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strojnjashka [21]
3 years ago
13

A 68-g sample of sodium is at an initial temperature of 42 °c. if 1840. joules of heat are applied to the sample, what is the fi

nal temperature of the sodium?
Chemistry
1 answer:
nadezda [96]3 years ago
5 0

Answer:

Final T = 64.0°C.

Explanation:

  • The amount of heat absorbed by Na (Q) can be calculated from the relation:

<em>Q = m.c.ΔT.</em>

where, Q is the amount of heat absorbed by Na (Q = 1840 J),

m is the mass of Na (m = 68.0 g),

c is the specific heat capacity of Na (c = 1.23 J/g °C),

ΔT is the temperature difference (final T - initial T) (ΔT = final T - 42.0°C).

∵ Q = m.c.ΔT.

∴ (1840 J) = (68.0 g)(1.23 J/g °C)(final T - 42.0°C)

(final T - 42.0°C) = (1840 J)/(68.0 g)(1.23 J/g °C) = 22.0°C.

<em>∴ final T</em> = 22.0°C + 42.0°C = <em>64.0°C.</em>

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<h3>Further explanation</h3>

Given

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0.3                           0                 0

2.239.10⁻⁴           2.239.10⁻⁴   2.239.10⁻⁴

0.3-2.239.10⁻⁴    2.239.10⁻⁴    2.239.10⁻⁴

\tt Ka=\dfrac{[H_3O^+][A^-]}{[HA]}\\\\Ka=\dfrac{(2.239.10^{-4}){^2}}{0.3-2.239.10^{-4}}\\\\Ka=1.671\times 10^{-7}

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