2C4H10 + 13O2 = 8CO2 + 10H2O

1. (2.06g C4H10)/(58.12 g/mol C4H10) = 0.035mol C4H10

2. (0.035molC4H10)(10 mol H2O/2mol C4H10) = 0.177mol H2O

3. (0.177mol H2O)(18.01g/mol H2O) = 3.19g H2O

8 0

3 years ago

Which of the following would be the best representation of input energy and then

Rom4ik [11]

A is the correct answer

4 0

2 years ago

What is the temperature shown on the thermometer below?

vaieri [72.5K]

Answer:

- 13 * c

Explanation:

hope it helps u.....

6 0

3 years ago

the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel

shtirl [24]

<u>Answer:</u> The entropy change of the process is $2.0\times 10^2J/K$

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=n\times \frac{\Delta H_{f}}{T}$

where,

$\Delta S$ = Entropy change

n = moles of acetone = 6.3 moles

$\Delta H_{f}$ = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol (Conversion factor: 1 kJ = 1000 J)

T = temperature of the system = $-94.7^oC=[273-94.7]=178.3K$

Putting values in above equation, we get:

$\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K$

Hence, the entropy change of the process is $2.0\times 10^2J/K$

4 0

3 years ago