Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
The answer is b because that’s it
Hydrochloric acid is usually purchased in a concentrated form that is 37.0% HCl by mass and has a density of 1.20g/mL. Calculate the molarity of the concd HCl.
1.20 g/mL x 1000 mL x 0.37 x (1/36.5) = about 12 M or so but you do it exactly.
Then mL x M = mL x M
mL x 12 M = 2800 mL x 0.475
Solve for mL of the concd HCl solution.
For radioactive decay, we can relate current amount, initial amount, decay constant and time using:
N = No x exp(-λt)
Half-life = ln(2)/λ
λ = ln(2) / 5730
N/No = 80% = 0.8
0.8 = exp( -ln(2)/5730 x t)
t = 1844 years