Answer:
water is 9/10
chemical would be 1/10
Explanation:
180/200 would be water concentration in solution
and 20/200 would be chemical solution concentration in solution (if the chemical were to be polar and mix)
Answer:
Explanation:
Reducing agent
Reducing agent is an element or compound that loses or donates an electron to another chemical species in a redox chemical reaction. Since the reducing agent is losing electrons,it is said to have been oxidized.
Answer:
Option(II) and option (IV) are correct.
Explanation:
Here products are
and
and reactant is 
(I) According to Le-chatelier principle,increase in
will shift the equilibrium towards reactant side to keep the equilibrium constant unchanged.
(II) According to Le-chatelier principle,decrease in
will shift the equilibrium towards product side to keep the equilibrium constant unchanged.
(III) Adding a catalyst will not change position of equilibrium. Catalyst only helps to achieve equilibrium in a lesser time.
(IV) As this reaction is an endothermic reaction therefore heat is consumed in formation of product. Therefore increase in temperature will lead to formation of more product to consume excess heat added.
(V) Pure solids and liquids do not affect position of equilibrium as their concentrations remain unchanged.
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!