Why is air cooled before nitrogen and oxygen are obtained

Chemistry

1 answer:

lozanna [386]1 year ago

8 0

Answer:

The main reason why air is cooled before nitrogen and oxygen are obtained is because these two gases are much more soluble in cold air than in warm air. As a result, if air were not cooled before these gases were separated, they would simply mix back together again.

You might be interested in

Given that cao(s) + h2o(l) → ca(oh)2(s), δh°rxn = –64.8 kj/mol, how many grams of cao must react in order to liberate 525 kj of

USPshnik [31]

Answer:

454.3 g.

Explanation:

From the given data:

1.0 mol of CaO liberates → – 64.8 kJ.

??? mol of CaO liberates → - 525 kJ.

∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.

∴ The no. of grams of CaO needed = no. of moles x molar mass = (8.1 mol)(56.077 g/mol) = 454.3 g.

8 0

3 years ago

Why is a chemical equilibrium described as dynamic?

Katen [24]

Because when equilibrium is reached, the reaction is still occurring in both directions, it's just that rate(forward) =rate(reverse) so there is no net change in the concentrations of the reactants or products.

8 0

3 years ago

Read 2 more answers

You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr

defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

$W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t$

The power needed to process 50 ton/hor is

$P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP$

2) The density of the packed bed can be expressed as

$\rho=f_v*\rho_v+f_s*\rho_s$

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).

Then we can rearrange

$\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37$