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2 years ago

Given 700 ml of oxygen at 7 ºC and 106.6 kPa pressure, what volume does it take at 27ºC and 66.6 kPa pressure

Chemistry

1 answer:

o-na [289]2 years ago

6 0

Answer:

The volume is 1.2L

Explanation:

Initial volume (V1) = 700mL = 0.7L

Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K

Initial pressure = 106.6kPa = 106600Pa

Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K

Final pressure (P2) = 66.6kPa = 66600Pa

Final volume (V2) = ?

To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.

(P1 × V1) / T1 = (P2 × V2) / T2

solve for V2 by making it the subject of formula,

P1 × V1 × T2 = P2 × V2 × T1

V2 = (P1 × V1 × T2) / (P2 × T1)

V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)

V2 = 22397193 / 18657990

V2 = 1.2L

The final volume of the gas is 1.2L

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Using the formula M1V1 = M2V2 , how many milliliters of a 2. 50 M hydrochloric acid solution is required to make 100. 0 mL of a

gladu [14]

Dilution of the solution can be calculated by the formula of the molarity and volume. The initial volume of 2.50 M solution was 30 mL.

<h3>What is the relationship between molar concentration and dilution?</h3>

Molar concentration or the dilution factor is in an inverse relationship and with an increase in the dilution, the molarity of the solution decreases.

Given,

Initial molarity = 2.50 M

initial volume = ?

Final molarity = 0.750 M

Final volume = 100.0 ml

Substituting values in the formula:

$\begin{aligned}\rm V_{1} &= \rm \dfrac{M_{2}V_{2}}{M_{1}}\\\\&= \dfrac{0.750 \times 100}{2.50}\\\\&= 30 \;\rm mL\end{aligned}$

Therefore, 30 mL was the initial volume of the solution before it was diluted.

Learn more about dilution here:

brainly.com/question/26896011

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1 year ago

One litre of hydrogen at STP weight 0.09gm of 2 litre of gas at STP weight 2.880gm. Calculate the vapour density and molecular w

expeople1 [14]

Answer:

we know, at STP ( standard temperature and pressure).

we know, volume of 1 mole of gas = 22.4L

weight of 1 Litre of hydrogen gas = 0.09g

so, weight of 22.4 litres of hydrogen gas = 22.4 × 0.09 = 2.016g ≈ 2g = molecular weight of hydrogen gas.

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weight of 2L of a gas = 2.88gm

so, weight of 22.4 L of the gas = 2.88 × 22.4/2 = 2.88 × 11.2 = 32.256g

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hence, vapor density of the gas is 16.128g.

Explanation:

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