<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>
<h3>Answer:</h3>
Strontium (Sr)
<h3>Explanation:</h3>
The condition given in statement is the presence of two valence electron. Hence, first we found the electronic configuration of given atoms as follow;
Rubidium [Kr] 5s¹
Strontium [Kr] 5s²
Zirconium [Kr] 4d² 5s²
Silver [Kr] 4d¹⁰ 5s¹
From above configurations it is cleared that only Strontium and Zirconium has two electrons in its valence shell.
We also know that s-block elements are more reactive than transition elements due to less shielding effect in transition elements hence, making it difficult for transition metals to loose electrons as compared to s-block elements. Therefore, we can conclude that Strontium present in s-block with two valence electrons is the correct answer.
Answer:
285g of fluorine
Explanation:
To solve this problem we need to find the mass of Freon in grams. Then, with its molar mass we can find moles of freon and, as 1 mole of Freon, CCl₂F₂, contains 2 moles of fluorine, we can find moles of fluorine and its mass:
<em>Mass Freon:</em>
<em>2.00lbs * (454g / 1lb) = </em>908g of Freon
<em>Moles freon -Molar mass: 120.91g/mol- and moles of fluorine:</em>
908g of Freon * (1mol / 120.91g) =
7.5 moles of freon * (2moles Fluorine / mole Freon): 15 moles of fluorine
<em>Mass fluorine -Atomic mass: 19g/mol-:</em>
15 moles F * (19g / mol) =
<h3>285g of fluorine</h3>
The primary components of rocks are minerals.
