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Yuliya22 [10]
2 years ago
5

4.

Chemistry
2 answers:
Rama09 [41]2 years ago
7 0

Answer:

\boxed {\boxed {\sf 426 \ mL}}

Explanation:

We are asked to find the volume of ammonia gas given a change in pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure of a gas. The formula is:

P_1V_1= P_2V_2

The ammonia gas originally occupies a volume of 450 milliliters at a pressure of 720 millimeters of mercury. Substitute the values into the formula.

450 \ mL * 720 \ mm \ Hg = P_2V_2

The pressure is changed to standard atmospheric pressure, which is 760 millimeters of mercury. The new volume is unknown.

450 \ mL * 720 \ mm \ Hg = 760 \ mm \ Hg * V_2

We are solving for the volume at standard pressure. We will need to isolate the variable V₂. It is being multiplied by 760 millimeters of mercury. The inverse of multiplication is division. Divide both sides of the equation by 760 mm Hg.

\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= \frac{760 \ mm \ Hg * V_2}{760 \ mm \ Hg}

\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= V_2

The units of millimeters of mercury (mm Hg) cancel.

\frac {450 \ mL * 720  }{760} = V_2

\frac {324,000}{760} \ mL = V_2

426.3157895 \ mL =V_2

The original values of volume and pressure have 3 significant figures. Our answer must have the same. For the number we calculated, that is the ones place. The 3 in the tenths place tells us to leave the 6 in the ones place.

426 \ mL \approx V_2

The volume at standard atmospheric pressure is approximately <u>426 milliliters. </u>

True [87]2 years ago
3 0
P1V1 = P2V2

P1 = 720 mmHg
V1 = 450. mL
P2 = 760 mmHg (this is the pressure at STP)

Use these to solve for V2:
(720)(450) = 760V2

V2 = 426 mL
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<h3>Further explanation</h3>

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  • 155 g of butane

relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol

tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]

\large mole\:=\:\large \frac{155}{58}

mole = 2,672

Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:

23.1 kJ / mol x 2,672 mol = 61,723 kJ

<h3>Learn more</h3>

the heat of vaporization

brainly.com/question/11475740

The latent heat of vaporization

brainly.com/question/10555500

brainly.com/question/4176497

Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic

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