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2 years ago

The atmosphere is about 21% oxygen. At a pressure of 749 torr, what is the partial pressure of O2?

Chemistry

1 answer:

Kitty [74]2 years ago

3 0

Answer:

157.29 Torr

Explanation:

This question may be solved by the mole fraction.

Mole fraction → Moles of gas / Total moles = Partial pressure of gas / Total pressure

If the atmosphere is about 21% you assume that the mole fraction is 0.21 so:

0.21 = Partial pressure of gas / Total pressure

0.21 = Partial pressure of gas / 749 Torr

Partial pressure of the gas = 749 Torr . 0.21 → 157.29 Torr

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In-s [12.5K]

Answer:

60 g/100 g water

Explanation:

Find 5 °C on the horizontal axis.

Draw a line vertically from that point until you reach the solubility curve for CaCl₂.

Then draw a horizontal line from there to the vertical axis.

The solubility of CaCl₂ is 60 g/100 g water.

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2 years ago

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Gre4nikov [31]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

A 7.12 L cylinder contains 1.21 mol of gas A and 4.94 mol of gas B, at a temperature of 28.1 °C. Calculate the partial pressure

GenaCL600 [577]

Answer:

$P_A=4.20atm\\\\P_B=17.1atm$

Explanation:

Hello!

In this case, since the equation for the ideal gas is:

$PV=nRT$

For each gas, given the total volume, temperature (28.1+273.15=301.25K) and moles, we can easily compute the partial pressure as shown below:

$P_A=\frac{n_ART}{V} =\frac{1.21mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_A=4.20atm\\\\P_B=\frac{n_BRT}{V} =\frac{4.94mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_B=17.1atm$

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8 0

2 years ago

A hydrate is typically a crystalline compound in which water molecules are chemically bound to another compound or an element.

dedylja [7]

Answer:

Subtract the mass of the CuSO4⋅ 5H2O from the mass of CuSO4 is the right one

7 0

2 years ago

At 524 mm Hg and 35 °C, a sample of gas occupies a volume of 275 ml. The gas is transferred to a 325-ml flask and the temperatur

Deffense [45]

Answer:

400.197mmHg

Explanation:

P1V1 / T1 = P2V2 / T2

Where P1=524 mm Hg V1 =275 ml T1 = 35°C +273 = 308k

V2= 325-ml T2= 5°C+273 = 278k , P2= ?

Substituting the values into the formula.

524 mm Hg ×275 ml /308k = P2×325-ml/278k

Cross multiply

524 mm Hg ×275 ml×278k=308k×P2×325-ml

40059800= 100100×P2

P2 = 40059800/100100

P2= 400.197mmHg

Hence, the second pressure will be 400.197mmHg

6 0

2 years ago