<span>let x=gallons of current mixture to be drained
and replaced with pure antifreeze.
4-x=gallons of current mixture remaining in the car.</span>
<span>
0.15(4-x)+1.00x=0.50 x 4
0.6-.15x+x=2
0.85x=1.4
x=1.4/0.85 =1.65 gal
Thus, 1.65 gallons of current mixture to be drained and replaced with pure
antifreeze.</span>
Answer:
5.56 × 10⁻⁸
Explanation:
Step 1: Given data
- Concentration of the weak acid (Ca): 0.187 M
Step 2: Calculate the concentration of H⁺
We will use the following expression.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%20%7D%7BCa%7D%20%3D%20%5Cfrac%7B%281.02%20%5Ctimes%2010%5E%7B-4%7D%29%5E%7B2%7D%20%7D%7B0.187%7D%20%3D%205.56%20%5Ctimes%2010%5E%7B-8%7D)
n=20 mol
(NH)4 SO4
Atomic masses :
N- 14
H- 1
S- 32
O- 16
Therefore M= 14×2 + 1×8 + 32 + 16×4
= 132
m= nM
= 20×132
= 2640g