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3 years ago

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A 7.11g sample of potassium chlorate was decomposed according to the following equation: 2KCLO3 -->2KCL + 3O2 How many moles

of oxygen are formed?

1 answer:

Ivan3 years ago

3 0

Answer:

The answer to your question is 0.087 moles of Oxygen

Explanation:

Data

mass of KClO₃ = 7.11 g

moles of O₂ = ?

Balanced chemical reaction

2KClO₃ ⇒ 2KCl + 3O₂

Process

1.- Calculate the molar mass of KClO₃

KClO₃ = (1 x 39.1) + (1 x 35.45) + (3 x 16)

= 39.1 + 35.45 + 48

= 122.55 g

2.- Convert the mass to KClO₃ to moles

122.55 g ------------------- 1 mol

7.11 g ------------------- x

x = (7.11 x 1)/122.55

x = 0.058 moles

3.- Calculate the moles of Oxygen

2 moles of KClO₃ ---------------- 3 moles of Oxygen

0.058 moles ---------------- x

x = (0.058 x 3) / 2

x = 0.087 moles of Oxygen

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1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?

Ratling [72]

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1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

<u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L

V (final volume) = 473 mL → 0.473 L

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)

P (final pressure) = ? (in atm)

<u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

$P_0*V_0 = P*V$

$1*1 = P*0.473$

$1 = 0.473\:P$

$0.473\:P = 1$

$P = \dfrac{1}{0.473}$

$\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}$

<u><em>Answer: </em></u>

<u><em>The new pressure of the gas is 2.11 atm </em></u>

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3 years ago

Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T

shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

C = $\frac{P}{K_{h}}$

where, pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at $25^{o}C$ are as follows.

$p(O_{2})$ = 0.21 atm = 0.21 bar

$p(N_{2})$ = 0.78 atm = 0.78 bar

$K_{h}$ for $O_{2}$ = $7.9 \times 10^{2} bar/mol$

$K_{h}$ for $N_{2}$ = $1.6 \times 10^{3} bar /mol$

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

$C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L$

$C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L$

Now, we will calculate the number of moles as follows.

$n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3$ mol

$n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3$ mol

As the molar mass of $O_2$ = 32 g/mol

Hence, mass of oxygen will be as follows.

Mass of $O_2 = 32 \times 1.995 \times 10^-3 \times 1000$

= 63.84 mg

As the molar mass of $N_{2}$ = 28

Mass of $N_{2} = 28 \times 3.66 \times 10^-3$ = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg and nitrogen is 102.5 mg.

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