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Sloan [31]
2 years ago
14

How much mass of sodium chloride ( ) should be dissolved in water to make 1.5 L of 0.75 M aqueous solution? The molar mass of is

58.5 g.
Chemistry
1 answer:
iogann1982 [59]2 years ago
4 0

Answer:

Mass = 65.8 g

Explanation:

Given data:

Mass of sodium chloride = ?

Volume of solution = 1.5 L

Molarity of solution = 0.75 M

Solution:

Number of moles of sodium chloride:

Molarity = number of moles / volume in L

By putting values,

0.75 M = number of moles = 1.5 L

Number of moles = 0.75 M × 1.5 L

Number of moles = 1.125 mol

Mass of sodium chloride:

Mass = number of moles × molar mass

Mass = 1.125 mol × 58.5 g/mol

Mass = 65.8 g

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0.5 m

Explanation:

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Group 17 elements (for example, chlorine) in the periodic table are known as
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The answer is halogens

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3 years ago
how many grams of calcium chloride can be prepared from 60.4 G of calcium oxide and 69.0 G of hydrochloric acid and a double dis
Ksenya-84 [330]
Balanced equation: 
<span>CaO + 2 HCl --> CaCl2 + H2O </span>
<span>Calculate moles of each reactant: </span>
<span>60.4 g CaO / 56.08 g/mol = 1.08 mol CaO </span>
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<span>Identify the limiting reactant: </span>
<span>Moles CaO needed to react with all HCl: </span>
<span>1.89 mol HCl X (1 mol CaO / 2 mol HCl) = 0.946 mol CaO </span>
<span>Because you have more CaO than that available, HCl is the limiting reactant. </span>

<span>Calculate moles and mass CaCl2: </span>
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8 0
3 years ago
Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di
pychu [463]

Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)

Moles of calcium nitrate = \frac{31.3 g}{164 g/mol}=0.1908 mol

Moles of ammonium fluoride = \frac{38.7 g}{37 g/mol}=1.046 mol

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

\frac{1}{2}\times 1.046 mol=0.523 mol calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

\frac{2}{1}\times 0.1908 mol=0.3816 molof dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

8 0
3 years ago
How did Neils Bohr change the model of the atom
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