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3 years ago

What is the new volume when the pressure of 20 milliliters of an ideal gas decreases from 202 kPA to 101 kPA?

Chemistry

1 answer:

Lana71 [14]3 years ago

6 0

Ideal Gas Law Formula.

Brainliest?

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3 years ago

Read 2 more answers

Be sure to answer all parts. A mixture of CO2 and Kr weighs 47.9 g and exerts a pressure of 0.751 atm in its container. Since Kr

ololo11 [35]

Answer:a) Mass of $CO_2$ in mixture = 26.3 grams

b) Mass of $Kr$ in mixture that can be recovered= 21.5 grams

Explanation:

Given :

Total Pressure = pressure of krypton + pressure of carbon dioxide = 0.751 atm

pressure of krypton = 0.231 atm

Thus pressure of carbon dioxide = 0.751 - 0.231 =0.52 atm

As we know:

$p=x\times P$

where,

$p$ = partial pressure

$P$ = total pressure = 0.751 atm

$x$ = mole fraction

For $CO_2$

$x_{CO_2}=\frac{p_{CO_2}}{P}=\frac{0.52}{0.751}=0.70$

${\text {Mass of} CO_2}=moles\times {\text {Molar mass}}=0.70\times 44=30.8g$

For $Kr$

$x_{Kr}=\frac{p_{Kr}}{P}=\frac{0.231}{0.751}=0.30$

${\text {Mass of} Kr}=moles\times {\text {Molar mass}}=0.30\times 84=25.2g$

Total mass = Mass of $CO_2$ + Mass of krypton = 30.8 + 25.2 = 56 g

Percentage of $CO_2=\frac{30.8}{56}\times 100=55\%$

a) Thus Mass of $CO_2$ in mixture = $\frac{55}{100}\times 47.9=26.3g$

Percentage of $Kr=\frac{25.2}{56}\times 100=45\%$

b) Thus Mass of $Kr$ in mixture that can be recovered= $\frac{45}{100}\times 47.9=21.5g$

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3 years ago

Aqueous potassium iodate and potassium iodide react in the presence of dilute hydrochloric acid, as shown below. kio3(aq) + 5ki(

Musya8 [376]

<span>0.0797 g Looking at the formula, 1 mole of KIO3 and 5 moles of KI will react and produce moles of iodine molecules or 6 moles of iodine atoms. So first, determine the number of moles of KIO3 and KI provided moles KIO3 = 0.0121 * 0.097 = 0.0011737 mol moles KI = 0.0308 * 0.017 = 0.0005236 mol The limiting reactant is KI at 0.0005236 mol so divide by 5 and multiply by 6 to get the number of moles of iodine atoms. 0.0005236 / 5 * 6 = 0.00062832 mol Lookup the atomic weight of iodine which is 126.90447 And multiply that by the number of moles of iodine produced 126.90447 g/mol * 0.00062832 mol = 0.079736617 g Rounding to 4 decimal places gives 0.0797 g</span>

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3 years ago