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3 years ago

Suggest a method to liquefy atmospheric gases.

Chemistry

2 answers:

kicyunya [14]3 years ago

8 0

Answer:

WASSUP BRO

Explanation:

The gases can be converted into liquids by bringing its particles closer so atmospheric either by decreasing temperature or by increasing pressure

mixer [17]3 years ago

3 0

Answer:

The gases can be converted into liquids by bringing its particles closer so atmospheric gases can be liquefied either by decreasing temperature or by increasing pressure.

Explanation:

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

Calculate the standard emf for the following reaction:

krek1111 [17]

In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
E = +1.47

Br(l) + 2e- = 2Br-
E = +1.065

We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V

4 0

3 years ago

Your grandfather is a farmer. You hear him complain about how his crops are not doing well. You know that we have the same amoun

Firdavs [7]

Answer:

You may need better soil and a more plentiful amount of water coming from another source, and maybe find another way to contain the rain water

Explanation:you may need to draw it on paper sorry bout how it looks

4 0

3 years ago

How does doubling the volume of solvent affect the concentration of the solution?

Alex17521 [72]

Answer:

Molarity is halved when the volume of solvent is doubled.

Explanation:

Using the dilution equation (volume 1)(molarity 1)=(volume 2)(molarity 2), we can demonstrate the effects of doubling volume.

Suppose the starting volume is 1 L and the starting molarity is 1 M, and doubling the volume would make the final volume 2 L.

Plugging these numbers into the equation, we can figure out the final molarity.

(1 L)(1 M)=(2 L)(X M)

X M= (1 L x 1 M)/(2 L)

X M= 1/2 M

This shows that the molarity is halved when the volume of solvent is doubled.

8 0

3 years ago

SOMEONE PLEASE HELP I WILL GIVE YOU BRAINLIEST IF YOU GET IT RIGHT.

VladimirAG [237]

Answer:

A (H)

Explanation:

The key uptop shows all the red are non metal and "H: is pink

6 0

3 years ago