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ira [324]
3 years ago
8

A student collects 350 mL of a vapor at a temperature of 67°C. The atmospheric pressure at the time of collection is 0.900 atm.

How many moles of the gas does the student collect?
0.011 moles.
0.053 moles
0.11 moles
0.53 moles
Chemistry
1 answer:
castortr0y [4]3 years ago
8 0

Answer: First option 0.011 moles

Explanation:

You need to use the formula for Ideal gases which is:

PV = nRT

Where:

P = Pressure (in atm)

V = Volume (In Liters)

n = moles

R = gas constant (this depends of the units; in this case is 0.082 L atm/K)

T = temperature (In K)

From the formula above, we can solve for n:

n = PV / RT

Let's convert first the temperature and volume to K and L respectively:

T = 67 + 273 = 340 K

V = 350 mL / 1000 = 0.35 L

Finally, let's put all the values in the formula above to solve for the value of n:

n = 0.9 * 0.35 / 0.082 * 340

n = 0.315 / 27.88

n = 0.011 moles

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r = 8917 Km

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Read 2 more answers
How many moles of aqueous potassium ions and sulfate ions are formed when 63.7 g of K2SO4 dissolves in water
Viefleur [7K]

Answer:

WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.

Explanation:

Equation for the reaction:

K2SO4 + H20 ------->2 K+ + SO4^2-

When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.

1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-

At STP, 1 mole of K2SO4 will be the molar mass of the substance

Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol

Molar mass = 174 g/mol

So therefore;

1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion

When 63.7 g is used; we have:

174 g = 2 moles of K+

63.7 g = ( 63.7 * 2 / 174) moles of K+

= 0.73 moles of K+

Forr sulfate ion, we have:

174 g = 1 mole ofSO4^2-

63.7 g = (63.7 * 1 / 174) moles of SO4^2-

= 0.366 moles of SO4^2-

In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.

4 0
3 years ago
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