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3 years ago

2. What is the molarity of a HCl solution containing 1.2 grams of hydrochloric acid in 0.025 L of solution?

Chemistry

1 answer:

alisha [4.7K]3 years ago

3 0

Answer:

1.3M

Explanation:

Convert from grams to moles:

molar mass of HCl = 1.01g(molar mass of H) + 35.45g(molar mass of Cl) = 36.46g HCl

1.2g HCl (1mol HCl/36.46g HCl)

= .0329 mol HCl

Molarity = mol/L (important formula for concentration)

Plug your values in:

Molarity = .0329mol/.025L

1.317M - but you used two significant figures in the question, so:

1.3M

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Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl

anzhelika [568]

Answer: ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

$CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]$

$\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]$

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Now put all the given values in this expression, we get

$\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]$

$\Delta H=-277.4kJ$

Therefore, the enthalpy change for this reaction is, -277.4 kJ

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