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2 years ago

For the oxidation of ammonia

Chemistry

1 answer:

OverLord2011 [107]2 years ago

5 0

The answer is 0.405 M/s

- (1/3) d[O2]/dt = 1/2 d[N2]/dt

- d[O2]/dt = 3/2 d[N2]/dt

- d[O2]/dt = 3/2 × 0.27

- d[O2]/dt = 0.405 mol L^(-1) s^(-1)

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At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

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

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

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

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

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

8 0

3 years ago

Which of the following equations are correctly balanced?

AlladinOne [14]

i think it's I

I was confused by IV then search on gg and it said ZnSO4 should be Zn2SO4 instead but still im not sure Zn2SO4 is real

3 0

3 years ago

Which statement is true of a chemical change? (Plato)

OleMash [197]

Answer:

Explanation:

The answer A is the best answer because it contains the most general characteristic of a chemical change.

5 0

3 years ago

Read 2 more answers

Physical properties can be observed

Alika [10]

Answer:

Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others. That is your answer! Thanks! :)

Explanation:

4 0

3 years ago

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How many grams of Ca(OH)2are required to make 1.5 L of a 0.81 M solution?

FrozenT [24]

Answer:

Mass = 90.28 g

Explanation:

Given data:

Mass of Ca(OH)₂ = ?

Volume of solution= 1.5 L

Molarity of solution = 0.81 M

Solution:

First of all we will calculate number of moles.

Molarity = number of moles / volume in L

by putting values,

0.81 M = Number of moles / 1.5 L

Number of moles = 0.81 M × 1.5 L

Number of moles = 1.22 mol

Mass of Ca(OH)₂ in gram:

Mass = number of moles × molar mass

Mass = 1.22 mol × 74.09 g/mol

Mass = 90.28 g

5 0

3 years ago