The valence of helium is 0.
If water is deionized and it is consumed, it may cause people to urinate more and eliminate more electrolytes from the body.
Answer:
E.) +5
Explanation:
Oxygen always has -2 oxidation number.
Because there are 3 oxygen atoms present, this means oxygen is contributing a -6 charge (-2 x 3 = -6).
Therefore, since the overall molecule is -1, chlorine must have an oxidation number of +5 to cancel all of the negative charges but 1.
You can also think of the problem like an equation. In this equation, "x" is the oxidation number of chlorine, (-2) is the oxidation number of oxygen, (3) is the number of oxygen atoms present, and the equation is set equal to (-1) because that is the overall charge of the molecule.
x - 2(3) = -1
x - 6 = -1
x = 5
Answer:
pH = 2.21
Explanation:
Hello there!
In this case, according to the reaction between NaF and HCl as the latter is added to the buffer:
It is possible for us to see how more HF is formed as HCl is added and therefore, the capacity of this HF/NaF-buffer is diminished as it turns acid. Therefore, it turns out feasible for us to calculate the consumed moles of NaF and the produced moles of HF due to the change in moles induced by HCl:
Next, we calculate the resulting concentrations to further apply the Henderson-Hasselbach equation:
![[HF]=\frac{0.450mol}{1.0L} =0.450M](https://tex.z-dn.net/?f=%5BHF%5D%3D%5Cfrac%7B0.450mol%7D%7B1.0L%7D%20%3D0.450M)
![[NaF]=\frac{0.050mol}{1.0L} =0.050M](https://tex.z-dn.net/?f=%5BNaF%5D%3D%5Cfrac%7B0.050mol%7D%7B1.0L%7D%20%3D0.050M)
Now, calculated the pKa of HF:
We can proceed to the HH equation:
![pH=pKa+log(\frac{[NaF]}{[HF]} )\\\\pH=3.17+log(\frac{0.05M}{0.45M} )\\\\pH=2.21](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BNaF%5D%7D%7B%5BHF%5D%7D%20%29%5C%5C%5C%5CpH%3D3.17%2Blog%28%5Cfrac%7B0.05M%7D%7B0.45M%7D%20%29%5C%5C%5C%5CpH%3D2.21)
Best regards!
Answer:
c
Explanation:
somehow i didnt have to look this up to help u lol i learned this is 6th grade
