Answer:
The presence of neutrons
Explanation:
Based on gold foil experiments, Rutherford proposed a atomic model known as Rutherford'd model or nuclear model or planetary model of the atom.
As per this model, most of the mass of the atom is concentrated in the center of the atom known as nucleus.
Electrons revolve around it same as planet revolve around sun.
Since, neutrons had not been discovered at the time of Rutherford'd model, therefore, as per the Rutherford'd model, nucleus is only made of protons.
Therefore, the correct option is option B
You must first calculate the total weight of 1 mole of the substance. You do this by finding the molar weights on the periodic table of each of the elements in the compound:

So now we take the amount of each element that is given in the compound and divide it by the total mass of the compound:
Carbon: 96.08g

Carbon = 49.48%
Nitrogen: 56.028g

Nitrogen = 28.85%
Oxygen: 31.998g

Oxygen = 16.48%
Answer: 88.2 g
Solution:
1) Chemical equation:
<span>2Al (s) + 3Fe(NO3)2 (aq) → 3Fe (s) + 2Al(NO3)3 (aq)
2) Theoretical molar ratios
2 mol Al : 3 mol Fe(NO3)2 : 3 mol Fe : 2 mol Al(NO3)3
3) Starting mass of pure iron nitrate
% = (mass of iron nitrate / mass of solution) * 100 = 87.5
=> mass of iron nitrate = 87.5 * mass of solution / 100
mass of solution = 325 g
=> mass of iron nitrate = 87.5 * 325 g / 100 = 284.375 g
4) moles of iron nitrate
moles = mass in grams / molar mass
molar mass of Fe(NO3)2 = 179.85 g/mol
moles = 284.375 g/ 179.85 g/mol = 1.58 moles Fe(NO3)2
5) proportion:
x 3 mol Fe
--------------------------- = ----------------------
1.58 mol Fe(NO3)2 3 mol Fe(NO3)2
Clear x:
x = 1.58 mol Fe
6) Convert 1.58 mol Fe into grams
mass = number of moles * atomic mass
atomic mass of iron = 55.845 g / mol
mass = 1.58 moles * 55.845 g/mol = 88.24 g
Rounded to 3 significant figures: 88.2 grams of Fe.
</span>
Answer:
(A). C6H5Br + Mg(in ether) -----------> C6H5MgBr.
(B). C6H5MgBr + O = C = O -----------> C6H5-COO^- Mg^+ Br.
(C). C6H5-COO^- Mg^+ Br + HCl --------> C6H5-COOH + Mg^+Br(OH).
PRODUCTS=> C6H5-COOH and Mg^+Br(OH).
Explanation:
A Grignard reagent is a reagent that/which is an organometallic compound that is R -Mg- X. The R = alkyl, vinyl or allyl and the X = halogens.
It must be noted that an important reaction of Grignard reagent is its reaction with compounds containing the Carbonyl that is -CO functional group and this kind of Reaction is known as a Grignard Reaction.
So, in this question we are told that;
=> "1-bromo-benzene andits subsequent reaction with solid carbon dioxide (CO2) followed by acidic workup (using HCl asthe acid). "
Thus;
(A). C6H5Br + Mg(in ether) -----------> C6H5MgBr.
(B). C6H5MgBr + O = C = O -----------> C6H5-COO^- Mg^+ Br.
(C). C6H5-COO^- Mg^+ Br + HCl --------> C6H5-COOH + Mg^+Br(OH).