Answer:
phosphates compounds is described as insoluble
Answer:
The answer is the explanation.
Explanation:
The acetic acid, CH3COOH, reacts with calcium hydroxide, Ca(OH)2, producing acetate ion, CH3COO- and water as follows:
2CH3COOH + Ca(OH)2 → 2CH3COO- + 2H2O + Ca²⁺
That means the moles of acetic acid decrease whereas the moles of acetate ion are increased. The ratio of [CH3COOH] / [CH3COO-] is different.
As a base is added, the concentration of H3O+ decreases increasing the pH.
That means:
<em>TRUE </em> A. The number of moles of CH3COOH will decrease.
<em>FALSE</em> B. The number of moles of CH3COO- will decrease.
<em>TRUE </em>C. The equilibrium concentration of H3O will decrease.
<em>FALSE </em>D. The pH will decrease.
<em>FALSE</em> E. The ratio of [CH3COOH] / [CH3COO-] will remain the same.
Answer:
I think it's B
Explanation:
apologies if I get this wrong
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.
