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Colt1911 [192]
3 years ago
15

The combustion of 1.00 mol of glucose, C6H12O6, releases 2820 kJ of heat. If 2.0 g of glucose is burned in a calorimeter contain

ing 1.0 kg of water, and the temperature increases by 3.5 oC, what is the heat capacity of the calorimeter?
Chemistry
1 answer:
DochEvi [55]3 years ago
3 0

Answer:

The heat capacity of the calorimeter is 4.76 kJ/°C

Explanation:

Step 1: Data given

1.00 mol of glucose releases 2820 kJ of heat

Mass of glucose = 2.0 grams

Mass of water = 1000 grams

The temperature increases with 3.5 °C

Step 2: Calculate moles

moles glucose = mass glucose / molar mass glucose

moles glucose = 2.0 grams / 180.16 g/mol

moles glucose = 0.0111 moles

Step 3: Calculate heat produced by the combustion

Heat produced = 2820 kJ/mol * 0.0111 moles

Heat produced = 31.302 kJ = 31302 J

Step 4: Calculate heat absorbed by the water

Q = m*c*ΔT

⇒ with m = the mass of water = 1000 grams

⇒ with c = the specific heat of water = 4.184 J/g°C

⇒ with ΔT = The change in temperature = 3.5 °C

Q = 1000 * 4.184 *3.5

Q = 14644 J absorbed by the water

Step 5: Calculate heat basorbed by the calorimeter

Q = 31302 - 14644 = 16658 J absorbed by the calorimeter

Step 6: Calculate the heat capacity of the calorimeter

c= 16658 J / 3.5 °C

c = 4759 J/°C = 4.76 kJ/°C

The heat capacity of the calorimeter is 4.76 kJ/°C

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3 0
3 years ago
: Starting with 0.3500 mol CO(g) and 0.05500 mol COCl2(g) in a 3.050-L flask at 668 K, how many moles of Cl2(g) will be present
Mrac [35]

Answer:

The number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

Explanation:

The reaction is:

CO(g) + Cl₂(g) ⇄ COCl₂(g)  

The equilibrium constant of the above reaction is:

K = 1.2x10³

To find the moles of Cl₂ present at equilibrium, let's evaluate the reverse reaction:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

The equilibrium constant for the reverse reaction is:

K_{r} = \frac{1}{1.2 \cdot 10^{3}} = 8.3 \cdot 10^{-4}

Now, we need to calculate the concentration of CO and COCl₂:

C_{CO} = \frac{\eta_{CO}}{V} = \frac{0.3500 moles}{3.050 L} = 0.115 M

C_{COCl_{2}} = \frac{\eta_{COCl_{2}}}{V} = \frac{0.05500 moles}{3.050 L} = 0.018 M

Now, from the reaction we have:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

0.018 - x       0.115+x   x    

The concentration of Cl₂ is:

K_{r} = \frac{[CO][Cl_{2}]}{[COCl_{2}]}

8.3 \cdot 10^{-4} = \frac{(0.115 + x)(x)}{0.018 - x}  

8.3 \cdot 10^{-4}*(0.018 - x) - (0.115 + x)(x) = 0  

By solving the above equation for x we have:

x = 1.29x10⁻⁴ M = [Cl₂]

Finally, the number of moles of Cl₂ present at equilibrium is:

\eta_{Cl_{2}} = C_{Cl_{2}}*V = 1.29 \cdot 10^{-4} mol/L*3.050 L = 3.94 \cdot 10^{-4} moles

Therefore, the number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

I hope it helps you!

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Explanation:

I hope I helped

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2 years ago
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