The answer is 0.62g.
Solution:
From year 1960 to year 2030, it has been
2030-1960 = 70 years
The half-life of the radioactive element is 28 years, then the sample will go through
70 years * (1 half-life/28 years) = 2.5 half-lives
Starting with a 3.5 gram sample, we will have
3.5*(1/2) after one half-life passes
3.5*(1/2) * (1/2) = 3.5*(1/4) after two half-lives pass
3.5*(1/4) * (1/2) = 3.5*(1/8) after three half-lives pass and so on
Therefore, we can write the remaining amount of the sample after the number n of half-lives have passed as
mass of sample = initial mass of sample/2^n
The mass of the remaining sample for n = 2.5half-lives can be now calculated as
mass of sample = 3.5 grams / 2^2.5 = 0.62 g
Answer:- 3.84 grams
Solution:- Volume of the sample is 44.8 mL and the density is 1.03 gram per mL.
From the density and volume we calculate the mass as:
mass = volume*density
= 46.1 g
From given info, potassium bromide solution is 8.34% potassium bromide by mass. It means if we have 100 grams of the solution then 8.34 grams of potassium bromide is present in. We need to calculate how many grams of potassium bromide are present in 46.1 grams of the solution.
The calculations could easily be done using dimensional analysis as:
= 3.84 g KBr
Hence, 3.84 grams of KBr are present in 44.8 mL of the solution.
7.2 * 10^23 molecules CO2 1 mole CO2
x ___________________
6.02* 10^23 molecules CO2
=1.196
=1.2 moles of CO2
Answer:
0.319 L
Explanation:
M(MgSO4) = 120 g/mol
46.1 g * 1 mol/120 g = 0.384 mol MgSO4
0.384 mol * 1 L/1.20 mol = 0.319 L
