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2 years ago

Calculate the missing variables in each experiment below using Avogadro’s law.

Chemistry

1 answer:

blagie [28]2 years ago

5 0

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml n1 = 9.11 mol

V2 = 123 ml n2 = ?

Formula

$\frac{V1}{n1} = \frac{V2}{n2}$

$n2 = \frac{n1V2}{V1}$

$n2 = \frac{(9.11)((123)}{(612)}$

n2 = 1.83 mol

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Electrons are transferred from atoms of to atoms of . This transfer makes the sodium atoms and the phosphorus atoms . As a resul

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Answer:

Explanation:

Electrons are transferred from atoms of sodium to atoms of phosophorus. This transfer makes the sodium atoms positive and the phosphorus atoms negative. As a result, the sodium and phosphorus atoms strongly attract each other.

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3 years ago

What is the significance of "Er" in the diagram?

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Answer:

The significance of "Er" in the diagram is :

B.) Threshold energy for reaction

Explanation:

Threshold energy : It is total amount of energy required by the reactant molecule to reach the transition state .

Activation energy : It is the excess energy absorbed by the molecules to reach the transition state.

Activation Energy = Threshold Energy - Average Kinetic Energy

This means Activation energy decreases on increasing kinetic energy

On increasing Temperature average kinetic energy of the molecule increases which reduces the activation energy and the reaction occur faster in that case.

Catalyst also reduces the Activation energy.

Er = Threshshold energy for reaction at 30 degree

Ea = Activation Energy

The given figure shows that the threshold energy decreases on increasing the temperature

Only the molecule having energy greater than Er can react to form product

3 0

3 years ago

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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

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

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

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

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

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

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

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3 years ago

Consider the reaction below. At 500 K, the reaction is at equilibrium with the following concentrations. [PCI5]= 0.0095 M [PCI3]

Dmitriy789 [7]

Answer: The equilibrium constant for the given reaction is 0.0421.

Explanation:

$PCl_5\rightleftharpoons PCl_3+Cl_2$

Concentration of $[PCl_5]$ = 0.0095 M

Concentration of $[PCl_3]$ = 0.020 M

Concentration of $[Cl_2]$ = 0.020 M

The expression of the equilibrium constant is given as:

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{0.020 M\times 0.020 M}{0.0095 M}$

$K_c=0.0421$ (An equilibrium constant is an unit less constant)

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vodomira [7]

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3 years ago

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