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klemol [59]
3 years ago
14

what is the specific heat of a substance if 1560 cal are required to raise the temperature of a 312-g sample by 15°C

Chemistry
1 answer:
jasenka [17]3 years ago
7 0

Answer:

0.33 cal⋅g-1°C-1  

Explanation:

The amount of heat required is determined from the formula:

q= mcΔT

To see more:

https://api-project-1022638073839.appspot.com/questions/what-is-the-specific-heat-of-a-substance-if-1560-cal-are-required-to-raise-the-t#235434

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Answer:

4190.22 L = 4.19 m³.

Explanation:

  • For the balanced reaction:

<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>

It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>

  • Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:

no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.

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<u><em>Using cross multiplication:</em></u>

5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.

  • Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).

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n is the no. of moles of the gas in mol (n = 60.95 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.

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