Answer:
[OH]- = 501.187 M
Explanation:
First, the given Ca(OH)2 is a base and not an acid. Therefore, we cannot do the calculations based on the pH vale. We need to get the pOH value.
pH + pOH = 14
11.3 + pOH = 14
pOH = 14 - 11.3 = 2.7
Now, pOH is calculated as follows:
pOH = -log [OH]
This means that:
2.7 = - log [OH]
[OH]- = 10^(2.7)
[OH]- = 501.187 M
Hope this helps :)
Answer is: this is isotope of barium, because barium has atomic number 56 (number of protons), mass number of this isotope of barium is 130 (56 protons + 76 neutrons). This isotope is cation with oxidation number +2, because barium has two protons more than electrons (56 - 54 = 2).
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.
Answer:
I dont know the answer for that question it's hard question isn't it