can be rearranged to show the relationship between the neutrons and protons leading to the atomic mass. Simply add N to both sides of the equation.
M = N + P
This shows that atomic mass is equivalent to the sum of protons and neutrons in an atom's nucleus.
Using the data: C2H4(g), = +51.9 kJ mol-1, S° = 219.8 J mol-1 K-1 CO2(g), = ‑394 kJ mol-1, S° = 213.6 J mol-1 K-1 H2O(l), = ‑286.0 kJ mol-1, S° = 69.96 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = 205 J mol-1 K-1 calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)
Answer:
The correct answer is 1332 KJ.
Explanation:
Based on the given information,
ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.
Now the balanced equation is:
C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)
ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2
ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)
ΔH°rxn = -1411.9 KJ
Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.
Now based on the balanced chemical reaction,
ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2
ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)
ΔS°rxn = -267.68 J/K or -0.26768 KJ/K
T = 25 °C or 298 K
Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get
ΔG = -1411.9 - 298.0 × (-0.2677)
ΔG = -1332 KJ.
Thus, the maximum work, which can obtained is 1332 kJ.
Determine which set of properties correctly describes copper (Cu)?
A. Giant structure, conducts electricity, high melting point, soluble in water, malleable
B. Malleable, brittle, soluble in oil or gasoline, high melting point, simple structure
C. Ionic lattice, conducts electricity, soluble in oil or gasoline, low melting point, ductile
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Answer:
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Explanation:
Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).
All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.
Testbank Question 47 Consider the molecular orbital model of benzene. In the ground state how many molecular orbitals are filled with electrons?
Answer:
There are fifteen molecular orbitals in benzene filled with electrons.
Explanation:
Benzene is an aromatic compound. Let us consider the number of bonding molecular orbitals that should be present in the molecule;
There are 6 C-C σ bonds, these will occupy six bonding molecular orbitals filled with electrons.
There are 6 C-H σ bonds, these will occupy another six molecular orbitals filled with electrons
The are 3 C=C π bonds., these will occupy three bonding molecular pi orbitals.
All these bring the total number of bonding molecular orbitals filled with electrons to fifteen bonding molecular orbitals.
Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH4Cl with 200.0 mL of 0.12 M NH3. The Kb for NH3 is 1.8 × 10-5.
Answer:
The pH of the solution is 9.06.
Explanation:
The reaction of the dissociation of NH₃ in water is:
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq) (1)
[NH₃] - x [NH₄⁺] + x x
The concentration of NH₃ and NH₄⁺ is:
[tex] [NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M [/tex]
[tex] [NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M [/tex]
From equation (1) we have:
[tex]Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/tex]
[tex] 1.8 \cdot 10^{-5} = \frac{(0.083 + x)*x}{0.053 - x} [/tex]
[tex] 1.8 \cdot 10^{-5}(0.053 - x) - (0.083 + x)*x = 0 [/tex]
By solving the above equation for x we have:
x = 1.15x10⁻⁵ = [OH⁻]
The pH of the solution is:
[tex] pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94 [/tex]
[tex] pH = 14 - pOH = 14 - 4.94 = 9.06 [/tex]
Therefore, the pH of the solution is 9.06.
I hope it helps you!
2NH3 → N2 + 3H2 If 2.22 moles of ammonia (NH3) decomposes according to the reaction shown, how many moles of hydrogen (H2) are formed? A) 2.22 moles of H2 B) 1.11 moles of H2 C) 3.33 moles of H2 D) 6.66 moles of H2
Answer:
C
Explanation:
According to the mole ratio, using 2NH3 will give you 3H2. Which means in order to find the moles of H2 you would only need to divide 2 and multiply 3 to get the amount of moles of H2 produced.
Answer:
I think it's C
Explanation:
Please, tell me if I'm incorrect.
Beginning with Na, record the number of energy levels, number of protons, and atomic radius for each element in period 3.
Answer:
Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.
Explanation:
There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm. The atomic radius of aluminium atom is 143 ppm. The atomic radius of silicon atom is 111 ppm. The atomic radius of phosphorus atom is 98 ppm. The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.
What are some geographic features that could be found in the hydrosphere?
Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.
hydro = water
Answer:
Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!
Explanation:
The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)
If one pound is the same as 454 grams, then convert the mass of 78 grams to pounds.
Answer:
0.17 lb
Explanation:
78 g * (1 lb/454 g)=0.17 lb
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There are 2.4g of calcium hydroxide reacted with nitric acid. Calculate the number of moles of calcium hydroxide used. Write your answer using proper significant digits and units. Show all your work.
Answer:
0.032 moles
Explanation:
no of moles =
[tex] \frac{mass \: in \: grams}{relative \: molecular \: mass} [/tex]
=
[tex] \frac{2.4}{40 + 32 + 2} [/tex]
= 0.032
Calcium hydroxide reacted with nitric acid the total number of moles will be 0.032 moles.
What is a mole?
A mole is Avogadro's number of particles, which is exactly 6.02214076×1023.
The mole is widely used in chemistry as a convenient way to express amounts of reactants and products of chemical reactions. For example, the chemical equation 2H2 + O2 → 2H2O can be interpreted to mean that for each 2 mol dihydrogen (H2) and 1 mol dioxygen (O2) that react 2 mol of water (H2O) form.
Number of moles = Mass of substance / Mass of one mole Number of moles
mass of substance = 2.4g
molar mass of calcium hydroxide is (1 ×40.078g/mol Ca) +(2 × 15.999g/mol O) + (2 × 1.008g/mol H) = 74.092 g/mol Ca (OH)2
substituting the value,
number of moles = 2.4 / 74.029
= 0.032 moles
Therefore, moles of calcium hydroxide will be 0.032 moles
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The intermolecular forces present in CH 3NH 2 include which of the following? I. dipole-dipole II. ion-dipole III. dispersion IV. hydrogen bonding
Answer:
I. dipole-dipole
III. dispersion
IV. hydrogen bonding
Explanation:
Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.
London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.
Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.
Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.
Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.
Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.
Answer:
The intermolecular forces present in CH_3NH_2 includes
II. (ion-dipole) and IV. (hydrogen bonding)Explanation:
The intermolecular forces present in CH_3NH_2 includes II. (ion-dipole) and IV. (hydrogen bonding)
It is a polar molecule due to NH polar bond and it can form Hydrogen bond also due to NH bond.
Interaction will be dipole- dipole and Hydrogen dispersion forces can always be taken into account.
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A balloon has an initial volume of 2.954 L containing 5.50 moles of helium. More helium is added so that the balloon expands to 4.325 L. How much helium (moles) has been added if the temperature and pressure stay constant during this process.
Answer:
8.05 moles
Explanation:
5.50 / 2.954 = x / 4.325
x = 8.05
According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added so that the balloon expands to 4.325 L.
What is ideal gas equation?The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.
In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314 so n=V/R=4.325/8.314=0.520 moles.
Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.
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A 25.00 mL sample of unknown concentration of HNO3 solution requires 22.62 mL of 0.02000 M NaOH to reach the equivalence point. What is the concentration of the unknown HNO3 solution
Answer:The concentration of the unknown HNO3 solution = 0.01809 M
Explanation:
For the acid-base reaction, HNO3 + NaOH-----> NaN03 + H20
we have that
C1 V1 = C2 V2
Where ,
C1 = concentration of HNO3=?
V1 = volume of HNO3 = 25.00 mL,
V2 = volume of NaOH = 22.62 mL,
C2 = concentration of NaOH = 0.02000 M
Therefore ,
25.00 mL x C1 = 22.62 mL x 0.02000 M
= (22.62 mL / 25.00 mL) x 0.02000 M = 0.01809 M
The concentration of the unknown HNO3 solution = 0.01809 M
How many atoms of oxygen are in one molecule of water (H2O)? one two four three
Answer:
there is one atom of oxygen and two atoms of hydrogen
Explanation:
Please Help! Use Hess’s Law to determine the ΔHrxn for: Ca (s) + ½ O2 (g) → CaO (s) Given: Ca (s) + 2 H+ (aq) → Ca2+ (aq) + H2 (g) ΔH = 1925.9 kJ/mol 2 H2 (g) + O2 (g) → 2 H2O (l) ΔH = −571.68 kJ/mole CaO (s) + 2 H+ (aq) → Ca2+ (aq) + H2O (l) ΔH = 2275.2 kJ/mole ΔHrxn =
Answer:
ΔHrxn = -635.14kJ/mol
Explanation:
We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:
(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol
(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole
(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole
Reaction (1) - (3) produce:
Ca(s) + H2O(l) → H2(g) + CaO(s)
ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol
Now this reaction + 1/2(2):
Ca(s) + ½ O2(g) → CaO(s)
ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)
ΔHrxn = -635.14kJ/molWhat is the mass number of an element
Answer:
A (Atomic mass number or Nucleon number)
Explanation:
The mass number is the total number of protons and nucleons in an atomic nucleus.
