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3 years ago

A reaction generates hydrogen gas (H) as a product. The re-

Chemistry

1 answer:

Amiraneli [1.4K]3 years ago

6 0

<u>Answer:</u> The average rate of the reaction is $7.82\times 10^{-3}M/min$

<u>Explanation:</u>

To calculate the molarity of hydrogen gas generated, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of hydrogen gas = $3.91\times 10^{-2}mol$

Volume of solution = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$\text{Molarity of }H_2=\frac{3.91\times 10^{-2}mol}{0.250L}=0.1564M$

Average rate of the reaction is defined as the ratio of concentration of hydrogen generated to the time taken.

To calculate the average rate of the reaction, we use the equation:

$\text{Average rate of the reaction}=\frac{\text{Concentration of hydrogen generated}}{\text{Time taken}}$

We are given:

Concentration of hydrogen generated = 0.1564 M

Time taken = 20.0 minutes

Putting values in above equation, we get:

$\text{Average rate of the reaction}=\frac{0.1564M}{20.0min}\\\\\text{Average rate of the reaction}=7.82\times 10^{-3}M/min$

Hence, the average rate of the reaction is $7.82\times 10^{-3}M/min$

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If 1.08 g of sodium sulfate reacts with an excess of phosphoric acid, how much sulfuric acid is produced?

vfiekz [6]

Answer: 0.745 g of $H_2SO_4$ will be produced from 1.08 g of sodium sulfate

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Na_2SO_4=\frac{1.08g}{142.04g/mol}=0.0076moles$

$3Na_2SO_4+2H_3PO_4\rightarrow 2Na_3PO_4+3H_2SO_4$

$Na_2SO_4$ is the limiting reagent as it limits the formation of product and $H_3PO_4$ is the excess reagent.

According to stoichiometry :

3 moles of $Na_2SO_4$ produce = 3 moles of $H_2SO_4$

Thus 0.0076 moles of $Na_2SO_4$ will require= $\frac{3}{3}\times 0.0076=0.0076moles$ of $H_2SO_4$

Mass of $H_2SO_4=moles\times {\text {Molar mass}}=0.0076moles\times 98.1g/mol=0.745g$

Thus 0.745 g of $H_2SO_4$ will be produced from 1.08 g of sodium sulfate

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3 years ago

However, when the temperature increases, the solubility of gaseous solutes will _______?

babunello [35]

Answer:

Decrease

Explanation:

Since the speed in which the gas molecules are faster as they are heated, they fly around in the container and logically, it is harder to insert a moving object into water than something more stationary or slower.

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3 years ago

1. Why is it necessary to equalize the pressure(i.e, have the water level the same in each tube) before taking a volume reading?

Anna11 [10]

Answer:

If you contact water with a gas at a certain temperature and (partial) pressure, the concentration of the gas in the water will reach an equilibrium ('saturation') according to Henry's law.

Explanation:

This means: if you increase the pressure (e.g. by keeping the vial closed), the CO2 concentration will increase. So it simply depends what concentration you need for your assay: 'CO2-saturated' water at low pressure or 'CO2-saturated' water at high pressure.

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3 years ago

An atom of chromium has 24 protons, 28 neutrons, and 24 electrons. A chromium atom has_____Subatomic particles in the necleus

navik [9.2K]

3 subatomic particles in the nucleus
which is proton(24),neutron (28) and electrons(24)

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3 years ago

Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty

anastassius [24]

Answer :

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression of $HC_2H_3O_2$ will be:

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression of $Co(H_2O)_6^{3+}$ will be:

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression of $CH_3NH_3^+$ will be:

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

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3 years ago