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CaHeK987 [17]
3 years ago
10

The decomposition of Ca(OH)2(s) intoCaO(s) and H2O(g) at constant pressure requires the addition of 109 kJ of heat per mole of C

a(OH)2.
A. Write a balanced thermochemical equation for the reaction.
B. Draw an enthalpy diagram for the reaction, showing the activation energy, Ea, and the enthalpy change for the reaction.

Chemistry
1 answer:
larisa [96]3 years ago
4 0

Answer:

A. Ca(OH)2 ----> CaO + H2O; ∆H = +109KJ

B. Check attached document below for the enthalpy diagram

Explanation:

A thermochemical equation is a balanced chemical equation in which enthalpy change is variable. It tells about the nature of the reaction.

Enthalpy change is the difference between the heat content of the products and reactants in a thermochemical equation.

∆H is negative for an exothermic reaction (a reaction where heat is given out), while it is positive for an endothermic reaction ( a reaction in which heat is added).

The activation energy, Ea, is the minimum amount of energy reactant particles must possess in order for a reaction to proceed towards product formation.

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if 10 moles of A and 7.5 moles of B are taken in a flask, identify the value of moles of A4B3 produced in a reaction.​
natta225 [31]

The number of moles of A4B3 produced in a reaction is obtained  by stoichiometry.

The term stoichiometry refers to mass-mole relationship. The first step in solving this problem is by writing the correct reaction equation as follows;

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3 years ago
6. A sample of a gas has a mass of 0.527 g. Its volume is 0.35 L at a temperature of 88 degree Celsius and a pressure of 945 mm
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<u>Answer:</u> The molar mass of the gas is 35.87 g/mol.

<u>Explanation:</u>

To calculate the mass of gas, we use the equation given by ideal gas:

PV = nRT

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PV=\frac{m}{M}RT

where,

P = Pressure of gas = 945 mmHg

V = Volume of the gas = 0.35 L

m = Mass of gas = 0.527 g

M = Molar mass of gas = ? g/mo

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of gas =  88^oC=[88+273]=361K

Putting values in above equation, we get:

945mmHg\times 0.35L=\frac{0.527g}{M}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 361K\\\\M=35.87g/mol

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O=2  x13/2=13                                  O = 8+5=13

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3 years ago
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