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Semmy [17]
3 years ago
11

Using the following data, determine the percent composition (by mass) of the NaCl (salt) in a binary mixture. The mass of the un

known binary mixture is 23.76g and the mass of the SiO2 (sand) alone is 15.01g. Record your answer to 3 significant figures, and include the percentage sign ('%'), or 'percent'.
Chemistry
1 answer:
son4ous [18]3 years ago
4 0

Answer: The percent composition (by mass) of the NaCl (salt) in a binary mixture is 36.8%

Explanation:

Binary mixture is a mixture containing two components which are mixed together in any proportion and do not combine chemically with each other.

Mass of mixture=Mass of SiO_2+Mass of NaCl = 23.76 g

Mass of SiO_2 = 15.01 g

Mass of NaCl = 23.76 g - 15.01 g =8.75 g

To calculate the mass percentage ,we use the formula:

\text{Mass percent}=\frac{\text{Mass of} NaCl}{\text{Mass of mixture}}\times 100

Putting values in above equation, we get:

{\text{Mass percent of} NaCl}=\frac{8.75g}{23.76g}\times 100=36.8\%

Thus the percent composition (by mass) of the NaCl (salt) in a binary mixture is 36.8%

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A 0.43g samle of KHP required 24.11cm of NaOH for neutralization. Calculate the molarity of NaOH
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1 mole of KHP reacted with 1 mole of NaOH.

Therefore, 0.002 mole of KHP will also react with 0.002 mole of NaOH.

Next, we shall convert 24.11 cm³ to L. This can be obtained as follow:

1000 cm³ = 1 L

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Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.00 2 mole

Volume = 0.02411 L

Molarity of NaOH =?

Molarity = mole /Volume

Molarity of NaOH = 0.002 / 0.02411

Molarity of NaOH = 0.083 M

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