Question

This table has information about the heat of fusion and the heat of vaporization of different substances. A 3-column table with 10 rows. Column 1 is labeled substance. Column 2 is labeled delta H fusion in kilojoules per mole. Column 3 is labeled delta H vaporization in kilojoules per mole. Ten rows are as follows: H 2 O, 6.01, 40.7. C O 2, 7.94, 25.2. O 2, 0.443, 6.81. N 2, 0.719, 5.58. F e, 14.9, 354. A l, 10.7, 255. C u, 13.0, 304. N a C l, 30.2, 171. C H 4, 0.936, 8.53. H 2 S, 2.37, 18.7. Which substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes? Use q equals n delta H.. a.CH4 b.H2S c.CO2 d.NaCl

Answer 1

H₂S

Further explanation

Given

ΔH fusion and ΔH vaporization  of different substances

Required

The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes

Solution

We can use the formula :

$\tt \Delta H=\dfrac{Q}{n}$

Q=heat/energy absorbed

n = moles

The heat absorbed : 58.16 kJ

moles = 3.11

so ΔH vaporization :

$\tt \Delta H_{vap}=\dfrac{58.16~kJ}{3.11~mol}\\\\\Delta H_{vap}=18.7~kJ/mol$

The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S

(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)

Answer 2

Answer:

b

Explanation: