D b c c b hope this helps
Answer: C) 1.6
Explanation: I got it right
2LiOH + H₂SO₄ = Li₂SO₄ + 2H₂O
v(H₂SO₄)=0.025 L
v(LiOH)=0.0836 L
c(LiOH)=0.12 mol/L
n(LiOH)=v(LiOH)c(LiOH)
n(H₂SO₄)=n(LiOH)/2=v(LiOH)c(LiOH)/2
c(H₂SO₄)=n(H₂SO₄)/v(H₂SO₄)=v(LiOH)c(LiOH)/(2v(H₂SO₄))
c(H₂SO₄)=0.0836*0.12/(2*0.025)=0.20064 mol/L
≈0.20 M
pH = 2.1
Let 
resembles the acid; as a weak acid (a small value of 
) 
would partially dissociate to produce protons 
and 
, its conjugate base. Let the final proton concentration (i.e., ![[H^{+}]](https://tex.z-dn.net/?f=%20%5BH%5E%7B%2B%7D%5D%20)
) be 
. (Apparently 
) Construct the following RICE table:
By definition, (all concentrations are under equilibrium condition)
![\left\begin{array}{ccc}K_{a}&=&[H^{+}] \cdot [A^{-}] / [HA]\\&=&x^{2} /(0.14 - x)\end{array}\right](https://tex.z-dn.net/?f=%20%20%5Cleft%5Cbegin%7Barray%7D%7Bccc%7DK_%7Ba%7D%26%3D%26%5BH%5E%7B%2B%7D%5D%20%5Ccdot%20%5BA%5E%7B-%7D%5D%20%2F%20%5BHA%5D%5C%5C%26%3D%26x%5E%7B2%7D%20%2F%280.14%20-%20x%29%5Cend%7Barray%7D%5Cright%20)
It is given that
Equating and simplifying the two expressions gives a quadratic equation; solve the equation for gives:
The pH of a solutions equals the opposite of the logarithm of its proton concentration to base 10; thus for this particular solution
![\text{pH} = -\text{ln(}[H^{+}]\text{)} / \text{ln(}10\text{)} = 2.1](https://tex.z-dn.net/?f=%20%5Ctext%7BpH%7D%20%3D%20-%5Ctext%7Bln%28%7D%5BH%5E%7B%2B%7D%5D%5Ctext%7B%29%7D%20%2F%20%5Ctext%7Bln%28%7D10%5Ctext%7B%29%7D%20%3D%202.1%20)
At summer season, snow will melt and will produce water using generators we can produce electric energy.
And in spring and winter can produce electric energy also using the wind
