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ladessa [460]
3 years ago
11

In the chemical reaction below, calcium (Ca) and water (H2O) react to form calcium hydroxide (Ca(OH)2) and hydrogen gas (H2). Ca

() + 2H2O() → Ca(OH)2() + H2() Calcium and water are the reactants in this chemical reaction. The products are calcium hydroxide and hydrogen gas. Which statement correctly describes the total mass of the products when the reaction is complete?
A. The total mass of the products will be greater than the original mass of the reactants.
B. The total mass of the products will be less than the original mass of the reactants.
C. The total mass of the products will be equal to the original mass of the reactants.
D. The total mass of the products cannot be found from the original mass of the reactants.
Chemistry
1 answer:
lakkis [162]3 years ago
8 0

Answer:

C.

Explanation:

The mass of the reactants should not change, in fact it would be equal because the only thing that changes is the form in which your products are in. The reactants will still have the same amount of mass from the products as no products were removed or added, the structure changed, the mass did not.

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All of the following conditions of STP are true except A. 101.3 kPa B. 273.15 K C. 22.4 L D. 3.81 kPa
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3 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

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We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

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As we are given :

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Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

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q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

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