Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
The answer is B. the occurrence of huge events in Earth's natural history
The geologic time scale is a system of chronological dating that relates geological strata to time. It is used by geologists, paleontologists, and other Earth scientists to describe the timing and relationships of events that have occurred during Earth's history.
Answer:
d. 1600 calories
Explanation:
The heat of fusion of water, L, is the amount of heat per gram required to melt the ice to water, a process which takes place at a constant temperature of 0 °C. The specific heat of water, c, is the amount of heat required to change the temperature of 1 gram of water by 1 degree Celsius.
We will convert the units of c from Jg⁻¹°C⁻¹ to cal·g⁻¹°C⁻¹ since the answers are provided in calories. The conversion factor is 4.18 J/cal.
(4.18 Jg⁻¹°C⁻¹)(cal/4.18J) = 1 cal·g⁻¹°C⁻¹
First we calculate the heat required to melt the ice, where M is the mass:
Q = ML = (15 g)(80 cal/g) = 1200 cal
Then, we calculate the heat required to raise the temperature of water from 0 °C to 25 °C.
Q = mcΔt = (15 g)(1 cal·g⁻¹°C⁻¹)(25 °C - 0 °C) = 380 cal
The answer is rounded so that there are two significant figures
The total heat required for this process is (1200 cal + 380 cal) = 1580 cal
The rounded answer is 1600 calories.
Answer:
21.28 grams solute can be added if the temperature is increased to 30.0°C.
Explanation:
Solubility of solute at 20°C = 32.2 g/100 grams of water
Solute soluble in 1 gram of water = 
Mass of solute in soluble in 56.0 grams of water:
Solubility of solute at 30°C = 70.2g/100 grams of water
Solute soluble in 1 gram of water = 
Mass of solute in soluble in 56.0 grams of water:
If the temperature of saturated solution of this solute using 56.0 g of water at 20.0 °C raised to 30.0°C
Mass of solute in soluble in 56.0 grams of water 20.0°C = 18.032 g
Mass of solute in soluble in 56.0 grams of water at 30.0°C = 39.312 g
Mass of of solute added If the temperature of the saturated solution increased to 30.0°C:
39.312 g - 18.032 g = 21.28 g
21.28 grams solute can be added if the temperature is increased to 30.0°C.
