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Over [174]
3 years ago
10

Approximate temperature of a .7 mold sample of a gas at 1.2 ATM and volume of .170 L

Chemistry
1 answer:
faust18 [17]3 years ago
4 0
Assuming you meant a 0.7 mole sample of a gas, we can approximate this by assuming that the gas is acting ideally, and use the ideal gas law PV=nRT. Using the following values:
P = 1.2 Atm
V = 0.170 L
n = 0.7 mol
R = 0.08206 L-atm/mol-K
We can rewrite the equation as: PV/nR = T
Plug in our values:
(1.2 atm)(0.170 L)/(0.08206*0.7 moles) = approximately 3.55 Kelvin = T
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Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

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Step 2: The balanced equation

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Step 3: Calculate the concentration of the acetic acid

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⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

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⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.258 M

⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The concentration of acetic acid is 8.36 M

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