Hope this helps.
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what is the lewis structure for OP(N3)3
Explanation:
this is the ans
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A solution containing a unknown ionic compound, vigorously bubbles when hydrochloric acid (HCl) is added to the solution. This might indicate that the solution contains which anion?
Answer:
CO3^2-
Explanation:
In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.
When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.
Experiment:
Part I: Voltaic Cell
Assume that you are provided with the following materials:
Strips of metallic zinc, metallic copper, metallic iron
1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine (I2)
Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos, identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.
For each cell created, include the following details.
Which electrode was the anode and which was the Cathode?
The anode and cathode half reactions.
Balanced equation for each cell you propose to construct.
Calculated Eocell
Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed)
Answer:
Here are four possible voltaic cells.
Explanation:
1. Standard reduction potentials
E°/V
I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Fe²⁺(aq) + 2e⁻ ⟶ Fe(s); -0.41
Zn²⁺(aq) + 2e⁻ ⟶ Zn(s); -0.76
2. Possible Voltaic cells
(a) Zn/I₂
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cell: Zn(s) + I₂(s) ⟶ Zn²⁺(aq) + 2I⁻(aq); 1.30
Zn(s)|Zn²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)
Zn is the anode; graphite is the cathode.
(b) Zn/Cu²⁺
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Cell: Zn(s) + Cu²⁺(s) ⟶ Zn²⁺(aq) + Cu(s); 1.10
Zn(s)|Zn²⁺(aq)∥Cu²⁺(aq)|Cu(s)
Zn is the anode; Cu is the cathode.
(c) Zn/Fe²⁺
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: Fe²⁺(aq) + 2e⁻ ⟶ Fe(s); -0.41
Cell: Zn(s) + Fe²⁺(s) ⟶ Zn²⁺(aq) + Fe(s); 0.35
Zn(s)|Zn²⁺(aq)∥Fe²⁺(aq)|Fe(s)
Zn is the anode; Fe is the cathode.
(d) Fe/I₂
E°/V
Anode: Fe(s) ⟶ Fe²⁺(aq) + 2e⁻; 0.41
Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cell: Zn(s) + I₂(s) ⟶ Zn²⁺(aq) + 2I⁻(aq); 0.95
Fe(s)|Fe²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)
Fe is the anode; graphite is the cathode.
If 2.9g of water is heated from 23.9C to 98.9C, how much heat (in calories) was added to the water?
Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
Br(g)
Cl2(g)
I2(g)
F2(g)
B. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2S(g)
H2O(g)
H2O2(g)
C. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous)
C(s, diamond)
C(s, graphite)
Answer:
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy
I2(g)>Br2(g)>Cl2(g)>F2(g)
B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2O2(g)>H2S(g) >H2O(g)
C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous) >C(s, graphite)>C(s, diamond)
Explanation:
Hello,
In this case, we can apply the following principles to explain the order:
- The greater the molar mass, the larger the standard molar entropy.
- The greater the molar mass and the structural complexity, the larger the standard molar entropy.
- The greater the structural complexity, the larger the standard molar entropy.
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy
I2(g)>Br2(g)>Cl2(g)>F2(g)
This is due to the fact that the greater the molar mass, the larger the standard molar entropy.
B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2O2(g)>H2S(g) >H2O(g)
This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.
C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous) >C(s, graphite)>C(s, diamond)
Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).
Regards.
