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zheka24 [161]
4 years ago
11

Hydrogen is a gas at a room temperature, but it is found with the alkali metals in group 1 (1A) on the periodic table. Which is

the most likely reason hydrogen is grouped with alkali metals?
Chemistry
1 answer:
bearhunter [10]4 years ago
5 0
Hydrogen is grouped with the alkali metals because they are grouped based on their properties and behavior, and like the other elements in Group 1 (1A), it only has one valence.
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At a certain temperature, the solubility of N2 gas in water at 4.07 atm is 95.7 mg of N2 gas/100 g water . Calculate the solubil
sertanlavr [38]

Answer: Thus the solubility of N_2 gas in water, at the same temperature, if the partial pressure of gas is 10.0 atm is 235mg/100g.

Explanation:-

The Solubility of N_{2} in water can be calculated by Henry’s Law. Henry’s law gives the relation between gas pressure and the concentration of dissolved gas.

Formula of Henry’s law,  C=k_{H}P.

k_{H}= Henry’s law constant = ?

The partial pressure (P) of N_{2} in water = 4.07 atm

\C= k_{H}\times P\\95.7mg=k_{H}\times 4.07

k_{H}=23.5

At pressure of 10.0 atm

C= k_{H}\times P\\C=23.5\times 10.0=235mg/100mg

Thus the solubility of N_2 gas in water, at the same temperature, is 235mg/100g

6 0
3 years 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
frozen [14]

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

0.75 = 2x + y

(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

6 0
3 years ago
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