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Vinil7 [7]
3 years ago
12

What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. m

l of solution?
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻

M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol

15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2


Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol                               2 mol
0.100 mol                           0.200 mol

We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M

Concentration of NO3⁻ is 0.667 M.



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Answer:

0.51M

Explanation:

Given parameters:

Initial volume of NaBr = 340mL

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Final volume  = 1000mL

Unknown:

Final molarity = ?

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This is a dilution problem whereas the concentration of a compound changes from one to another.

In this kind of problem, we must establish that the number of moles still remains the same.

    number of moles initially before diluting = number of moles after dilution

Number of moles  = Molarity x volume

Let us find the number of moles;

          Number of moles  = initial volume x initial molarity

Convert mL to dm³;

                  1000mL  = 1dm³

                     340mL gives \frac{340}{1000}   = 0.34dm³

Number of moles  = initial volume x initial molarity  = 0.34 x 1.5 = 0.51moles

Now to find the new molarity/concentration;

               Final molarity  = \frac{number of moles}{Volume}    = \frac{0.51}{1}    = 0.51M

We can see a massive drop in molarity this is due to dilution of the initial concentration.

6 0
3 years ago
40.0L of N₂ gas are in a sealed container at STP.How many moles of N₂ are present?9 mol
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Explanation:

We have to find the number of moles of N₂ that are present in a sample that has a volume of 40.0 L at STP.

STP means Standard Conditions of Temperature and Pressure. These conditions are 273.15 K and 1 atm. We know that 1 mol of N₂ will occupy 22.4 L. We can use that ratio to find the answer to our problem.

1 mol of N₂ = 22.4 L

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Answer: 1.79 moles of nitrogen are present.

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