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kompoz [17]
2 years ago
6

If I have 340 mL of a 1.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it?

Chemistry
1 answer:
ipn [44]2 years ago
5 0

Answer:

0.5667 M ≅ 0.57 M.

Explanation:

It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.

It can be expressed as:

(MV) before dilution = (MV) after dilution.

M before dilution = 1.5 M, V before dilution = 340 mL.

M after dilution = ??? M, V after dilution = 340 mL + 560 mL = 900 mL.

∴ M after dilution = (MV) before dilution/(V) after dilution = (1.5 M)(340 mL)/(900 mL) = 0.5667 M ≅ 0.57 M.

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Read 2 more answers
Ammonia, NH3; ammonium nitrate, NH4NO3; and ammonium hydrogen phosphate, (NH4)2HPO4, are all common fertilizers. Rank the compou
ValentinkaMS [17]

Answer:

Ammonia, NH₃ > ammonium nitrate, NH₄NO₃ > ammonium hydrogen phosphate, (NH₄)₂HPO₄

Explanation:

Mass percentage -

Mass percentage of A is given as , the mass of the substance A by mass of the total solution multiplied by 100.

i.e.

mass % A = mass of A / mass of solution * 100

Given,  

mass of nitrogen = 14 g/mol

mass of hydrogen = 1 g/mol

mass of oxygen = 16 g/mol

mass of phosphorus = 31 g/mol

1. Ammonia, NH₃

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 1 * 14 g/mol  + 3 * 1 g/mol  

mass of solution = 17 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 17 g/mol * 100

mass % N = 82.35 % .

2.   ammonium nitrate, NH₄NO₃

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 2 * 14 g/mol  + 4 * 1 g/mol  + 3 * 16 g/mol

mass of solution = 80 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 80 g/mol * 100

mass % N = 17.50 % .

3.   ammonium hydrogen phosphate, (NH₄)₂HPO₄

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 2 * 14 g/mol  + 9 * 1 g/mol  + 4 * 16 g/mol + 1 * 31 g/mol

mass of solution = 132 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 132 g/mol * 100

mass % N = 10.60 % .

Hence , the correct order is -

Ammonia, NH₃ > ammonium nitrate, NH₄NO₃ > ammonium hydrogen phosphate, (NH₄)₂HPO₄

3 0
2 years ago
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