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sashaice [31]
4 years ago
5

You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution

is known. The concentration of the KCN solution given in Part A corresponds to a mass percent of 0.180 %. What mass of a 0.180 % KCN solution contains 768 mg of KCN?
Chemistry
1 answer:
LUCKY_DIMON [66]4 years ago
8 0

<u>Answer:</u> The mass of solution having 768 mg of KCN is 426.66 grams.

<u>Explanation:</u>

We are given:

0.180 mass % of KCN solution.

0.180 %(m/m) KCN solution means that 0.180 grams of KCN is present in 100 gram of solution.

To calculate the mass of solution having 768 mg of KCN or 0.786 g of KCN   (Conversion factor:  1 g = 1000 mg)

Using unitary method:

If 0.180 grams of KCN is present in 100 g of solution.

So, 0.768 grams of KCN will be present in = \frac{100g}{0.180g}\times 0.768=426.66g of solution.

Hence, the mass of solution having 768 mg of KCN is 426.66 grams.

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Answer:

final mole fraction of O₂ = 58.84% , CO₂ = 17.64% , H₂O = 23.52% .

final partial pressure of O₂ = 2.942 atm , CO₂ = 0.882 atm , H₂O = 1.176 atm .

Explanation:

Assuming that propane is present as a gas , and also that the combustion is complete:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

then taking as a reference propane sample= 1 mol , then

initial moles of O₂ = 3* 5 moles = 15 moles

final moles of O₂= 3* 5 moles - 5 moles = 10 moles

final moles of CO₂ = 3 moles

final moles of H₂O = 4 moles

total number of moles = 10 + 3 + 4 = 17 moles

final mole fraction of O₂ = 10/17 = 0.5884 = 58.84%

final mole fraction of  CO₂ = 3/17 = 0.1764 = 17.64%

final mole fraction of  H₂O = 4/17 = 0.2352 = 23.52%

From Dalton's law for ideal gases , the partial pressure p=P*x then

final partial pressure of O₂ = 5 atm * 10/17 = 2.942 atm

final partial pressure of CO₂ = 5 atm * 3/17 = 0.882 atm

final partial pressure of H₂O = 5 atm * 1/17 = 1.176 atm

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3 years ago
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Luden [163]

Answer:

A) M = 100X

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C) M = 178.88X

Explanation:

Given data:

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we know that number of grain per square inch is given as

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where M is magnification, n is grain size

therefore we have

64 = 2^{7-1}(\frac{100}{M})^2

solving for M we get

M = 100 X

B)  total grain per inch^2 = 500 grain/inch^2

we know that number of grain per square inch is given as

Nm = 2^{n-1} (\frac{100}{M})^2

where M is magnification, n is grain size

therefore we have500 = 2^{7-1}(\frac{100}{M})^2

solving for M we get

M = 36 X

C) Total grain per inch^2 = 20 grain/inch^2

we know that number of grain per square inch is given as

Nm = 2^{n-1} (\frac{100}{M})^2

where M is magnification, n is grain size

therefore we have20 = 2^{7-1}(\frac{100}{M})^2

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Explanation:

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