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faust18 [17]
3 years ago
12

2.5 moles of sodium chloride is dissolved to make 0.050 liters of solution

Chemistry
1 answer:
Hitman42 [59]3 years ago
7 0

The answer is:

the molarity = 50 moles/liters

The explanation:

when the molarity is = the number of moles / volume per liters.

and when the number of moles =2.5 moles

and the volume per liters = 0.05 L

so by substitution:

the molarity = 2.5moles/0.05L

                    = 50 moles /L

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Bonding electrons are involved in chemical bonding these electrons have their valnce shell incomplete
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An atom has a negative charge. Which of the following must have occurred?
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It should be A)It lost a neutron.
5 0
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The expected value for a chemical equation is 47g of water, after an experiment you find that you have 2.58 moles of water. What
strojnjashka [21]
<h3><u>Answer</u>;</h3>

Actual yield = 46.44 g

<h3><u>Explanation;</u></h3>

1 mole of water = 18 g/mol

Therefore;

The experimental yield = 2.58 moles

equivalent to ; 2.58 × 18 = 46.44 g

The theoretical value is 47 g

Percentage yield = 46.44/47 × 100%

                             = 98.8%

The questions asks for actual yield = 46.44 g

6 0
3 years ago
when a solution of sodium chloride is added to a solution of copper(ii) nitrate, no precipitate is observed. Write the molexular
zhuklara [117]

Explanation:

1.

Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)

2.

Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)

A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.

3.

Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)

2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)

Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)

2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)

4.

The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.

Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.

6 0
3 years ago
Calculate the molar solubility of CaF2 in a 0.25 m solution of NaF(aq).
Kipish [7]

Answer:

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

8 0
3 years ago
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