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Stolb23 [73]
2 years ago
6

A reaction is thermodynamically unstable (spontaneous) but no change is observed. The reaction is probably Select the correct an

swer below: kinetically unstable. kinetically stable. thermodynamically stable but kinetically unstable. None of the above
Chemistry
1 answer:
mart [117]2 years ago
7 0

Answer:

kinetically stable.

Explanation:

When we say that a system is thermodynamically unstable, it means that there is still a state in which the system is expected to have lower energy than it currently has. A thermodynamically unstable system is yet to attain equilibrium hence it can still undergo further chemical processes in order to attain thermodynamic stability.

When we say that a system is kinetically stable, it means that the activation energy or energy barrier for the reaction system is high. Thus reactants are not easily converted into products. The reaction system remains the same for a long while without change.

Finally, when a reaction is thermodynamically unstable (spontaneous) but no change is observed, the reaction is kinetically stable.

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The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ mol−1. Given the following information: he
Alex17521 [72]

Explanation:

It is given that total energy is -443 kJ/mol and formula to calculate the lattice energy is as follows.

       Total energy = heat of sublimation + bond dissociation energies + ionization energy for Cs + EA of Cl^{-} + lattice energy

       -443 kJ/mol = 76 + 121 + 376 - 349 + Lattice energy

   Lattice energy = (-443 - 76 -121 - 376 + 349) kJ

       Lattice energy = -667 kJ

Therefore, we can conclude that -667 kJ is the magnitude of the lattice energy for CsCl.

8 0
3 years ago
As long as there is some residual liquid present after equilibrium is reached, the vapor pressure of a liquid at any given tempe
stepan [7]

Answer:

True

Explanation:

Every material in made up of intensive or extensive property. Intensive property of a system does not depend on the system size or the amount of material in the system. But extensive property on the other hand depends on the amount of material present in the system.

Examples of intensive properties include temperature, density, vapor pressure and viscosity.

Assuming that there is some residual liquid left after equilibrium is reached, no matter how much liquid is present, at any given temperature, the vapor pressure will be the same because it is an intensive property.

8 0
3 years ago
Question 4 (2 points)<br> CuO(s) + H2(g)<br> Cu(s) +<br> H2O(1)<br> Balance the equation
CaHeK987 [17]

Answer:

CuO(s) + H₂(g) --> Cu(s) + H₂O(l)

Explanation:

It is already balanced. You can see that the values of the elements of the reactants are equal to the values of the elements of the products.

7 0
3 years ago
2. Using the following data, calculate the average atomic mass of magnesium (give your answer to the nearest
Savatey [412]

Answer: 24.309amu

Explanation:

4 0
3 years ago
Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t
myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol

Moles of methane gas = n_1=\frac{8.00 g}{16 g/mol}=0.5 mol

Moles of ethane gas  =n_2=\frac{18.0 g}{30 g/mol}=0.6 mol

Moles of propane gas = n_3

n=n_1+n_2+n_3

2.22=0.5 mol +0.6 mol+ n_3

n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol

Mole fraction of methane =\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}

\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252

Similarly, mole fraction of ethane and propane :

\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703

\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045

Partial pressure of each gas can be calculated by the help of Dalton's' law:

p_i=P\times \chi_1

Partial pressure of methane gas:

p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm

Partial pressure of ethane gas:

p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm

Partial pressure of propane gas:

p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = n_1=\frac{37.8 g g}{28g/mol}=1.35 mol

Moles of oxygen gas  =n_2=\frac{62.2 g}{32 g/mol}=1.94 mol

Mole fraction of nitrogen=\chi_1=\frac{n_1}{n_1+n_2}

\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103

Similarly, mole fraction of oxygen

\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897

Partial pressure of each gas can be calculated by the help of Dalton's' law:

p_i=P\times \chi_1

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg

Partial pressure of oxygen gas:

p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg

3 0
3 years ago
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