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bulgar [2K]
2 years ago
11

What is the maximum mass of glucose (C6H12O6) that can be burned in 10 g of oxygen?

Chemistry
1 answer:
Dahasolnce [82]2 years ago
5 0
C_6H_{12}O_6 + 6 \ O_2 \to 6 \ CO_2 + 6 \ H_2O
1 mole of glucose : 6 moles of oxygen

First calculate the number of moles of oxygen in 10 g:
M=32 \ \frac{g}{mol} \\
m=10 \ g \\
n=\frac{10 \ g}{32 \ \frac{g}{mol}}=0.3125 \ mol

1 mole of glucose reacts with 6 moles of oxygen
x moles of glucose reacts with 0.3125 moles of oxygen
x=\frac{1 \ mol \cdot 0.3125 \ mol}{6 \ mol} = \frac{3125}{60000} \ mol= \frac{5}{96} \ mol

Now calculate the mass of 5/96 moles of glucose.
M=180 \ \frac{g}{mol} \\
n=\frac{5}{96} \ mol \\
m=180 \ \frac{g}{mol} \cdot \frac{5}{96} \ mol=\frac{900}{96} \ g=9.375 \ g

The maximum mass of glucose that can be burned in 10 g of oxygen is 9.375 g.
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mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

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Answer:

Explanation has been given below.

Explanation:

  • Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
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