Equation for Half life :
A = a(0.5)^(t/h)
A is current amount, "a" is initial amount, h is halflife, t is time
5 = 40(0.5)^(t/1.3x10^9)
5/40 = (0.5)^(t/1.3x10^9)
take the log of both sides , power rule
Log(5/40) = (t/1.3x10^9) * Log(0.5)
(1.3x10^9) * Log(5/40) / Log(0.5) = t
3.9x10^9 years = t
And if you think about what a half life is, the time it take for the amount to reduce to half.
40/2 = 20
20/2 = 10
10/2 = 5
It went through 3 half-lifes
3 * 1.3x10^9 = 3.9x10^9 years
Answer:
For eacht 4 moles Fe consumed, we will produce 2 moles Fe2O3
The mole ration is 4:2 (option 1)
Explanation:
Step 1: The unbalanced equation
Fe + O2 → Fe2O3
Step 2: Balancing the equation
Fe + O2 → Fe2O3
On the left side we have 2x O (in O2) and on the right side we have 3x O (in Fe2O3) . To balance the amount of O on both sides, we have to multiply O2 by 3 and Fe2O3 by 2.
Fe + 3O2 → 2Fe2O3
On the left side we have 1x Fe, On the right side we have 4x (in 2Fe2O3). To balance the amount of Fe we have to multiply Fe (on the left side) by 4.
Now the equation is balanced.
4Fe + 3O2 → 2Fe2O3
For eacht 4 moles Fe consumed, we will produce 2 moles Fe2O3
The mole ration is 4:2 (option 1)
Answer:
392.97 litres
Explanation:
From the equation of reaction, we can see that 1 mole of methane yielded 1 mole of carbon iv oxide. Hence, 15.9 moles of methane will yield 15.9 moles of carbon iv oxide.
At s.t.p one mole of a gas occupies a volume of 22.4L ,hence 15.9 moles of a gas will occupy a volume of 22.4 × 15.9 which equals
356.16L.
Now, we can use the general gas equation to get the volume produced at the values given.
We have the following values;
V1 = 356.16L P1= 1 atm ( standard pressure) T1 = 273K ( standard temperature) V2 = ? T2 = 23.7 + 273 = 296.7K P2 = 0.985 atm
The general form of the general gas equation is given as :
(P1V1)T1 = (P2V2)/T2
After substituting the values , we get V2 to be 392.97Litres
Answer:
Explanation:
The half-life of carbon (5730 y) is the time it takes for half the carbon to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table.
<u>No of half-lives</u> <u>Fraction remaining</u>
1
2
3
The general formula is
where <em>n</em> = the number of half-lives.
Thus, of the original carbon remains after 17 190 y.