Answer:
Explanation:
(a) Part 1: 
reaction. This is a nucleophilic substitution reaction in which we have two steps. Firstly, chlorine, a good leaving group, leaves the carbon skeleton to form a relatively stable secondary carbocation. This carbocation is then attacked by the hydroxide anion, our nucleophile, to form the final product.
To summarize, this mechanism takes places in two separate steps. The mechanism is attached below.
Part 2: 
reaction. This is a nucleophilic substitution reaction in which we have one step. Our nucleophile, hydroxide, attacks the carbon and then chlorine leaves simultaneously without an intermediate carbocation being formed.
The mechanism is attached as well.
(b) The rate determining step is the slow step. Formation of the carbocation has the greatest activation energy, so this is our rate determining step for . For , we only have one step, so the rate determining step is the attack of the nucleophile and the loss of the leaving group.
Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
Answer:One carbon atom forms a double bond with an oxygen atom and two single bonds with two hydrogen atoms
Explanation:
Answer:
It kind of is logical so my answer is yes
The uranium within these items is radioactive and should be treated with care. Uranium's most stable isotope, uranium-238, has a half-life of about 4,468,000,000 years. It decays into thorium-234 through alpha decay or decays through spontaneous fission.
