Beryllium is a chemical element with symbol Be and atomic number 4. It is a relatively rare element in the universe, usually occurring as a product of the spallation of larger atomic nuclei that have collided with cosmic rays.
Answer:
Precipitation
Explanation:
There are four main stages in the water cycle. They are evaporation, condensation, precipitation and collection. Let's look at each of these stages. Evaporation: This is when warmth from the sun causes water from oceans, lakes, streams, ice and soils to rise into the air and turn into water vapour (gas).
Answer:
The balanced chemical equation: NH₃ + 2 HF → NH₄⁺ + HF₂⁻
Explanation:
According to the Brønsted–Lowry acid–base theory, the acid- base reaction is a type of chemical reaction between the acid and base to give a conjugate acid and a conjugate base.
In this reaction, a Brønsted–Lowry acid loses a proton to form a conjugate base. Whereas, a Brønsted–Lowry base accepts a proton to form a conjugate acid.
Acid + Base ⇌ Conjugate Base + Conjugate Acid
The acid dissociation constant (Kₐ) <em>signifies the acidic strength of a chemical species.</em>
∵ pKₐ = - log Kₐ
Thus for a strong acid, Kₐ value is large and pKₐ value is small.
pKₐ (HF) = 3.2 → strong acid
pKₐ (NH₃) = 38 → weak acid
<u>The chemical reaction involved in the dissolution process:</u>
NH₃ + 2 HF → NH₄⁺ + HF₂⁻
In this acid-base reaction, the acid HF reacts with NH₃ base to give the conjugate base HF₂⁻ and conjugate acid NH₄⁺.
<u>HF (acid) donates a proton to form the conjugate base, HF₂⁻ ion. NH₃ (base) accepts a proton to form the conjugate acid. </u>
<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq)
+ Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ +
H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>)
= 0.021 M.
Ka(HCN) = 4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷
4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻]
/ [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] =
x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M
- x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>
