The rate of Formation of Carbocation mainly depends on two factors'
1) Stability of Carbocation: The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.
2) Ease of detaching of Leaving Group: The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,
R-I > R-Cl > R-F
B > C > A
solution:
Hydration is the addition of water; hydrogenation is the addition of hydrogen.
desire rxn: _C4H6(g) + 2 H2(g)-----> C4H10(g)___dHhy = ??
knowns:
__________C4H6 + 11/2 O2 --------> 4CO2 + 3H2O______dHox = -2540.2 kJ/mole
__________4CO2 + 5H2O -----------> C4H10 + 13/2 O2___-dHox = 2877.6 kJ/mole
___________2(1/2 O2 + H2 -------------> H2O)___________2*dHox = 2(-285.8 kJ/mole)
Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem
<span>0.127 moles
The formula for nitroglycerin is C3H5N3O9 so let's first calculate the molar mass of it.
Carbon = 12.0107
Nitrogen = 14.0067
Hydrogen = 1.00794
Oxygen = 15.999
C3H5N3O9 = 3 * 12.0107 + 5 * 1.00794 + 3 * 14.0067 + 9 * 15.999 = 227.0829
Now calculate the number of moles of nitroglycerin you have by dividing the mass by the molar mass
2.50 ml * 1.592 g/ml / 227.0829 g/mol = 0.017527 mol
The balanced formula for when nitroglycerin explodes is
4 C3H5N3O9 => 12 CO2 + 10 H2O + O2 + 6 N2
Since all of the products are gasses at the time of the explosion, there is a total of 29 moles of gas produced for every 4 moles of nitroglycerin
Now multiply the number of moles of nitroglycerin by 29/4
0.017527 mol * 29/4 = 0.12707075 moles
Round to 3 significant figures, giving 0.127 moles</span>
Answer:
See explanation and image attached
Explanation:
This reaction is known as mercuric ion catalyzed hydration of alkynes.
The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.
Problem One (left)
This is just a straight mc deltaT question
<em><u>Givens</u></em>
m = 535 grams
c = 0.486 J/gm
tf = 50
ti = 1230
Formula
E = m * c * (ti - tf)
Solution
E = 535 * 0.486 * ( 1230 - 50)
E = 535 * 0.486 * (1180)
E = 301077
Answer: A
Problem Two
This one just requires that you multiply the two numbers together and cut it down to 3 sig digits.
E = H m
H = 2257 J/gram
m = 11.2 grams
E = 2257 * 11.2
E = 25278 to three digits is 25300 Joules. Anyway it is the last one.
Three
D and E are both incorrect for the same reason. The sun and stars don't contain an awful lot of Uranium (1 part of a trillion hydrogen atoms). It's too rare. The other answers can all be eliminated because U 235 is pretty stable in its natural state. It has a high activation complex.
Your best chance would be enriched Uranium (which is another way of saying refined uranium). That would be the right environment. Atomic weapons and nuclear power plants (most) used enriched Uranium. You can google "Little Boy" if you want to know more.
Answer: B
Four
The best way to think about this question is just to get the answer. Answer C.
A: incorrect. Anything sticking together implies a larger and larger result. Gases don't work that way. They move about randomly.
B: Wrong. Heat and Temperature especially depend on movement. Stopping is not permitted. If a substance's molecules stopped, the substance would experience an extremely uncomfortable temperature drop.
C: is correct because the molecules neither stop nor do they stick. The hit and move on.
D: Wrong. An ax splitting something? That is not what happens normally and not with ordinary gases. It takes more energy that mere collisions or normal temperatures would provide to get a gas to split apart.
E: Wrong. Same sort of comment as D. Splitting is not the way these things work. They bounce away as in C.
Five
Half life number 1 would leave 0.5 grams behind.
Half life number 2 would leave 1/2 of 1/2 or 1/4 of the number of grams left.
Answer: 0.25
Answer C
