Answer: -
The first step involves protonation of the carbonyl oxygen.
After protonation, the Alcohol oxygen now attacks the carbon of the carbonyl.
Thus a six membered ring is formed with 5 carbon atoms and 1 oxygen atom. The 1st position carbon atom has 2 OH groups.
One of these gets again protonated.
This leaves as water. With the loss of the H+, there results a carbonyl at 1 position.
Thus 5-hydroxypentanoic acid forms a lactone or 2-oxanone in presence of acid.
Answer:
1.8 × 10² cal
Explanation:
When 0.32 g of a walnut is burned, the heat released is absorbed by water and used to raise its temperature. We can calculate this heat (Q) using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity of water
m: mass of water
ΔT: change in the temperature
Considering the density of water is 1 g/mL, 58.1 mL = 58.1 g.
Q = c × m × ΔT
Q = (1 cal/g.°C) × 58.1 g × 3.1°C
Q = 1.8 × 10² cal
Answer:
The isotope notation for an ion of silver-109 with a charge of positive 1 is 
.
Explanation:
The elements with same atomic number with different mass numbers are called isotopes.
The isotopes of silver is as follows.
From the above two isotopes have same atomic number and different mass number.
From the given,the isotope notation for an ion of silver-109 with a charge of positive 1
It can be represented is as follows.
General representation of isotope is as in attachment.
Answer:
72.6 mL
Explanation:
A quick way to solve this titration problem when you have a monoprotic acid is to use the Dilution equation, M1V1=M2V2.
.589(x)=.821(52.1)
X=72.6 mL
