Answer:
We need 17.2 L of Ca(OH)2
Explanation:
Step 1: Data given
Concentration of Ca(OH)2 = 1.45 M
Moles of H2SO4 = 25.0 moles
Step 2: The balanced equation
Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4
Step 3: Calculate moles Ca(OH)2
For 1 mol Ca(OH)2 we need 1 mol H2SO4 to produce 2 moles H2O and 1 mol CaSO4
For 25.0 moles H2SO4 we'll need 25.0 moles Ca(OH)2 to produce 50 moles H2O and 25.0 moles CaSO4
Step 4: Calculate volume of Ca(OH)2
Volume Ca(OH)2 = moles Ca(OH)2 / concentration Ca(OH)2
Volume Ca(OH)2 = 25.0 moles / 1.45 M
Volume Ca(OH)2 = 17.2 L
We need 17.2 L of Ca(OH)2
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After 25 days, it remains radon 5.9x10^5 atoms.
Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.
N(Ra) = 5.7×10^7; initial number of radon atoms
t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days
n = 25 days / 3.8 days
n = 6.58; number of half-lifes of radon
N1(Ra) = N(Ra) x (1/2)^n
N1(Ra) = 5.7×10^7 x (1/2)^6.58
N1(Ra) = 5.9x10^5; number of radon atoms after 25 days
The half-life is independent of initial concentration (size of the sample).
More about half-life: brainly.com/question/1160651
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