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Answer:
Part A: 36 MBq; Part B: 18 MBq
Explanation:
The half-life is the time it takes for half the substance to disappear.
The activity decreases by half every half-life
A =Ao(½)^n, where n is the number of half-lives.
Part A
3.0 da = 1 half-life
A = Ao(½) = ½ × 72 MBq = 36 MBq
Part B
6.0 da = 2 half-lives
A = Ao(½)^2 = ¼ × 72 MBq = 18 MBq
Answer: secondary structure
Explanation:
Answer:
The hydrogen produces the smaller amount of ammonia.
Step-by-step explanation:
We are given the masses of two reactants, so this is a <em>limiting reactant problem</em>.
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 28.02 2.016 17.03
N₂ + 3H₂ ⟶ 2NH₃
Mass/g: 70.0 7.00
1. Calculate the moles of N₂ and H₂
Moles N₂ = 70.0 × 1/28.02
Moles N₂ = 2.498 mol N₂
Moles H₂ = 7.00 × 2.016
Moles H₂ = 3.472 mol N₂
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2. Calculate the moles of NH₃ from each reactant
<em>From</em> N₂:
The molar ratio is 2 mol NH₃/1 mol N₂
Moles of NH₃ = 2.498 × 2/1
Moles of NH₃ = 4.996 mol NH₃
<em>From</em> H₂:
The molar ratio is 2 mol NH₃/3 mol H₂
Moles of NH₃ = 3.472 × 2/3
Moles of NH₃ = 4.139 mol NH₃
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3. Identify the limiting reactant
The limiting reactant is H₂, because it produces fewer moles of NH₃.
solubility will increase, since it is basic in nature, on acidifying solubility increases.
being basic on acidification, the solubility increases.
is alkaline in nature, it gets neutralized and there is no increase in acidity.
is a salt and it shows no effect on acidification.
is a salt but insoluble in water and shows no effect on acidification.
is strongly basic in nature, on acidifying the solubility increases.