Answer:
see explanation
Explanation:
Write the balanced COMPLETE ionic equation for the reaction when Na₂CO₃ and AgNO₃ are mixed in aqueous solution. If no reaction occurs, simply write only NR.
Ag (+1) + NO3(-1) + 2 Na(+1) + Co3 (-2)--> Ag2CO3 (s) + 2 Na (+1) + 2NO3(-1)
Answer:
pH = 13.09
Explanation:
Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15
Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15
K = Ksp X Kf
= 3*2*10^-15 * 10^15
= 6
Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M
Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)
Initial: 0 0.3 0
Change: -2x +x
Equilibrium: 0.3 - 2x x
K = Zn(OH)₄²⁻/[OH⁻]²
6 = x/(0.3 - 2x)²
6 = x/(0.3 -2x)(0.3 -2x)
6(0.09 -1.2x + 4x²) = x
0.54 - 7.2x + 24x² = x
24x² - 8.2x + 0.54 = 0
Upon solving as quadratic equation, we obtain;
x = 0.089
Therefore,
Concentration of (OH⁻) = 0.3 - 2x
= 0.3 -(2*0.089)
= 0.122
pOH = -log[OH⁻]
= -log 0.122
= 0.91
pH = 14-0.91
= 13.09
The matter will be consumed by other living organisms and the blood will settle to the bottom of the body
Answer:
Option A. KCl (aq)
Option D. Mg(OH)₂(s
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
MgCl₂(aq) + KOH(aq) —>
In solution, MgCl₂(aq) and KOH(aq) will dissociate as follow:
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
KOH(aq) —> K⁺(aq) + OH¯(aq)
MgCl₂(aq) + KOH(aq) —>
Mg²⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + OH¯(aq) —> 2K⁺(aq) + 2Cl¯(aq) + Mg(OH)₂ (s)
MgCl₂(aq) + KOH(aq) —> 2KCl (aq) + Mg(OH)₂(s)
Thus, the products of the above reaction are: KCl(aq) and Mg(OH)₂(s)
Thus, option A and D gives the correct answer to the question.
