Answer:
the solubility of CaCO3 is 0.015g/l 25 °C
is favored at equilibrium
Explanation:
The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?
solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)
CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)
in partial dissociation , we can say
2.25x 10^-8=
let Ca^2+=CO3^-2=S
2.25x10^-8=S*S
S^2=2.25x10^-8
S=0.00015mol/L
Converting that to g/l
the relative molecular mass of CaCO3=100g/mol
0.00015*100g/mol
0.015g/l
the solubility of CaCO3 is 0.015g/l @room temperature
is favored at equilibrium
Answer:
a process that involves rearrangement of the molecular or ionic structure of a substance, as distinct from a change in physical form or a nuclear reaction.
Or
Chemical reaction, a process in which one or more substances, the reactants, are converted to one or more different substances, the products. ... A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products.
Answer:
2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.
Explanation:
<em>∵ pH = - log[H₃O⁺]</em>
∴ 4.6 = - log[H₃O⁺].
∴ log[H₃O⁺] = - 4.6.
∴ [H₃O⁺] = 2.51 x 10⁻⁵.
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
[H₃O⁺] = 2.51 x 10⁻⁵ M.
∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.51 x 10⁻⁵ M) = 3.98 × 10⁻¹⁰ M ≅ 4.0 × 10⁻¹⁰ M.
<em>So, the right choice is: 2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.</em>
Answer:
Explanation:
To answer this question successfully, we need to remember that atoms are neutral species, since the number of protons, the positively charged particles, is equal to the number of electrons, the negatively charged particles. That said, we may firstly find an atom which has 3 electrons (and, as a result, 3 protons, as it should be neutral).
The number of protons is equal to the atomic number of an element. We firstly may have an atom with 3 protons and 3 electrons (atomic number of 3, this is Li).
Similarly, we may take the atomic number of 4, beryllium, and remove 1 electron from it. Upon removing an electron, it would become beryllium cation, 
.
We may use the same logic going forward and taking the atomic number of 5. This is boron. In this case, we need to remove 2 electrons to have a total of 3 electrons. Removal of 2 electrons would yield a +2-charged cation: 
.
