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JulsSmile [24]
3 years ago
15

The volume of a gas at 2.0 atm is 3.0 L. What is the volume of the gas at 1.5 atm at the same temperature?

Chemistry
1 answer:
pshichka [43]3 years ago
6 0

Answer:

4 L

Explanation:

Ideal gas law is P1V1T2=P2V2T1

V2=P1V1/P2

T is not necessary to add since it is constant.

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Mg(ClO2)2 and Al(ClO2)3 contain the same number of atoms
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7 0
3 years ago
If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate
NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

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Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

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I have attached the answer

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Answer:

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Explanation:

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