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2 years ago

What is the pressure of 1.50 moles of a gas in a 30.0L tank at a temperature of 285K?

Chemistry

1 answer:

julia-pushkina [17]2 years ago

7 0

The pressure of the gas : 1.1685 atm

<h3>Further explanation</h3>

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

n=moles=1.5

V=volumes = 30 L

T=temperature=285 K

The pressure :

$\tt P=\dfrac{nRT}{V}\\\\n=\dfrac{1.5\times 0.082\times 285}{30}\\\\P=1.1685~atm$

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If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate

andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

Volume of NaOH = 32.40 mL = 0.03240 L

Molarity of NaOH = 0.258 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of the acetic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 1

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

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒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

6 0

3 years ago

Read the given equation:

Assoli18 [71]

Answer:

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3 years ago

A little boy has placed his pet squirrel into a balloon with a volume of 3L at a pressure of 1.0 atm. If the boy takes his pet s

s2008m [1.1K]

Answer: The squirrel's balloon will be 0.86 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

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$P_1\text{ and }V_1$ are initial pressure and volume.

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We are given:

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Putting values in above equation, we get:

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3 years ago

A generic weak acid with formula HA has a Ka = 2.76 x 10-8. Calculate the Kb for the conjugate base of the acid.

Reika [66]

Answer:

3.62x10⁻⁷ = Kb

Explanation:

The acid equilibrium of a weak acid, HX, is:

HX + H₂O ⇄ X⁻ + H₃O⁺

Where Ka = [X⁻] [H₃O⁺] / [HX]

And basic equilibrium of the conjugate base, is:

X⁻ + H₂O ⇄ OH⁻ + HX

Where Kb = [OH⁻] [HX] / [X⁻]

To convert Ka to Kb we must use water equilibrium:

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Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

Thus, we can obtain:

Kw = Ka*Kb

Solving for Kb:

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