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3 years ago

Calculate the amount in moles of 125 g of CaSO4.

Chemistry

1 answer:

Leto [7]3 years ago

7 0

Answer:

Number of moles = 0.92 mol

Explanation:

Given data:

Mass of CaSO₄ = 125 g

Number of moles of CaSO₄ = ?

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass of CaSO₄:

Molar mass of CaSO₄ = 40 + 32+ 16×4

Molar mass of CaSO₄ = 40 + 32+ 64

Molar mass of CaSO₄ = 136 g/mol

Number of moles:

Number of moles = mass/ molar mass

Number of moles = 125 g/ 136 g/mol

Number of moles = 0.92 mol

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Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

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Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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The intermediate balanced chemical reaction will be,

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(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

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(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

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