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Romashka-Z-Leto [24]
3 years ago
11

How can the major components of air be separated

Chemistry
1 answer:
WARRIOR [948]3 years ago
4 0
Air can be separated into its components by means of distillation in special units. So-called air fractionating plants employ a thermal process known as cryogenic rectification to separate the individual components from one another in order to produce high-purity nitrogen, oxygen and argon in liquid and gaseous form.
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What is the value for AG at 100 Kif AH = 27 kJ/mol and AS = 0.09 kJ/(mol-K)?
Elina [12.6K]

Answer:

ΔG =   18KJ/mol

Explanation:

Given data:

ΔS = 0.09 Kj/mol.K

ΔH = 27 KJ/mol

Temperature  = 100 K

ΔG = ?

Solution:

Formula:

ΔG = ΔH - TΔS

ΔH = enthalpy

ΔS = entropy

by putting values,

ΔG =  27 KJ/mol - 100K(0.09 Kj/mol.K)

ΔG =  27 KJ/mol - 9 KJ/mol

ΔG =   18KJ/mol

7 0
3 years ago
At Jim's auto shop, it takes him minutes to do an oil change and minutes to do a tire change. Let be the number of oil changes h
Juli2301 [7.4K]

Answer:

The total time that Jim needs to change x oil changes and y tire changes is less than 180 min.

The time needed for x oil changes is 12 * x.

The time needed for y tire changes is 18 * y.

The total time is the sum of the above times and needs to be less than 180 that is

12 * x + 18 * y < 180 divide both sides of equation by 6

12/6 * x + 18/6*y < 180/6

2*x + 3*y < 30

2*x < 30 - 3*y divide both sides by 2 to get the inequality for x

x < 30/2 - 3/2*y = 15 - 1.5 y < 15 that is x < = 15

2*x + 3*y < 30

3*y < 30 - 2*x divide both sides by 3 to get the inequality for y

y < 30/3 - 2/3 *x = 10 - 2/3*x < 10 that is y < = 10

Also we can write x + y < x+ 3/2 * y < 15.

Explanation:

Jim's can do not more then 5 oil changes and not more then 10 tire changes or all together she can do not more then 15 total of oil and tire changes.

5 0
3 years ago
The decomposition of in solution in carbon tetrachloride is a first-order reaction: The rate constant at a given temperature is
Elan Coil [88]

Answer:

see below

Explanation:

The rate constant is missing in question, but use C(final) = C(initial)e^-kt = 0.200M(e^-k·10). Fill in k and compute => remaining concentration of reactant

6 0
4 years ago
Read 2 more answers
What volume in liters of carbon monoxide will be required to produce 18.9 L of nitrogen in the reaction below
mina [271]

Answer:

37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.

Explanation:

Equation for the reaction:

2 CO + 2 NO ------> N2 + 2 CO2

2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen

At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.

So therefore, we can say:

2 * 22.4 L of CO produces  22.4 L of N2

44.8 L of CO produces 22.4 L of N2

Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:

44.8 L of CO = 22.4 L of N

x L = 18.9 L

x L = 18.9 * 44.8 / 22.4

x L = 18.9 * 2

x = 37.8 L

The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L

8 0
3 years ago
How are mitosis and meiosis similar? How are they different? Provide 1 way the two processes are the same and 2 ways the 2 proce
swat32

Answer:

Similar: both processes of cell division; both processes take place in the nucleus of the cell

Different: mitosis divides into 2, meiosis divides into 4; one is the division of body cells and the other is of specifically sex cells

Explanation:

https://byjus.com/biology/mitosis-and-meiosis/

^ this has more info!

3 0
3 years ago
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