Answer:

Explanation:
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In this case, since the buffer is not given, we assume it is based off ammonia, it means the ammonia-ammonium buffer, whereas the ammonia is the weak base and the ammonium ion stands for the conjugate acid. In such a way, when adding HI to the solution, the base of the buffer, NH3, reacts with the former to promote the following chemical reaction:

Because the HI is totally ionized in solution so the iodide ion becomes an spectator one.
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Answer:
Solving by the method of exponential growth.
bacteria = 2
after one hr = 2² = 4
after 2nd hr = 2³ = 8
after 3rd hr = 2⁴ = 16
after 4th hr = 2⁵ = 32
uniquely identifies a chemical element. yupppppp
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
Answer:
3.59x10^21 molecules
Explanation:
1mole of a substance contains 6.02x10^23 molecules.
Therefore, 1mole of C8H18 will also contain 6.02x10^23 molecules
1mole of C8H18 = (12x8) +(18x1) = 96 + 18 = 114g.
1mole (i.e 114g) oh C8H18 contains 6.02x10^23 molecules.
Therefore, 0.68g of C8H18 will contain = (0.68 x 6.02x10^23)/114 = 3.59x10^21 molecules