Answer:47.05% is the percent yield of nitrogen in the reaction.
Explanation:
heoretical yield of nitrogen gas = x
Moles of ammonia =
According to reaction,2 moles of ammonia gives 1 mol of nitrogen gas.
Then 2.3529 mol of ammonia will give:
of nitrogen gas
Mass of 1.1764 moles of nitrogen gas,x = 1.1764 mol × 28 g/mol=32.94 g
Experiential yield of nitrogen gas = 15.5 g
Percentage yield:
hope that help
47.05% is the percent yield of nitrogen in the reaction.
D. In columns 3-12 in the centre of the table
Explanation:
No. Isotopes are atoms of the same element with different atomic masses (due to the different number of neutrons)
For example, carbon exists as carbon-12 and carbon-14, which both have 6 protons but have 6 and 8 neutrons respectively.
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don't know because this is the question which I never heard
This question is missing the part that actually asks the question. The questions that are asked are as follows:
(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.
(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.
We can use the equation for a first order rate law to find the amount of material remaining after 4 days:
[A] = [A]₀e^(-kt)
[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.
(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.
4 days x 1 year/365 days = 0.0110
A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg
The decay of americium is so slow that no noticeable change occurs over 4 days.
(b) We can simply plug in the information of iodine-125 and solve for A:
A = (1.00)e^(-0.011 x 4)
A = 0.957 mg
Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.
