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Katarina [22]
3 years ago
8

Heat of fusion (?Hfus) is used for calculations involving a phase change between solid and liquid, with no temperature change. F

or H2O, ?Hfus=6.02 kJ/mol.Specific heat capacity (C) is used for calculations that involve a temperature change, but no phase change. For liquid water, C=4.184 J/(g??C).Heat of vaporization (?Hvap) is used for calculations involving a phase change between liquid and gas, with no temperature change. For H2O, ?Hvap=40.7 kJ/mol.Part AHow much heat is required to raise the temperature of 92.0g of water from its melting point to its boiling point?Express your answer numerically in kilojoules.
Chemistry
1 answer:
Butoxors [25]3 years ago
6 0

Answer:

q = 38,5 kJ

Explanation:

In its melting point, at 0°C, water is liquid. The boiling point of water is 100°C. It is possible to estimate the heat you required to raise the temperature of water from 0°C to 100°C using:

q = C×m×ΔT

Where C is specific heat of water (4,184J/g°C), m is mass of water (92,0g) and ΔT is change in temperature (100°C-0°C = 100°C)

Replacing:

q = 4,184J/g°C×92,0g×100°C

q = 38493 J, in kilojoules:

<em>q = 38,5 kJ</em>

<em></em>

I hope it helps!

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The mass spectra of alcohols often fail to exhibit detectable M peaks but instead show relatively large ________ peaks.
nirvana33 [79]

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6 0
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Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma
ohaa [14]
<span>Answer: 0.094%


</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />

<span>Only the ionization of the formic acid is the important part.
</span><span />

<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />

<span>2) Mass balance:
</span><span />

<span>                   HCOOH(aq)     HCOO⁻(aq)     H⁺(aq).

Start             0.311                 0.189

Reaction       - x                      +x                   +x

Final             0.311 - x          0.189 + x            x


3) Acid constant equation:
</span><span />

<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />

<span>= (0.189 + x )x / (0.311 - x) = 0.000177


4) Solve the equation:


You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />

<span>With that approximation the equation to solve becomes:


</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M


5) With that number, the percent of ionization (alfa) is:
</span><span />

<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
8 0
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