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hichkok12 [17]
3 years ago
11

A solution is made by mixing 37.g of thiophene C4H4S and 72.g of heptane C7H16. Calculate the mole fraction of thiophene in this

solution. Round your answer to 2 significant digits.
Chemistry
1 answer:
guajiro [1.7K]3 years ago
4 0

Answer:

0.38

Explanation:

Molar mass of thiophene= 84g/mol

Mass of thiophene = 37g

Number of moles= 37/84= 0.44 moles

Molar mass of heptane= 100 g/mol

Mass of heptane = 72g

Number of moles = 72/100= 0.72 moles

Total number of moles= 0.44 + 0.72= 1.16 moles

mole fraction of thiophene = 0.44/1.16= 0.38

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Which of the following makes up the Earth's hydrosphere?
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The substances below are listed by increasing specific heat capacity value. Starting at 30 Celsius, they absorb 100 kJ of therma
Georgia [21]

Answer:

Silver.

Explanation:

To obtain the right answer to question, let us calculate the change in temperature for each substance assuming they all have the same mass as 100g.

This is illustrated below:

1. For Siver:

Mass (M) = 100g

Specific heat capacity (C) = 0.239J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 0.239)

ΔT = 4184°C

2. For Aluminium:

Mass (M) = 100g

Specific heat capacity (C) = 0.921J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 0.921)

ΔT = 1086°C

3. For Lithium:

Mass (M) = 100g

Specific heat capacity (C) = 3.56J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 3.56 )

ΔT = 281°C

4. For water:

Mass (M) = 100g

Specific heat capacity (C) = 4.184J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 4.184)

ΔT = 239°C

Summary

Temperature change of each substance is given below

1. Silver => 4184°C

2. Aluminum => 1086°C

3. Lithium => 281°C

4. Water => 239°C

From the calculations made above, Silver has the highest rise in temperature.

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Consider the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) Using standard thermodynamic data at 298K, calculate the entropy chang
balandron [24]

Answer:

the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.

Explanation:

3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)

∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants

∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]

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For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,

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