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.
Answer:
c. -1020.9 kJ
Explanation:
4Fe (s) + 3 O₂ (g) --> 2 Fe₂O₃(s) ΔH = -826.0 kJ/mol.
atomic weight of iron = 56
69.03 g = 69.03 / 56
= 1.23268 moles
Heat released by 1.23268 moles
= 1.23268 x 826.0
= -1020.9 kJ .
Answer:
the volume is 12.79ml
Explanation:
12.8+12.7+12.78+12.88=51.16
51.16/4
=12.79
Answer:
The product of aerobic respiration is Carbon dioxide.
Explanation:
- The process of breaking down glucose to produce energy and waste products is called respiration. Livings beings need respiration process to generate energy so that they can survive.
- The types of respiration are : Anaerobic and aerobic respiration.
- Aerobic respiration takes place in presence of oxygen and produces large amount of energy.
- The final product of aerobic respiration are carbon dioxide, water and 38 ATP of energy.
