If the Kelvin
temperature of a gas is doubled, the volume of the gas will increase by two. It
follows Charles law where in for a mixed gas of mass, the volume is directly
proportional to the temperature at constant pressure.
18 D
19 B
20 D
21 Anabolic steroid abuse can lead to kidney and liver failure, enlarged heart or high blood pressure and changes in blood cholesterol.
Answer: The hydroboration of an alkene occurs in TWO CONCERTED STEP which places the boron of the borane on the LESS SUBSTITUTED carbon of the double bond. The oxidizing agent then acts as a nucleophile, attacking the electrophilic BORON and resulting in the placement of a hydroxyl group on the attached carbon. Thus, the major product of the hydroboration oxidation reaction DOES NOT follow Markovnikov's rule.
Explanation:
Hydroboration is defined as the process which allows boron to attain the octet structure. This involves a two steps pathway which leads to the production of alcohol.
--> The first step: this involves the initiation of the addittion of borane to the alkene and this proceeds as a concerted reaction because bond breaking and bond formation occurs at the same time.
--> The second step: this involves the addition of boron which DOES NOT follow Markovnikov's rule( that is, Anti Markovnikov addition of Boron). This is so because the boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon.
Note: The Markovnikov rule in organic chemistry states that in alkene addition reactions, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component adds to the carbon atom with more hydrogen atoms bonded to it.
Answer:
Real gas behaves like ideal gas at high temperature and low pressure.
The balanced equation for the above reaction is as follows;
2Ca + O₂ ---> 2CaO
stoichiometry of Ca to O₂ is 2:1
we first need to find the limiting reactant
number of Ca moles - 6.84 mol
number of O₂ moles - 4.00 mol
if Ca is the limiting reactant
if 2 mol of Ca reacts with 1 mol of O₂
then 6.84 mol of Ca reacts with - 6.84 / 2 = 3.42 mol of O₂
this means that Ca is the limiting reactant and O₂ is in excess
therefore amount of CaO produced depends on amount of limiting reactant present
stoichiometry of Ca to CaO is 2:2
number of moles of Ca reacted = number of CaO moles formed
number of moles of CaO formed - 6.84 mol
answer is 6.84 mol
