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frez [133]
3 years ago
15

The specific heat of gold is 0.031 calories/gram°C. If 10.0 grams of gold were heated and the temperature of the sample changed

by 20.0°C, how many calories of heat energy were absorbed by the sample?
Chemistry
2 answers:
Vitek1552 [10]3 years ago
6 0

Answer : The heat energy absorbed by the sample was 6.2 calories.

Explanation :

Formula used :

Q=m\times c\times \Delta T

where,

Q = heat absorbed = ?

m = mass of gold = 10.0 g

c = specific heat of gold = 0.031cal/g^oC

\Delta T = change in temperature  = 20.0^oC

Now put all the given value in the above formula, we get:

Q=10.0g\times 0.031cal/g^oC\times 20.0^oC

Q=6.2cal

Therefore, the heat energy absorbed by the sample was 6.2 calories.

Phoenix [80]3 years ago
3 0
10 grams of gold would have a specific heat of 0.31 calories ie 10 x 0.031 and if heated up by 20.0 degrees centigrade then 20.0 x 0.31 would be a 6.2 calorie increase in heat. In other words, the specific heat means that it is the calories produced in the gram of gold by a 1 degree rise in degrees centigrade.
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Answer:

1) 0.05 mol.

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<em>1) Number of moles of the compound:</em>

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<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) =  <em>0.1 mol.</em>

<em>3) Number of moles of carbonate ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

<em>4) Number of moles of hydrogen atoms:</em>

  • Every 1.0 mol of (NH₄)₂CO₃  contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>

<em>5) Number of hydrogen atoms:</em>

  • It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

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

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

<em>∴ The no. of atoms in  0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>

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