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2 years ago

B. Determine the percent composition of each element in NaCl. (1 point) Must show work to

Chemistry

1 answer:

andrey2020 [161]2 years ago

8 0

Answer:

Sodium (Na): (.5 point)

22.99+35.453=58.433. 22.99/58.433= 39%

Chlorine (Cl): (.5 point)

22.99+35.453=58.433. 35.453/58.433= 61%

Explanation:

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Answer:

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3 years ago

What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?

mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

$C=\frac{n}{V}$ .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is $K_{2}SO_{4}$ thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass

Putting the values,

$n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol$

Thus, number of moles of $K_{2}SO_{4}$ will be $0.1061\times 0.5=0.053 mol$ .

The volume of solution is 225 mL, converting this into L,

$1 mL=10^{-3}L$

Thus,

$225 mL=0.225 L$

Putting the values in equation (1),

$C=\frac{(0.053 mol}{0.225 L}=0.236 M$

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3 years ago

Draw the structures of methyl oleate and propylene glycol. Which one is more polar, and how can you tell

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3 years ago

What is the maximum mass of P2I4 that can be prepared from 8.80g of P4O6 and 12.37g of iodine according to the reaction:

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The balanced chemical reaction would be as follows:

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8.80 g P4O6 (1 mol / 219.88 g) = 0.04 mol P4O6
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Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:

0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (569.57 g / 1 mol) = 14.24 g P2I4

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3 years ago

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What is the molarity of 6 moles (mol) of NaCl dissolved in 2 L of water?

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Answer:

C: $\frac{6 mol}{2 L}$