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2 years ago

A sample of sulfurous acid (H2SO3) has a mass of 1.31 g.

Chemistry

1 answer:

lakkis [162]2 years ago

6 0

Answer:

a) 2 (H+) ions

b) 1 (SO3²-) ions

c) 1.36 × 10^-22 grams.

Explanation:

According to this question, sulfurous acid has a chemical formula; H2SO3. It is made up of hydrogen and sulfite ion. Hydrogen ion (H+) is the cation while sulfite ion (SO32-) is the anion.

Based on the chemical formula, there are 2 moles of hydrogen ions that reacts with 1 mole of sulfite ion as follows:

2H+ + SO3²- → H2SO3

Hence;

- there are 2 hydrogen ions (2H+) present in H2SO3.

- there is 1 sulfite ion (SO3²-) present in H2SO3.

c) The mass of one formula unit of H2SO3 is calculated thus:

= 1.008 (2) + 32.065 + 15.999(3)

= 2.016 + 32.065 + 47.997

= 82.08 a.m.u

Since, 1 gram is = 6.02 x 10^23 a.m.u

82.08 a.m.u = 82.08/6.02 × 10^-23

= 13.6 × 10^-23

= 1.36 × 10^-22 grams.

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Answer:

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1 year ago

A methanol-water mixture is to be flash distilled at 1 atm. If the feed is 25 mole %methanol, what are the liquid and vapor comp

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Answer:

Explanation:

Given that:

The distillation is carried out at a pressure of 1 atm

The feed harbors 25% mole of methanol (z)

The total moles of feed is usually 100 moles

In the system, we have both methanol and water

Using the total mole balance for the distillation column.

Fz = Lx + Vy

where;

F = amount of feed

z = mole fraction of ethanol (in feed)

L = amount of liquid product out of the column

V = amount of vapor product out of the column

x = mole fraction of methanol out of the liquid

y = mole fraction of methanol out of the vapor.

SO;

(a)

If all the feed is vaporized, then the vapor will likely have the same composition as the feed.

(b)

If no vaporization of the feed takes place, then the bottoms moving out of the column contains the same composition as the feed.

(c)

If 1/3 of the feed is vaporized; then 2/3 of the feed is at the bottom.

The balance equation would be:

$Fz = (\dfrac{2}{3}F) x + (\dfrac{1}{3}F)y \\ \\ z = \dfrac{2}{3}x+\dfrac{1}{3}y$

Replacing z = 0.25; we have:

$0.25 = \dfrac{2}{3}x+\dfrac{1}{3}y$

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(d)

If 2/3 of the feed is vaporized;

Then:

$Fz = (\dfrac{1}{3}F) x + (\dfrac{2}{3}F)y \\ \\ z = \dfrac{1}{3}x+\dfrac{2}{3}y$

replacing z = 0.25

$0.25 = \dfrac{1}{3}x+\dfrac{2}{3}y$

0.75 = x + 2y

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