Answer:
The concentrations are :
![[HAsc^-]=0.000702 M](https://tex.z-dn.net/?f=%5BHAsc%5E-%5D%3D0.000702%20M)
![[Asc^{2-}]=5.92\times 10^{-8} M](https://tex.z-dn.net/?f=%5BAsc%5E%7B2-%7D%5D%3D5.92%5Ctimes%2010%5E%7B-8%7D%20M)
The pH of the solution is 3.15.
Explanation:
Initial
c 0 0
Equilibrium
c-x x x
![K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}](https://tex.z-dn.net/?f=K_%7Ba1%7D%3D%5Cfrac%7B%5BHAs%5E-%5D%5BH%5E%2B%5D%7D%7B%5BH_2Asc%5D%7D)
Solving for x:
x = 0.000702 M
Initially
x 0 0
At equilibrium ;
(x - y) y y
![K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}](https://tex.z-dn.net/?f=K_%7Ba2%7D%3D%5Cfrac%7B%5BAs%5E%7B2-%7D%5D%5BH%5E%2B%5D%7D%7B%5BHAsc%5E-%5D%7D)
Putting value of x = 0.000702 M
Total concentration of ![[H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%2By%3D0.000702%20M%2B5.92%5Ctimes%2010%5E%7B-8%7D%20M%3D7.0206%5Ctimes%2010%5E%7B-4%7D%20M)
The pH of the solution :
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
Its D because
Li = 6
Be =9
C= 12
O=15
A. 0.02 mol of O2
B. 0.1 mol of CI2
C. 1 mol of N2
D. 2 mol of H2
Bolded answer is correct.
Equation is as follow,
<span> 4 Na (s) + O</span>₂ <span>(g) → 2Na</span>₂<span>O (s)
According to equation,
91.92 g (4 moles) of Na produces = 123.92 g (2 moles) of Na</span>₂O
So,
17.4 g of Na will produce = X g of Na₂O
Solving for X,
X = (17.4 g × 123.92 g) ÷ 91.92 g
X = 23.45 g of Na₂O
Remember, 1 mole= 6.022x10^23 atoms, molecules, or formula units.
Answer is 1.42x10^24
