Answer:
solubility is 1.984x10⁻⁹M
Explanation:
When CaCO₃ is in water, the equilibrium that occurs is:
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Kps = [Ca²⁺] [CO₃²⁻] = 4.96x10⁻⁹
If you have a 0.250M solution of Na₂CO₃, [CO₃²⁻] = 0.250M:
[Ca²⁺] [0.250M] = 4.96x10⁻⁹
Assuming you are adding an amount of CaCO₃:
[X] [0.250 + X] = 4.96x10⁻⁹
<em>Where X is the amoun of CaCO₃ you can add, that means, solubility</em>
X² + 0.250X - 4.96x10⁻⁹ = 0
Solving for X:
X = -0.25M → False answer, there is no negative concentrations.
X = 1.984x10⁻⁹M.
That means, <em>solubility is 1.984x10⁻⁹M</em>
Answer:
1.63425 × 10^- 18 Joules.
Explanation:
We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.
We will be using the formula in the equation (1) below;
Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).
Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.
Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;
∆E= - R(H) × [1/nf^2 - 1/ni^2 ].
Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.
Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].
= -2.179 × 10^-18 Joules × (0.25 - 1).
= - 2.179 × 10^-18 × (- 0.75).
= 1.63425 × 10^- 18 Joules.
<u><em>Answer:</em></u>
- The correct answer is a.
- The ratio of their masses is always the same
<u><em>Explanation:</em></u>
- Dalton law was based on law of conservation of masses and law of conservation definite proportional. So here , it is cleared that the ratio of their masses should be always same.
Answer:
The nucleus is a collection of particles called protons, which are positively charged, and neutrons, which are electrically neutral. Protons and neutrons are in turn made up of particles called quarks. The chemical element of an atom is determined by the number of protons, or the atomic number, Z, of the nucleus.