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3 years ago

What is the volume occupied by 2.00 moles of Ar(g) at STP? Another one I am stuck on...please help!

1 answer:

3 0

Our only unknown is the volume of H2(g) . Our known variables are P,n,R, and T. At STP, the temperature is 273K and the pressure is 1 atm. Rearrange the equation to solve for V: n×R×TP. V=2.20mol ×0.0821L×atm mol×K ×273K 1atm. V=49.3L

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Deffinition of density

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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:

$Q=3,228 kJ/mol \times 0.004831 mol=15.5955 kJ=15,595.5 J$

Heat capacity of the calorimeter = C

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'

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

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

$\Delta T'=2.943^oC$

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

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3 years ago

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