Answer:also confused ?
Explanation:cant see full answer
Answer:
13.5 years
Explanation:
Initial Concentration [Ao] = 10g
Final Concentration [A] = 0.768g
Time t= 50 years
Half life t1/2 = ?
These quantities are related by the following equations;
ln[A] = ln[Ao] - kt ......(i)
t1/2 = ln(2) / k ...........(ii)
where k = rate constant
Inserting the values in eqn (i) and solving for k, we have;
ln(0.768) = ln(10) - k (50)
-0.2640 = 2.3026 - 50k
50k = 2.3026 + 0.2640
k = 2.5666 / 50 = 0.051332
Insert the value of k in eqn (ii);
t1/2 = ln(2) / k
t1/2 = 0.693 / 0.051332 = 13.5 years
Answer:
The mass is 0.855 grams (option A)
Explanation:
Step 1: Data given
aluminium sulfate = Al2(SO4)3
Numer of moles Al2(SO4)3 = 2.50 * 10^-3 moles
atomic mass Al = 26.99 g/mol
atomic mass S = 32.065 g/mol
Atomic mass O = 16 g/mol
Step 2: Calculate molar mass Al2(SO4)3
Molar mass = 2* 26.99 + 3*32.065 + 12*16
Molar mas = 342.175 g/mol
Step 3: Calculate mass Al2(SO4)3
Mass Al2(SO4)3 = moles Al2(SO4)3 * molar mass Al2(SO4)3
Mass Al2(SO4)3 = 2.5 *10^-3 moles * 342.175 g/mol
Mass Al2(SO4)3 = 0.855 grams
The mass is 0.855 grams (option A)
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