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inn [45]
3 years ago
8

How many kilojoules of heat are needed to completely vaporize 42.8 grams of c4h10o at its boiling point given that c4h10o has a

heat of vaporization (δhvap) of +26.5 kj/mol and a molar mass of 74.12 g/mol?
Chemistry
1 answer:
Darina [25.2K]3 years ago
6 0
Answer is: 15.30 kilojoules of heat are needed to completely vaporize C₄H₁₀<span>.
m(</span>C₄H₁₀) = 42.8 g.
M(C₄H₁₀) = 74.12 g/mol.
n(C₄H₁₀) = m(C₄H₁₀) ÷ M(C₄H₁₀).
n(C₄H₁₀) = 42.8 g ÷ 74.12 g/mol.
n(C₄H₁₀) = 0.577 mol.
Q = n(C₄H₁₀) · ΔHvap.
<span>Q = 0.577 mol </span>· 26.5 kJ/mol.
<span>Q = 15.30 kJ, heat of butane.

</span>
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To support this answer, we first calculate for the pKa value as the negative logarithm of the Ka value: 
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Hi there!

A solution that is 1000 ug/ ml  (or 1000 mg / l) is 1000 ppm.

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Step 2 (1:50 dilution):

Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)

Step 3 (1:4 dilution):

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