Equation for half-lives:
Nt = No x (1/2)^n
No = initial amount
Nt = final amount after t years
n = number of half lives = t/(single half-life)
t = years
Nt = 3/12 = 0.25
No = 12/12 = 1.00
n = t/(24400)
3/12 = (12/12) x (0.5)^(t/24400)
(0.25) = 1.00 x (0.5)^(t/24400)
0.25/1.00 = 0.5^(t/24400)
ln(0.25) = ln(0.5^(t/24400))
ln(0.25) = (t/24400)*ln(0.5)
ln(0.25)/ln(0.5) = (t/24400)
2 = t/24400
2*24400 = t
t = 48800 yrs
answer is <u>t = 48,800 yrs</u>
Answer : The mass of sucrose added to water will be, 189.0 grams.
Explanation :
As we are given that 9 % solution (mass per volume) that means 9 grams of sucrose present in 100 mL volume of solution.
Total given volume of solution = 2.1 L = 2100 mL (1 L = 1000 mL)
Now we have to determine the mass of sucrose in solution.
As, 100 mL of solution contains 9 grams of sucrose
So, 2100 mL of solution contains
grams of sucrose
Therefore, the mass of sucrose added to water will be, 189.0 grams.
Answer: -
Mass of Hydrated KAl(SO₄)₂ = 2.0 g
Molar mass of anhydrous KAl(SO₄)₂ = 258.20 g/ mol
Mass of of anhydrous KAl(SO₄)₂ = mass of the 2nd heating = A
The mass of water released = mass of the Aluminum Cup + 2.0 grams of KAl(SO₄)₂ - mass of the 2nd heating
= H g
Moles of water released = 
Moles of anhydrous KAl(SO₄)₂ = 
Required ratio = 
The answer is: each compound is composed of one type of molecule.
<span>Well, I have personally used both A and C (provided the light frequency takes it above the work function of the condutor).
D could also work - a good conductor subject to a small voltage will carry a very large current. For instance, a flashlightg bulb filament is hot enough to emit electrons.
B is a no go.</span><span>
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