It has only single bonds.So it is saturated
C, because vinegar ph scale is 2.5
Answer would be A)the atom
D) 59 (protons+neutrons=mass)
4.14x10^-3 per minute
First, calculate how many atoms of Cu-61 we initially started with by
multiplying the number of moles by Avogadro's number.
7.85x10^-5 * 6.0221409x10^23 = 4.7273806065x10^19
Now calculate how many atoms are left after 90.0 minutes by subtracting the
number of decays (as indicated by the positron emission) from the original
count.
4.7273806065x10^19 - 1.47x10^19 = 3.2573806065x10^19
Determine the percentage of Cu-61 left.
3.2573806065x10^19/4.7273806065x10^19 = 0.6890455577
The formula for decay is:
N = N0 e^(-λt)
where
N = amount left after time t
N0 = amount starting with at time 0
λ = decay constant
t = time
Solving for λ:
N = N0 e^(-λt)
N/N0 = e^(-λt)
ln(N/N0) = -λt
-ln(N/N0)/t = λ
Now substitute the known values and solve:
-ln(N/N0)/t = λ
-ln(0.6890455577)/90m = λ
0.372447889/90m = λ
0.372447889/90m = λ
0.00413830987 1/m = λ
Rounding to 3 significant figures gives 4.14x10^-3 per minute as the decay
constant.
