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adell [148]
2 years ago
12

If the same large amount of heat is added to a 250 g piece of aluminum and a 150 g piece of aluminum, what will happen? The 150

g Al will reach a higher temperature. The 250 g Al will reach a higher temperature. There will be no significant change in the temperature of either sample. Both samples will reach the same final temperature.
Chemistry
1 answer:
Inga [223]2 years ago
3 0

Answer:

The 150 g Al will reach a higher temperature.

Explanation:

  • The amount of heat added to a substance (Q) can be calculated from the relation:

<em>Q = m.c.ΔT.</em>

where, Q is the amount of heat added,

m is the mass of the substance,

c is the specific heat of the substance,

ΔT is the temperature difference (final T - initial T).

Since, Q and c is constant, ΔT will depend only on the mass of the substance (m).

∵ ΔT is inversely proportional to the mass of the substance.

<em>∴ The piece with the lowest mass (150.0 g) will reach a higher temperature than that of a higher mass (250.0 g).</em>

<em>So, the right choice is: The 150 g Al will reach a higher temperature.</em>

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A student has 70.5 mL of a 0.463 M aqueous solution of sodium bromide. The density of the solution is 1.22 g/mL. Find the follow
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a.) 86.01 g.

b.) 3.36 g.

c.) 0.394 m ≅ 0.40 m.

d.) 4.77%.

e.) 3.9%.

Explanation:

<em>a.) mass of the solution:</em>

The density of the solution is the mass per unit volume.

<em>∵ Density of solution = (mass of solution)/(volume of the solution).</em>

∴ Mass of the solution = (density of solution)*(volume of the solution) = (1.22 g/mL)*(70.5 mL) = 86.01 g.

<em>b.) grams of sodium bromide  :</em>

  • Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.

∵ M = (no. of moles of NaBr)/(Volume of the solution (L))

∴ no. of moles of NaBr = M*(Volume of the solution (L)) = (0.463 M )*(0.0705 L) = 0.0326 mol.

<em>∵ no. of moles of NaBr = (mass of NaBr)/(molar mass of NaBr)</em>

∴ mass of NaBr = (no. of moles of NaBr)*(molar mass of NaBr) = (0.0326 mol)*(102.894 g/mol) = 3.36 g.

<em>c.) molality of the solution:</em>

  • Molality (m) is defined as the no. of moles of solute dissolved per 1.0 kg of the solvent.

∵ m = (no. of moles of NaBr)/(mass of the soluvent (kg))

no. of moles of NaBr = 0.0326 mol,

mass of solvent = mass of the solution - mass of NaBr = 86.01 g - 3.36 g = 82.65 g = 0.08265 kg.

∴ m = (no. of moles of NaBr)/(mass of the soluvent (kg)) = (0.0326 mol)/(0.08265 kg) = 0.394 m ≅ 0.40 m.

<em>d.) % (m/v) of the solution:</em>

∵ (m/v)% = [(mass of solute) /(volume of the solution)]* 100

∴ (m/v)% = [(3.36 g)/(70.5 mL)]* 100 = 4.77%.

<em>e.) % (m/m) of the solution:</em>

∵ (m/m)% = [(mass of solute) /(mass of the solution)]* 100

∴ (m/m)% = [(3.36 g)/(86.01 g)] * 100 = 3.9 %.

4 0
3 years ago
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