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dem82 [27]
3 years ago
14

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

late the volume of the 1.475 M stock NaOH solution needed to prepare 250.0 mL of 0.1374 M dilute NaOH solution.
Chemistry
1 answer:
Svet_ta [14]3 years ago
8 0

Answer:

V_1=23.3~mL

Explanation:

In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:

C_1*V_1=C_2*V_2

Now we can identify the variables:

C_1=~1.475_M

V_1=~?

C_2=~0.1374_M

V_2=~250.0~mL

If we plug all the values into the equation:

1.475_M*V_1=0.1374_M*250.0~mL

And we solve for V_1:

V_1=\frac{0.1374_M*250.0~mL}{1.475_M}

V_1=23.3~mL

I hope it helps!

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Nitrogen gas (112 g) reacts with hydrogen gas to produce 40.8 g of ammonia according to the following
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Answer:

%yield of NH₃ = 30%

Explanation:

Actual yield of NH₃ = 40.8g

Theoretical yield = ?

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Mass = number of moles * molar mass

Mass = 1 * 28.00 = 28g of N₂ (the number of moles of N₂ from the equation is 1).

From the equation of reaction,

28g of N₂ produce (2 * 17)g of NH₃

28g of N₂ = 34g of NH₃

112g of N₂ = x g of NH₃

X = (112 * 34) / 28

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