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patriot [66]
3 years ago
15

Calculate the percent by mass of potassium nitrate in a solution made from 45.0 g kno3 and 295 ml of water. The density of water

is 0.997 g/ml.
Chemistry
1 answer:
Tanzania [10]3 years ago
6 0

Answer:  the percent by mass of potassium nitrate is 13.3%

Explanation:

<u>1) Data:</u>

  • Mass of solute, m₁ = 45.0 g
  • Volume of solvent, V₂ = 295 ml
  • density of water, d₂ = 0.997 g/ml

<u>2) Formulae</u>:

  • Percent by mass, % = (mass of solute / mass of solution) × 100

  • Density, d = mass / volume

<u>3) Solution</u>:

  • d = mass / volume ⇒ m₂ = d₂ × V₂ = 0.997 g/ml × 295 ml = 294. g
  • mass of solution = m₁ + m₂ = 45.0g + 294. g = 339. g
  • % = (45.0 g / 339. g) × 100 = 13.3 %
  • The answer has to be reported with 3 significant figures.
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Mass in grams of hydrochloric acid HCl = ?

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A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L
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Answer:

\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}

Explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

P_1V_1=P_2V_2

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

25.0 \ L * 2.05 \ atm = P_2V_2

The volume is decreased to 14.5 liters, but the pressure is unknown.

25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}

\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2

The units of liters cancel.

\frac {25.0  * 2.05 \ atm }{14.5 }=P_2

\frac {50.25\  atm }{14.5 }=P_2

3.53448276 \ atm = P_2

The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

  • 3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

3.53 \ atm \approx P_2

The new pressure is approximately <u>3.53 atmospheres.</u>

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