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GarryVolchara [31]
3 years ago
6

If you dilute 19.0 mL of the stock solution to a final volume of 0.310 L , what will be the concentration of the diluted solutio

n?
Chemistry
1 answer:
Dahasolnce [82]3 years ago
6 0

Answer:

M_2=0.613M_1

Explanation:

M_1 = Concentration of stock solution

M_2 = Concentration of solution

V_1 = Volume of stock solution = 19 mL

V_2 = Volume of solution = 0.31 L= 310 mL

We have the relation

M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{M_119}{310}\\\Rightarrow M_2=M_1\times\dfrac{19}{310}\\\Rightarrow M_2=0.613M_1

\boldsymbol{\therefore M_2=0.613M_1}

The concentration of the diluted solution will be 0.613 times the concentration of the stock solution.

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A compound that is usually used as a fertilizer can also be used as a powerful explosive. the compound has the composition 35.00
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Answer:

             Empirical Formula  =  NH₄NO₃ (Ammonium Nitrate)

Solution:

Step 1: Calculate Moles of each Element;

                      Moles of N  =  %N ÷ At.Mass of N

                      Moles of N  = 35.0 ÷ 14

                      Moles of N  =  2.5 mol


                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  = 59.96 ÷ 16

                      Moles of O  =  3.7475 mol


                      Moles of H  =  [100% - (%N + %O)] ÷ At.Mass of H

                      Moles of H  = [100% - (35.0 + 59.96)] ÷ 1.008

                      Moles of H  = [100% - 94.96] ÷ 1.008

                      Moles of H  = 5.04 ÷ 1.008

                      Moles of H  =  5 mol

Step 2: Find out mole ratio and simplify it;

                N                                        H                                     O

               2.5                                       5                                3.7475

            2.5/2.5                                5/2.5                          3.7475/2.5

                 1                                         2                                     1.5

Multiply Mole Ratio by 2,

                 2                                         4                                     3

Result:

         Empirical Formula  =  N₂H₄O₃

Or,

         Empirical Formula  =  NH₄NO₃

This empirical formula is also a Molecular Formula for Ammonium Nitrate a well known Fertilizer and often misused in the formation of Explosives.

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