Answer: the product is ketone or aldehyde
Explanation:
The first step is the conversion of acetal to hemiacetal in the presence of H3O+/ ROH, and then the final conversion of hemiacetal to ketone/aldehyde using
H3O+/ ROH...
Attached is the structural conversion
<u>Answer:</u> The moles of water produced are 1.54 moles.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:

Given mass of ethane = 15.42 g
Molar mass of ethane = 30.07 g/mol
Putting values in above equation, we get:

The chemical equation for the combustion of ethane follows:

By Stoichiometry of the reaction:
2 moles of ethane produces 6 moles of water
So, 0.513 moles of ethane will produce =
of water
Hence, the moles of water produced are 1.54 moles.
Answer:
Are basic:
[OH⁻] = 3.13x10⁻⁷M and [H₃O⁺] = 9.55x10⁻⁹M
Explanation:
A solution is basic when pH = - log [H₃O⁺] is higher than 7.
It is possible to convert [OH⁻] to [H₃O⁺] using:
[H₃O⁺] = 1x10⁻¹⁴ / [OH⁻]
a. [OH⁻] = 3.13x10⁻⁷M
[H₃O⁺] = 1x10⁻¹⁴ / [3.13x10⁻⁷M]
[H₃O⁺] = 3.19x10⁻⁸M
pH = - log [H₃O⁺] = 7.50
[OH⁻] = 3.13x10⁻⁷M is basic
b. pH = -log [H₃O⁺] = - log 0.000747M = 3.13.
This solution is not basic
c. [H₃O⁺] = 9.55x10⁻⁹M
pH = 8.02
This solution is also basic.
Answer:
prepared by mixing 20 ml of 0050 m hcl with 30 ml of 010 m ba oh 2 ... No. of milliequivalents of HCl = 2× 0.05 × 1. =1.
Explanation:
is this ?