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aleksley [76]
2 years ago
12

A pipet is used to transfer 5.00 mL of a 1.25 M stock solution in flask "S" to a 25.00 mL volumetric flask "B," which is then di

luted with DI H2O to the calibration mark. The solution is thoroughly mixed. Next, 2.00 mL of the solution in volumetric flask "A" is transferred by pipet to 50.00 mL volumetric flask "B" and then diluted with DI H2O to the calibration mark. Calculate the molarity of the solution in volumetric flask "B". How do I solve this?
Chemistry
1 answer:
Leto [7]2 years ago
5 0

Answer: the molarity of the solution in volumetric flask "B' is 0.0100 M

Explanation:

Given that;

the Molarity of stock solution M₁ = 1.25M

The molarity os solution in volumetric flask A (M₂) = M₂

Volume of stock solution pipet out (V₁) = 5.00mL

Volume of solution in volumetric flask A V₂ = 25.00mL

using the dilution formula

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

WE SUBSTITUTE

M₂ =  ( 1.25 × 5.00 ) / 25.00 mL

M₂ = 0.25 M

Now volume of solution pipet out from volumetric flask A V₂ = 2.00 mL

Molarity of solution in volumetric flask B (M₃) = M₃

Volume of solution in volumetric flask B V₃ = 50.00m L

Using dilution formula again

M₂V₂ = M₃V₃

M₃ = M₂V₂ / V₃

WE SUBSTITUTE

M₃ = ( 0.25 × 2.0) / 50.0

M₃ = 0.0100 M

Therefore the molarity of the solution in volumetric flask "B' is 0.0100 M

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bekas [8.4K]

Answer:

9.88

Explanation:

As higher is the Ksp, more soluble is the compound. So, Co(OH)₂ is the less soluble hydroxide.

The maximum concentration of it must be 1x10⁻⁶ M, and the reaction is:

Co(OH)₂(s) ⇄ Co⁺²(aq) + 2OH⁻(aq)

So, [Co⁺²] = 1x10⁻⁶M

Ksp =  [Co⁺²] *[OH⁻]²

[OH⁻]² = 5.9x10⁻¹⁵/1x10⁻⁶

[OH⁻] = √(5.9x10⁻⁹)

[OH⁻] = 7.6811x10⁻⁵

pOH = -log[OH⁻]

pOH = -log(7.6811x10⁻⁵)

pOH = 4.11

Knowing that pH + pOH = 14

pH = 14 - 4.11

pH = 9.88

7 0
1 year ago
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The atmosphere supports a column of mercury that is 748 mm in height. What is atmospheric pressure in torr? Convert this pressur
gulaghasi [49]

Answer:

748 torr

Explanation:

mmHg and torr are equivalent so, you'll have 748 torr.

7 0
1 year ago
Give two areas where the compressible nature of gas is applied​
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Answer:

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6 0
1 year ago
A vessel of capacity 400 cc filled with chlorine under 80 cm is connected by a narrow tube and stopcock with another vessel of c
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Answer:

The pressure when the stopcock is opened is opened is 87.783 cm

Explanation:

The given parameters of the question are;

The volume of the vessel of chlorine = 400 cc

The pressure of the vessel of chlorine = 80 cm

The volume of the vessel of chlorine = 250 cc

The pressure of the vessel of chlorine = 100 cm

Daltons law of Partial Pressures states that the total pressure exerted by a volume of a mixture of gases is equal to the partial pressures exerted by the individual gases in the mixture with respect to the given volume

Therefore;

The total volume of the mixture = 400 cc + 250 cc = 650 cc

The partial pressure exerted by the chlorine gas in the total volume is given by Boyles law as follows;

P₁·V₁ = P₂·V₂

P₂ = P₁·V₁/V₂

Where;

P₁ = 80 cm = The pressure in volume V₁ = 400 cc

P₂ₓ = The partial pressure of chlorine in volume V₂ = 650 cc

Substituting, we have;

P₂ₓ = 80 × 400/650 ≈ 49.321 cm

Similarly, the partial pressure exerted by the nitrogen gas in the total volume is given by Boyles Law as follows;

P₂ₐ = P₁·V₁/V₂

Where;

P₁ = 100 cm = The pressure in volume V₁ = 250 cc

P₂ₐ = The partial pressure of nitrogen in volume V₂ = 650 cc

Substituting, we have;

P₂ₐ = 100× 250/650 ≈ 38.462 cm

The pressure of the combined gas, P, when the stopcock is opened is opened is given by Dalstons Law of partial pressure as P = P₂ₐ + P₂ₓ

Therefore, the pressure, P when the stopcock is opened is opened = 49.321 cm + 38.462 cm = 87.783 cm

3 0
1 year ago
Which statement describes a chemical reaction?
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4 0
2 years ago
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