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Solnce55 [7]
3 years ago
10

What is the work required for the separation of air (21-mol-% oxygen and 79-mol-% nitrogen) at 25°C and 1 bar in a steady-flow p

rocess into product streams of pure oxygen and nitrogen, also at 25°C and 1 bar, of the thermodynamic efficiency of the process is 5% and if Tσ = 300 K
Chemistry
1 answer:
Ronch [10]3 years ago
4 0

Answer:

4oo

Explanation:

it

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How many moles of H2 are produced from 5.8 moles of NH3
vodomira [7]

Answer:

1. 8.7moles of H2

2. 2.25moles of O2

Explanation:

1. 2NH3 —> N2 + 3H2

From the equation,

2moles of NH3 produce 3 moles of H2.

Therefore, 5.8moles of NH3 will produce Xmol of H2 i.e

Xmol of H2 = (5.8x3)/2 = 8.7moles

2. C3H8 + 5O2 —> 3CO2 + 4H2O

From the equation,

5moles of O2 produced 4moles of H2O.

Therefore, Xmol of O2 will produce 1.8mol of H2O i.e

Xmol of O2 = (5x1.8)/4 = 2.25moles

4 0
4 years ago
All of the following pairs of ions are isoelectronic except which one?
mash [69]

Answer: Fe2+ & Mn3+

Explanation:

6 0
3 years ago
Name the unit being abbreviated in each measurement (spelling counts). Only write the name of the unit, do not include the numbe
givi [52]

Answer:

Litre (L) , Centimetre (cm) , Kilogram (Kg), Seconds (s) and Kelvin (K)

Explanation:

The units are used for the following measurement;

Litre = Volume

Centimetre = Length

Kilogram = Mass

Seconds = Time

Kelvin = Temperature

5 0
3 years ago
Given the equation representing a system at equilibrium in a sealed, rigid container:
LenaWriter [7]

Answer:

Choice 1. "HI to increase".

Explanation:

I found out the hard way.

4 0
3 years ago
Read 2 more answers
The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c
netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M

Hence, the solubility of oxygen gas at 628 torr is 4.58\times 10^{-3}M

4 0
3 years ago
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