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2 years ago

Calculate the standard entropy change for the following reaction cu(s) + 1/2 O2(g) —> cuo(s)

Chemistry

2 answers:

Evgen [1.6K]2 years ago

8 0

3433Explanation:

the awser sucks and its not RIGHT

IceJOKER [234]2 years ago

8 0

Answer:

The standard entropy change for the reaction -93.12 J/K.mol

Explanation:

The Standard entropy change:

$\Delta S^{0} = \Sigma nS^{0}(products) - \Sigma mS^{0}(reactants)$

ΔS° = [42.59] - [33.17 + (205.07/2)] J/K.mol

ΔS° = [42.59] - [33.17 + 102.535] J/K.mol

ΔS° = [42.59] - [135.705] J/K.mol

ΔS° = - 93.12 J/K.mol

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You have 17 liters of gas at STP. If the temperature rises to 94C and while the volume decreases to 12 liters, what will the ne

fenix001 [56]

Answer:

$P_2=1.90atm$

Explanation:

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In this case, according to the ideal gas equation ratio for two states:

$\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}$

Whereas both n and R are cancelled out as they don't change, we obtain:

$\frac{P_1V_1}{P_2V_2} =\frac{T_1}{T_2}$

Thus, by solving for the final pressure, we obtain:

$\frac{P_2V_2}{P_1V_1} =\frac{T_2}{T_1}\\\\P_2=\frac{T_2P_1V_1}{V_2T_1}$

Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:

$P_2=\frac{367K*1.00atm*17L}{12L*273.15K}\\\\P_2=1.90atm$

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2 years ago

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xxTIMURxx [149]

Answer:

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Explanation:

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GarryVolchara [31]

Answer:

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2 years ago

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Lunna [17]

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2 years ago

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Identify the limiting reactant when 1.22 g of O2 reacts with 1.05 g H2 to produce water.

kupik [55]

The reaction between oxygen, O2, and hydrogen, H2, to produce water can be expressed as,

2H2 + O2 --> 2H2O

The masses of each of the reactants are calculated below.

2H2 = 4(1.01 g) = 4.04 g
O2 = 2(16 g) = 32 g

Given 1.22 grams of oxygen, we determine the mass of hydrogen needed.
(1.22 g O2)(4.04 g H2 / 32 g O2) = 0.154 g of O2

Since there are 1.05 grams of O2 then, the limiting reactant is 1.22 grams of oxygen.

Answer: 1.22 g of oxygen

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3 years ago