Answer:
The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Explanation:
Given data:
[OH⁻] = 9.50 ×10⁻⁹M
pOH = ?
Solution:
pOH = -log[OH⁻]
Now we will put the value of OH⁻ concentration.
pOH = -log[9.50 ×10⁻⁹M]
pOH = 8
Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
B) It will accelerate to the right because 500 N> 300 N
Answer:
Just ask google for help
Explanation:
I always do it and get it right
The motivation to abstain from adding water to concentrated acids is that, with a few acids, amid weakening, a considerable measure of warmth is discharged, by adding the corrosive to the water, the generally extensive measure of water will retain the warmth. On the off chance that you added water to concentrated corrosive when you initially beginning pouring the water, it could get sufficiently hot for the little measure of water that was filled all of a sudden bubble and splatter corrosive on you. Concentrated sulfuric corrosive is most famous for doing this, not all acids get that hot on weakening, but rather in the event that you make a propensity for continually adding the corrosive to water for every one of them, you can't turn out badly.
The number of electrons in the outermost shell of an atom determines<span> its </span>reactivity<span>.</span><span>
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