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prisoha [69]
3 years ago
12

10. What is the atmospheric pressure if the partial pressures of nitrogen, oxygen, and

Chemistry
1 answer:
sashaice [31]3 years ago
7 0

Answer:

1.009 atm is the total pressure for the mixture

Explanation:

To determine the total pressure in atm, of the three gases (N₂, O₂ and Ar) we have to sum all the values.

Sum of partial pressures in a mixture = Total pressure

First of all, we need to convert the values from mmHg to atm

604.5 mmHg . 1atm / 760 mmHg = 0.795 atm

162.8 mmHg . 1atm / 760 mmHg = 0.213 atm

0.500 mmHg . 1atm / 760 mmHg = 6.58×10⁻⁴ atm

Partial pressure N₂ + Partial pressure O₂ + Partial pressure Ar = Total P

0.795 atm + 0.213 atm + 6.58×10⁻⁴ atm = 1.009 atm

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125.0 g of an unknown substance is heated to 97.0°C. It is then placed in a calorimeter than contains 250g of water with an init
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The heat released by the substance in the calorimeter is equal to  the heat absorbed by water which results to the decrease and increase in temperature, respectively.
We use m Cp ΔT to balance the heat involved 

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Answer:

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Explanation:

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Can you tell me about atoms
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3 years ago
Suppose of potassium sulfate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium
dimulka [17.4K]

Answer:

This question is incomplete, here's the complete question:

<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>

Explanation:

Reaction :-

K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4

Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol

Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol

Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L

Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L

Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol

Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.

0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+

Final concentration of potassium cation

= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M

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3 years ago
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