Answer is: the number of ions produced in the dissociation of aluminium fluoride is 4.
<span>
Chemical dissociation of aluminium fluoride in
water:
AlF</span>₃(aq) →
Al³⁺(aq) + 3F⁻(aq).<span>
There are four ions, one aluminium cation and
three fluoride anions.
Aluminium has oxidation +3, because it lost
three electrons, to have electron configuration as noble gas neon and fluorine has oxidation -1, because it gain one electron to </span>have electron configuration as noble gas neon.
Reactivity is a chemical
property of a substance. According to EPA regulations, it is normally unstable
and readily
undergoes violent change without
detonating. it can explode or violently react when exposed to water, when
heated, or under STP.
<h3>Answers:</h3>
1) 2 Units of Ozone
2) 3 Units of Ozone
3) 9 Units of Ozone
<h3>Solution:</h3>
1) From 6 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
6 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 2 Units of Ozone
2) From 9 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
9 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 3 Units of Ozone
3) From 27 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
27 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 9 Units of Ozone
This is false because males have 1 X and 1 Y chromosome. It's females who have 2 X chromosomes. So, it's false.
Answer:
Kb = 6.22x10⁻⁷
Explanation:
Triethanolamine, C₆H₁₅O₃N, is in equilibrium with water:
C₆H₁₅O₃N(aq) + H₂O(l) ⇄ C₆H₁₅O₃NH⁺(aq) + OH⁻(aq)
Kb is defined from concentrations in equilibrium, thus:
Kb = [C₆H₁₅O₃NH⁺] [OH⁻] / [C₆H₁₅O₃N]
The equilibrium concentration of these compounds could be written as:
[C₆H₁₅O₃N] = 0.486M - X
[C₆H₁₅O₃NH⁺] = X
[OH⁻] = X
pH is -log [H⁺], thus, [H⁺] = 10^-pH = 1.820x10⁻¹¹M
Also, Kw = [OH⁻] ₓ [H⁺];
1x10⁻¹⁴ = [OH⁻] ₓ [H⁺]
1x10⁻¹⁴ = [OH⁻] ₓ [1.820x10⁻¹¹M]
5.495x10⁻⁴M = [OH⁻], that means <em>X = 5.495x10⁻⁴M</em>
Replacing in Kb formula:
Kb = [5.495x10⁻⁴M] [5.495x10⁻⁴M] / [0.486M-5.495x10⁻⁴M]
<em>Kb = 6.22x10⁻⁷</em>
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