Answer:
Explanation:
we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly
Q=mcΔT
ΔT=T2-T1
ΔT=26.8-10.2=16.6
C for water is 4.184
therefore
Q=1.00*4.184*16.6
Q=69.4 j
now we have to covert joule into calorie
1 calorie =4.2 j
x calorie=69.4 j/2
so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius
The energy increases because the molecules in water move faster
Answer:
a) Attached below
b) The presence of racemic mixture found as product in both cases shows that products are identical ( i.e. they have same configuration
Explanation:
Diagrams of the products obtained from hydroboration-oxidation of cis-2-butene , hydroboration-oxidation of trans-2-butene.
attached below
The presence of racemic mixture found as product in both cases shows that products are identical ( i.e. they have same configuration )
<span>31.8/(31.8+225) *100 = 5.86%
12 moles * 22.4 L/
[email protected]= 26.9 L</span>