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Answer:

![[H^+]=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5x10%5E%7B-13%7DM)
![[OH^-]=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.02M)
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:

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]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%3D%5Cfrac%7Bx%7D%7B%5BMg%28OH%29_2%5D%7D)
Thus, x for this problem is:
![x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5BMg%28OH%29_2%5D%7D%7B3%7D%3D%5Cfrac%7B0.03M%7D%7B3%7D%5C%5C%5C%5Cx%3D%20%200.01M)
Now, according to an ICE table, we have that:
![[OH^-]=2x=2*0.01M=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2x%3D2%2A0.01M%3D0.02M)
Therefore, we can calculate the H^+, pH and pOH now:
![[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B1x10%5E%7B-14%7D%7D%7B0.02%7D%3D5x10%5E%7B-13%7DM)

Best regards!
Explanation:
2H2O2 => 2H2O + O2
Moles of hydrogen peroxide = 0.150dm³ * (0.02mol/dm³) = 0.003mol .
Moles of oxygen = 0.0015mol.
Volume of oxygen = 0.0015mol * (22.4dm³/mol) = 0.0336dm³.
The equator is the parallel
Answer: 0.72 litres of water is wasted in one day.
Explanation:
First you need to find out how many minutes are in a day. Do this by multiplying the number of minutes in an hour (60) by the number of hours in a day (24). 24 x 60 = 1440. If the faucet is dripping at 5 drops per minute, then multiply 5 by the number of minutes in a day (1440) to see how many drops drip in one day. 5 x 1440 = 7200. Now we need to figure out how many mL fo water that is. if 10 drops is 1 mL, then we need to divide the total number of drops (7200) by 10. 7200 divided by 10 is 720. That means 720 mL of water is dripping per day. Finally, we must convert mL to litres. There are 1000 mL in one litre, so divide 720 by 1000. The final answer is 0.72