Answer:
Gamma
Explanation:
I'm not sure how to do it without calculations but:
E=hv
7*10^7 J/mol=6.626*10^34 Js * v
v=1*10^41
Gamma rays.
More here: https://www.hasd.org/faculty/AndrewSchweitzer/spectroscopy.pdf
Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols
For every 1 mol of gas, there will be
24000 cm^3 of gas
Vol. = 1.14 x 24 dm^3
= 27.36 dm^3
Acetaminophen as a chemical formula of C8H9NO2. The molar
masses are:
C8H9NO2 = 151.163 g/mol
C = 12 g/mol
H = 1 g/mol
N = 14 g/mol
O = 16 g/mol
<span>TO get the mass percent, simply multiply the molar mass of
each elements with the number of the
element divide by the molar mass of acetaminophen, that is:</span>
%C = [(12 * 8) / 151.163] * 100% = 63.50%
%H = [(1 * 9) / 151.163] * 100% = 5.954%
%N = [(14 * 1) / 151.163] * 100% = 9.262%
<span>%O = [(16 * 2) / 151.163] * 100% = 21.17% </span>
<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
The equation used to calculate half life for first order kinetics:

We are given:

Putting values in above equation, we get:

Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 206.7 yrs
= initial amount of the reactant = ?
[A] = amount left after decay process = 1.40 g
Putting values in above equation, we get:
![0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}](https://tex.z-dn.net/?f=0.0101yr%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B206.7yrs%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B1.40%7D)
![[A_o]=11.3g](https://tex.z-dn.net/?f=%5BA_o%5D%3D11.3g)
Hence, the initial amount of Uranium-232 present is 11.3 grams.
Answer:
Here's what I get
Explanation:
1. Balanced equation
HQ⁻ + CH₃-Br ⟶ HQ-CH₃ + Br⁻
(I must use HQ because the Brainly Editor thinks the O makes a forbidden word)
2. Mechanism
HQ⁻ + CH₃-Br ⟶[HQ···CH₃···Br]⁻⟶ HQ-CH₃ + Br⁻
A C B
The hydroxide ion attacks the back side of the carbon atom in the bromomethane (A).
At the same time as the Q-H bond starts to form, the C-Br bond starts to break.
At the half-way point, we have a high-energy intermediate (C) with partially formed C-O and C-Br bonds.
As the reaction proceeds further, the Br atom drops off to form the products — methanol and bromide ion (B).
3. Energy diagram
See the diagram below.