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ella [17]
3 years ago
10

You are starting with 2 pounds of 30% acetic acid solution. You want to dilute

Chemistry
1 answer:
Tpy6a [65]3 years ago
8 0

Answer:

The mass of water to be added is 2 pounds

Explanation:

The given parameters are;

The mass of the given solution = 2 pounds

The concentration of the given solution = 30%

The desired concentration of the solution = 15%

The mass, m of the acetic acid in the given solution = 30% × 2 pounds

m = 30/100 × 2 pounds = 0.6 pounds

To make a 15% acetic acid solution of acetic acid, the mass X of the required volume, is given as follows;

15% of X = 0.6 pounds

15/100 × X = 3/20 × 0.6 pounds

∴ The mass of the solution required  X = 0.6 × 20/3 = 4 pounds

The mass of the solution that will contain 0.6 pounds of acetic acid giving a 15% acetic acid solution is 4 pounds

Therefore, the mass of water to be added to the original solution to make the a 15% acetic acid solution is 2 pounds.

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A gas has a volume of 490. mL at a temperature of -35.0 degrees C. What volume would the gas occupy at 42.0 degrees Celsius? Ple
miskamm [114]

Answer:

648.5 mL

Explanation:

Here we will assume that the pressure of the gas is constant, since it is not given or specified.

Therefore, we can use Charle's law, which states that:

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Mathematically:

\frac{V}{T}=const.

where

V is the volume of the gas

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The equation can be rewritten as

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where in this problem we have:

V_1=490 mL is the initial volume of the gas

T_1=-35.0^{\circ} + 273 = 238 K is the initial temperature

T_2=42.0^{\circ}+273=315 K is the final temperature

Solving for V2, we find the final volume of the gas:

V_2=\frac{V_1 T_2}{T_1}=\frac{(490)(315)}{238}=648.5 mL

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3 years ago
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When the temperature of the gas changes from cold to hot, the amount of pressure is ___________.
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