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Goshia [24]
3 years ago
10

The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the st

ems are used to make rhubarb pie. if pka1 = 1.23 and pka2 = 4.19, what is the ph of a 0.0269 m solution of oxalic acid?
Chemistry
1 answer:
VLD [36.1K]3 years ago
3 0
Chemical reaction (dissociation) 1: C₂O₄H₂(aq) ⇄ C₂O₄H⁻(aq) + H⁺(aq).
Chemical reaction (dissociation) 2: C₂O₄H⁻(aq) ⇄ C₂O₄²⁻(aq) + H⁺(aq).
c(C₂O₄H⁻) = c(H⁺) = x.
c(C₂O₄H₂) = 0.0269 M.
pKa₁ = 1.23.
Ka₁ = 10∧(-1.23) = 0.059.
Ka₁ = c(C₂O₄H⁻) · c(H⁺) / c(C₂O₄H₂).
0.059 = x² / (0.0269 M - x).
Solve quadratic eqaution: x = c(H⁺) = 0.02 M.
pH = -log(0.02 M) = 1.7.

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How many moles of NaCl are produced if 239.7 grams of Na2S reacts with plenty of AlCl3?
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Answer:

6.142 moles of NaCl

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2AlCl3 + 3Na2S —> Al2S3 + 6NaCl

Next, we determine the number of mole in 239.7 g of Na2S. This is illustrated below:

Mass mass of Na2S = 78.048g/mol

Mass of Na2S = 239.7g

Number of mole Na2S =..?

Mole = Mass /Molar Mass

Number of mole Na2S = 239.7/78.048 = 3.071 moles

Finally, we can obtain the number of mole of NaCl produced from the reaction as follow:

From the balanced equation above,

3 moles of Na2S reacted to produce 6 moles of NaCl.

Therefore, 3.071 moles of Na2S will react to produce = (3.071 x 6)/3 = 6.142 moles of NaCl

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Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.
goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}

In the first solution:

X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625

X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375

340torr = 0.625P^0_{A}+0.375P^0_{B} <em>(1)</em>

For the second equation:

X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700

X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300

370torr = 0.700P^0_{A}+0.300P^0_{B}<em>(2)</em>

Replacing (2) in (1):

340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})

340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}

-122.5torr = -0.250P°A

P^0_{A} = 490 torr

<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>

And for benzene:

370torr = 0.700*490torr+0.300P^0_{B}

P^0_{B}=90torr

<em>Vapour pressure of benzene at 50°C is 90torr</em>

3 0
3 years ago
When a skin diver is on the top of the water at 1.03 atm, her lung volume is 3.62L. How will the volume of her lungs change as s
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Answer:

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Explanation:

We'll begin by calculating the final volume. This can be obtained as follow:

Initial pressure (P₁) = 1.03 atm

Initial volume (V₁) = 3.62 L

Final pressure (P₂) = 2.68 atm

Final volume (V₂) =?

P₁V₁ = P₂V₂

1.03 × 3.62 = 2.68 × V₂

3.7286 = 2.68 × V₂

Divide both side by 2.68

V₂ = 3.7286 / 2.68

V₂ = 1.39 L

Finally, we shall determine the change in volume. This can be obtained as follow:

Initial volume (V₁) = 3.62 L

Final volume (V₂) = 1.39 L

Change in volume (ΔV) =?

ΔV = V₂ – V₁

ΔV = 1.39 – 3.62

ΔV = –2.23 L

Thus, the change in the volume of her lung is –2.23 L.

NOTE: The negative sign indicate that the volume of her lung reduced as she goes below the surface!

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