Answer:
<em>think</em><em> </em><em>it's</em><em> </em><em>B</em>
Explanation:
Gravity is a force of attraction that exists between any two masses, any two bodies, any two particles. Gravity is not just the attraction between objects and the Earth. It is an attraction that exists between all objects, everywhere in the universe.
<em>H</em><em>o</em><em>p</em><em>e</em><em> </em><em>it'll</em><em> </em><em>help</em><em>!</em>
<em>s</em><em>t</em><em>a</em><em>y</em><em> </em><em>safe</em><em>:</em><em>)</em><em> </em><em>^</em><em>-</em><em>^</em>
P₁V₁ = P₂V₂
Convert 473ml > L
(1atm)(1.00L) = (0.473L)P₂
Divide by (0.473L)
Solve for 2.11atm
Answer:
23892U=23490Th +42He
Explanation:
In alpha decay, the daughter nucleus is two units less than the parent in atomic number. The mass number also decreases by 4 units. The daughter is thus found two places before the parent in the periodic table.
Part 1)
when the balanced equation for this reaction is:
and by using ICE table
Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)
initial 0 0
change +X +2X
Equ X 2X
When KSp expression = [Mg2+][OH-]
when we have KSp = 5.61 x 10^-11
and when we assumed [Mg2+] = X
and [ OH-] = (2X)^2
when we assume X is the value of molar solubility of Mg(OH)2
so, by substitution:
5.61 x 10^-11 = 4X^3
∴ X = 2.4 x 10^-4 M
∴ molar solubility of Mg(OH)2 = X = 2.4 x 10^-4 M
Part 2)
the molar solubility of Mg(OH)2 in 0.16 m NaOH we assumed it = X
by using the ICE table:
Mg(OH)2(s) → Mg2+(aq) + 2OH-
initial 0 0.16m
change +X +2X
equ X (0.16+2x)
when Ksp = [mg2+][OH-]^2
5.61 x 10^-11 = X * (0.16+2X)^2 by solving for X
∴ X = 1.3 x 10^-5 M
∴ the molar solubility = X = 1.3 x 10^-5 M
