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Sergeeva-Olga [200]
3 years ago
11

You need to make an aqueous solution of 0.192 M barium sulfide for an experiment in lab, using a 500 mL volumetric flask. How mu

ch solid barium sulfide should you add?
Chemistry
1 answer:
shutvik [7]3 years ago
7 0

Answer:

Explanation:

Molecular weight of barium sulphide = 169

500 mL of .192 M barium sulfide = .5 x .192 moles of barium sulphide

= .096 moles of barium sulfide

= .096 x 169 gram of barium sulfide

= 16.22 grams of barium sulfide .

We shall have to add 16.22 gram .

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sertanlavr [38]

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Explanation:

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A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new
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When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

PV=nRT

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V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K   (0°C = 273 K)

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