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2 years ago

7

The partial pressure of a gas is a inversely proportional to the temperature of the gas. b. directly proportional to the volume

of the gas. c. always constant during a chemical reaction. d. directly proportional to the number of moles of the gas. e. proportional to the gas constant, R.

1 answer:

vlabodo [156]2 years ago

3 0

Answer:

d. directly proportional to the number of moles of the gas.

Explanation:

The formula for the partial pressure of a gas is written as -

P = n * P ( total )

P = partial pressure of the gas ,

n = mole fraction of the gas ,

P ( total ) = Total pressure of the mixture of gas .

Hence , from the above equation ,

the partial pressure is directly proportional to the mole fraction , i.e. , on increasing the number of moles of gas , the partial pressure increases .

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How much does the air weigh within a column that is 1 square inch in area that extends from sea level all the way to the top of

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Answer:

14.7 lbs

Explanation:

Air pressure is the weight of the air above us. It is approximately 14.7 pounds or lbs per square inch at sea level. It means that an air column weights 14.7 lbs, 1 square inch in diameter, reaching all the way up to the top of the atmosphere.

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3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

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3 years ago

7.) A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To

marysya [2.9K]

Answer:

T₂ = 721 k

Explanation:

Given data:

Initial volume = 285 mL

Initial pressure = 1.88 atm

Initial temperature = 355 K

Final temperature = ?

Final volume = 435 mL

Final pressure = 2.50 atm

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂

T₂ = P₂V₂ T₁ / P₁V₁

T₂ = 2.50 atm × 435 mL × 355 K / 1.88 atm × 285 mL

T₂ = 386062.5 atm. mL. K /535.8 atm. mL

T₂ = 721 k

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3 years ago

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3 years ago

Which of the following best defines Avogadro's number? (5 points)

Likurg_2 [28]

Answer:

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2 years ago

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