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grandymaker [24]
3 years ago
12

Which of the following equations violates the law of conservation of mass?. . 2Al + 6HCl----> 2AlCl3 + 3H2. . Na2C2O4 + Pb(NO

3)2----> PbC2O4 + 2NaNO3. . 3Mg + 2Cr(NO3)3 -----> 3Mg(NO3)3 + 2Cr. . AgNO3 + LiCl ----> AgCl + LiNO3
Chemistry
1 answer:
Kaylis [27]3 years ago
4 0
<span>3Mg + 2Cr(NO3)3 -----> 3Mg(NO3)3 + 2Cr violates the law of conservation of mass.</span>
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14. 60. g of NaOH is dissolved in enough distilled water to make 300 mL of a stock solution. What volumes of this solution and d
zepelin [54]

The question is incomplete, the complete question is attached below.

Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

Explanation : Given,

Mass of NaOH = 60 g

Volume of stock solution = 300 mL

Molar mass of NaOH = 40 g/mol

First we have to calculate the molarity of stock solution.

\text{Molarity}=\frac{\text{Mass of }NaOH\times 1000}{\text{Molar mass of }NaOH\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M

Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock solution.

M_2\text{ and }V_2 are the molarity and volume of diluted solution.

From data (A) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 20mL=1M\times V_2\\\\V_2=100mL

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (B) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 20mL=1M\times V_2\\\\V_2=100mL

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 60mL=1M\times V_2\\\\V_2=300mL

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 60mL=1M\times V_2\\\\V_2=300mL

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

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