100 degree celcius and 0 degree Celsius
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.
Deuterium is a relatively uncommon form of hydrogen, but can be created from water.
- Heavy hydrogen commonly known as deuterium
- stable isotopes of hydrogen
- gets its name from the Greek word deuterons means second.
- has only one proton and one neutron
- nucleus of the hydrogen's deuterium atom is known as a deuteron containing one proton and one neutron.
- Deuterium forms chemical bonds that are stronger than regular hydrogen
- gas deuterium is colorless
- Deuterated water is used in Magnetic Resonance Spectroscopy.
- used in the determination of the isotopologue of various organic compounds.
- used in Infrared Spectroscopy.
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