The reaction is shown below, Acid protonates the carbonyl oxygen and makes the carbonyl carbon more electrophilic. Water attacks on activated carbonyl group and forms a tetrahedral intermediate.
Intermediate: Structure of Intermediate is shown both in 2-D and 3-D (below attached).
Carbonyl group is regenerated with the elimination of ethanol.
Reaction is as below, The final product is carboxylic acid.
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Answer:
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Answer:
30%
Explanation:
<em>This is the chemical formula for zinc bromate: Zn(BrO₃)₂. Calculate the mass percent of oxygen in zinc bromate. Round your answer to the nearest percentage.</em>
Step 1: Determine the mass of 1 mole of Zn(BrO₃)₂
M(Zn(BrO₃)₂) = 1 × M(Zn) + 2 × M(Br) + 6 × M(O)
M(Zn(BrO₃)₂) = 1 × 65.38 g/mol + 2 × 79.90 g/mol + 6 × 16.00 g/mol
M(Zn(BrO₃)₂) = 321.18 g/mol
Step 2: Determine the mass of oxygen in 1 mole of Zn(BrO₃)₂
There are 6 moles of atoms of oxygen in 1 mole of Zn(BrO₃)₂.
6 × m(O) = 6 × 16.00 g = 96.00 g
Step 3: Calculate the mass percent of oxygen in Zn(BrO₃)₂
%O = mO/mZn(BrO₃)₂ × 100%
%O = 96.00 g/321.18 g × 100% ≈ 30%
Answer:
Half life is 6 years.
Explanation:
T½ = In2 / λ
Where λ = decay constant.
But N = No * e^-λt
Where N = final mass after a certain period of time
No = initial mass
T = time
N = 0.625g
No = 10g
t = 24 years
N = No* e^-λt
N / No = e^-λt
λ = -( 1 / t) In N / No (inverse of e is In. Check logarithmic rules)
λ = -(1 / 24) * In (0.625/10)
λ = -0.04167 * In(0.0625)
λ = -0.04167 * (-2.77)
λ = 0.1154
T½ = In2 / λ
T½ = 0.693 / 0.1154
T½ = 6.00 years.
The half life of radioactive cobalt-60 is 6 years
