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3 years ago

How many grams are contained in a 0.893 mol sample of methane, ch4?

1 answer:

6 0

We are given with a compound, Methane (CH4), with a molar mass of 0.893 mol sample. We are tasked to solve for it's corresponding mass in g. We need to solve first the molecular weight of Methane, that is

C=12 g/mol

H=1g/mol

CH4= 12 g/mol +1(4) g/mol = 16 g/mol

With 0.893 mol sample, its corresponding mass is

g CH4= 0.893 mol x 16g/mol =14.288 g

Therefore, the mass of methane is 14.288 g

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The combustion of 0.590 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

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Answer:

2.943 °C temperature change from the combustion of the glucose has been taken place.

Explanation:

Heat released on combustion of Benzoic acid; :

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.590 g

Moles of benzoic acid = $\frac{0.590 g}{122.12 g/mol}=0.004831 mol$

Energy released by 0.004831 moles of benzoic acid on combustion:

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Change in temperature of the calorimeter = ΔT = 2.125°C

$Q=C\times \Delta T$

$15,595.5 J=C\times 2.125^oC$

$C=7,339.05 J/^oC$

Heat released on combustion of Glucose: :

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=1.400 g

Moles of glucose = $\frac{1.400 g}{180.16 g/mol}=0.007771 mol$

Energy released by the 0.007771 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.007771 mol=21.6030 kJ=21,603.01 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$21,603.01 J=7,339.05 J/^oC\times \Delta T'$

$\Delta T'=2.943^oC$

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3 years ago

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Answer:

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100x 0.108x125= 1350 cal

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