Freidal craft reaction is the attack of a carbon or carbon chain on aromatic ring with the help of anhydrous AlCl3 to produce alkylated benzene ring.
Only ketone not be able to undergo friedal craft reaction as, it is not aromatic compound whereas all the given reactants are aromatic and gives friedal craft reaction.
Hi!
<u>The correct options would be: </u>
is an electron
is emitted from nucleus
has a -1 charge
Explanation:
A beta particle is a result of a neutron (a neutral particle) changing into two particles of opposite charges - a high energy electron (with a negative charge -1) and a positron (with a positive charge +1). Neutrons are present in the nucleus of an atom, and thus the beta particles are said to be emitted from the nucleus. They do have a charge, not zero, but it is not +2 and can only be either -1 or +1. This form of radiation is not electromagnetic energy because beta particles are massless, and do not travel at the speed of light (both being characteristics of electromagnetic radiation). Beta particles are not pure forms of energy.
Hope this helps.
This question is missing the part that actually asks the question. The questions that are asked are as follows:
(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.
(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.
We can use the equation for a first order rate law to find the amount of material remaining after 4 days:
[A] = [A]₀e^(-kt)
[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.
(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.
4 days x 1 year/365 days = 0.0110
A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg
The decay of americium is so slow that no noticeable change occurs over 4 days.
(b) We can simply plug in the information of iodine-125 and solve for A:
A = (1.00)e^(-0.011 x 4)
A = 0.957 mg
Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.
The heat required to vaporize 43.9 g of acetone at its boiling point is calculated as below
the heat of vaporization of acetone at its boiling point is 29.1 kj/mole
find the moles of acetone = mass/molar mass
= 43.9g /58 g/mol =0.757 moles
heat (Q) = moles x heat of vaporization
= 29.1 kj/mole x 0.757 moles = 22.03 kj
moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
[H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is [H2SO4] = 0.08955 M
