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Basile [38]
3 years ago
7

The Earth only has a limited supply of fresh water; eventually, we will have to use ocean water. We will purify ocean water thro

ugh a process called desalinization. Currently, it is not profitable to purify ocean water. How would this change in the future?
Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
6 0
I don't learn chemistry yet but I see where you are right now... I think this will change in the near future because we will only have the ocean water to survive... So we purify the water... Because this will almost be all we have, we'll have no choice BUT to make money on this.... Sorry if it's wrong I don't learn chemistry yet
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8 0
3 years ago
The crystal structure of an ionic compound depends on the
Murljashka [212]

Answer:

The ion's sizes and their radius ratio.

Explanation:

Ionic compounds take the form of three dimentional arrays by alternating anions and cations in order to minimize the potential energy of the system by maximizing the attractive force between opposite charges.

<em>The resultant geometric structures are known as crystal lattices and their arrangements depend on the ion's sizes and, since they are bound though electrostatic attraction, their radius ratio.</em>

I hope you find this information useful and interesting! Good luck!

4 0
3 years ago
Have any of you guys ever got suspended because you made a stink bomb in science class?
AleksAgata [21]

Answer:

No :s

Explanation:

3 0
3 years ago
Read 2 more answers
⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqu
lions [1.4K]

Answer:

pH=12.3\\\\pOH=1.7\\

[H^+]=5x10^{-13}M

[OH^-]=0.02M

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)

Thus, since the ionization occurs at an extent of 1/3, we can set  up the following relationship:

\frac{1}{3} =\frac{x}{[Mg(OH)_2]}

Thus, x for this problem is:

x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x=  0.01M

Now, according to an ICE table, we have that:

[OH^-]=2x=2*0.01M=0.02M

Therefore, we can calculate the H^+, pH and pOH now:

[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M

pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7

Best regards!

4 0
3 years ago
Help me out hear i need help
____ [38]

Answer:

Explanation:

i cant see the attachment

6 0
3 years ago
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