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Illusion [34]
2 years ago
7

In fractional distillation, liquid can be seen running from the bottom of the distillation column back into the distilling flask

. What effect does this returning condensate have on the fractional distillation?
Chemistry
1 answer:
shepuryov [24]2 years ago
7 0

Answer:

substances with a higher boiling point are returning back to the flask which allows another substances with the specific context temperature (lower boiling point) to boil over and be purified.

Explanation:

The reason it happens because the lower boiling point substance vaporizes and crosses over while the other substance is waiting for its boiling point to reach

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The concentration of hydrogen peroxide in a hair bleach is 0.050 mol dm–3. Use your answer from (b) to calculate the dilution fa
stellarik [79]

Answer:

0.03

Explanation:0.050 -

8 0
2 years ago
A cylinder contains 4.0 g of nitrogen gas. A piston compresses the gas to half its initial volume. Afterward, by what factor has
notka56 [123]

Answer:

The mass density will be doubled

Explanation:

  • Density is given by dividing the mass of a substance by its volume.
  • An increase in mass causes an increase in density and vice versa, while a  decrease in volume causes an increase in density and volume.
  • Therefore, when the volume is halved, then the density will be doubled if the mass is kept constant.
  • This has no effect on the number of moles as the mass is constant.
8 0
2 years ago
For the reaction: MgF2(s) ⇌ Mg2+(aq) + 2F- (aq), Ksp= 6.4 × 10-9, the addition of 0.10 M NaF to the solution cause what effect o
galina1969 [7]

Answer:

Shifts the equilibrium to the left. reduces solubility.

Explanation:

  • MgF2(s) ↔ Mg2+(aq) + 2F-(aq)

          S                   S              2S

∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²

⇒ 4S² * S = 6.4 E-9

⇒ 4S³ = 6.4 E-9

⇒ S³ = 1.6 E-9

⇒ S = 1.1696 E-3 M

  • NaF(s) → Na+(aq)  +  F-(aq)

        0.10M     0.10M        0.10M

  • MgF2(s) ↔ Mg2+(aq)  + 2F-(aq)

          S'                 S'              2S' + 0.10

⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²

If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:

⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'

⇒ S' = 6.4 E-7 M

∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption

We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility  is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).

8 0
2 years ago
An atom whose valence electrons conform to the octet rule is:
Readme [11.4K]

Answer: D. less likely to form any bond

Explanation:

3 0
2 years ago
If a 0.4856 gram sample of khp is dissolved in sufficient water to prepare 250 ml of solution, and 25 ml of the solution require
dangina [55]

Mass of potassium hydrogen pthalate KHP is 0.4856 g, its molar mass is 204.22 g/mol, number of moles of KHP can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass, putting the values,

n=\frac{0.4856 g}{204.22 g/mol}=0.00237 mol

This will be number of moles of NaOH at equivalent point.

Detailed calculations:

Molarity is defined as number of moles in 1 L of solution, for 250 mL of solution, molarity will be:

M=\frac{0.00237 mol}{250 \times 10^{-3}L}=0.009511 M

For 25 mL, apply dilution law as follows:

M_{1}V_{1}=M_{2}V_{2}

Putting the values,

0.009511\times 250=M_{2}\times 25 mL

On rearranging,

M_{2}=\frac{0.009511\times 250}{25}=0.09511 M

Convert molarity into number of moles,

n=M\times V=0.09511 mol/L\times 25\times 10^{-3}L=0.00237 mol

At equivalent point, number of moles of KHP will be equal to NaOH, thus, number of moles of NaOH will be 0.00237 mol.

Calculation for molarity:

Volume of NaOH is 18.75 mL, thus, molarity can be calculated as follows:

M=\frac{n}{V}

Putting the values,

M=\frac{0.00237 mol}{18.75\times 10^{-3}L}=0.1264 M

Therefore, molarity of NaOH is 0.1264 M

7 0
2 years ago
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