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weeeeeb [17]
2 years ago
11

A sample of an ideal gas at 1.00 atm and a volume of 1.02 l was placed in a weighted balloon and dropped into the ocean. as the

sample descended, the water pressure compressed the balloon and reduced its volume. when the pressure had increased to 65.0 atm, what was the volume of the sample? assume that the temperature was held constant.
Chemistry
1 answer:
melisa1 [442]2 years ago
8 0
<span>0.0157 L. For this equation use the combined gas law, which states P1*V1/T1 = P2*V2/T2, but in this case we can remove the T, because T1 = T2, because the temperature is constant. Then we solve for V2 because it is the new volume. So P1*V1/P2 =V2 Plug in the variables, P1 = 100 atm, V1 = 1.02 L, P2 = 65.0 atm, and V2 = 0.0157 atm.</span>
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How many grams of Na2SO4 should be weighed out to prepare 0.5L of a 0.100M solution?​
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Answer:

7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.

Explanation:

First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.

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1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4

1    ml of solution contain     14.208÷1000= 0.014 gram

0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.

 So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.

     

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B) 16 g

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