Answer:
Percentage of a sample remains after 60.0 min is 13.03%.
Explanation:
- It is known that the decay of isotopes of C-11 obeys first order kinetics.
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- Half-life time (t1/2) in first order reaction = 0.693/k, where k is the rate constant.
∴ k = 0.693/(t1/2) = 0.693/(20.4 min) = 0.03397 min⁻¹.
- The integrated law for first order reaction is:
<em>kt = ln[A₀]/[A],</em>
where, k is the rate constant (k = 0.03397 min⁻¹).
t is the time of the reaction (t = 60.0 min).
[A₀] is the initial concentration of C-11 ([A₀] = 100.0 %).
[A] is the remaining concentration of C-11 ([A] = ???%).
<em>∵ kt = ln[A₀]/[A]</em>
∴ (0.03397 min⁻¹)(60.0 min) = ln(100%)/[A]
∴ 2.038 = ln(100%)/[A]
∴ 7.677 = (100%)/[A]
<em>∴ [A] </em>= (100%)/(7.677) = <em>13.03%.</em>
<em>So, percentage of a sample remains after 60.0 min is 13.03%.</em>
Half life is 15 hours
so 30 hours would be 2 half lives
after 15 hours it would be at 50%
after 15 more hours it would be at 50% of 50%
so .5 x .5 = .25
it would be 25%
A compound with the formula C6H12 is not considered a Saturated hydrocarbon.
Why is C6H12 isn't considered a Saturated hydrocarbon?
The ring's presence demonstrates that it is unsaturated. Keep in mind that the general formula for aliphatic hydrocarbons, CnH2n+2, serves as the foundation for its saturation. A chemical is unsaturated if it does not meet this requirement.
Example:
Hexane (C6H14)
C = 6; H = 14 = 2(6) + 2
resulting in hexane becoming saturated.
Cyclohexane(C6H12)
C = 6 and H = 12 do not equal 14 (x)!
cyclohexane is an unsaturated molecule as a result.
Cycloalkanes have the general formula C2H2n as well.
Hence, the given statement is false.
Learn more about the hydrocarbons here,
brainly.com/question/17578846
# SPJ4
Answer:
1.78 M⁻²s⁻¹
Explanation:
Let's consider the following reaction:
A + B → C + D
The reaction order for A is 1 and the reaction order for B is 2.
The rate law is:
r = k.[A].[B]²
where,
r is the rate of the reaction
k is the rate constant
[A] and [B] are the molar concentrations of the reactants
Then, we can find the value of k.
![r=k.[A].[B]^{2} \\k=\frac{r}{[A].[B]^{2} } =\frac{0.060M.s^{-1} }{(0.400M).(0.290M)^{2} } =1.78M^{-2} s^{-1}](https://tex.z-dn.net/?f=r%3Dk.%5BA%5D.%5BB%5D%5E%7B2%7D%20%5C%5Ck%3D%5Cfrac%7Br%7D%7B%5BA%5D.%5BB%5D%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B0.060M.s%5E%7B-1%7D%20%7D%7B%280.400M%29.%280.290M%29%5E%7B2%7D%20%7D%20%3D1.78M%5E%7B-2%7D%20s%5E%7B-1%7D)
Answer:
the answer is 3.33⋅1023 molecules of Cl2. Note that 6.022⋅1023 is also known as Avogadro's number, and it can be referred to as the number of molecules in one mole of that substance. Also, I calculated 70.9 because 35.45 g/mol (the mass of one Chlorine atom) times 2 is 70.9 g/mol (Cl2 has two Chlorine atoms).
Explanation: