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kotegsom [21]
3 years ago
12

The ester below can be separated into phenol and an acetate salt, via saponification: heating the ester with a strong base, such

as sodium hydroxide, will produce phenol and sodium acetate. the two chemicals can then be separated by heat and filtration.
a. calculate the unsaturation number of the ester above.
b. calculate the molecular weight of the ester above

Chemistry
1 answer:
erastovalidia [21]3 years ago
4 0
I found a similar question online which will help me answer your incomplete question. To make it easier, show all the elements of the compound given. It is shown in the second picture attached.

a.) The formula for unsaturation number is shown in the 3rd picture attached. Following this, 
n = 8
m = 2(8) + 2 + 0 + 0 - 0 = 18
Thus,
x = Unsaturation number = (18 - 8)/2 = 5
<em>The unsaturation number is 5.</em>

b.) The molar mass of C is 12.01 g/mol; H is 1 g/mol; O is 16 g/mol. So, the molecular weight is:
Molecular weight = 12.01(8) + 8(1) + 2(16) = 136.08 g/mol
<em>The molecular weight of the ester is 136.08 g/mol.</em>

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2 years ago
what is the mole fraction of neon in a mixture that contains 0.628 g of helium, 11.491 g of neon, and 7.613 g of argon?​
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Answer:

0.2

Explanation:

Given parameters:

Mass of helium = 0.628g

Mass of neon = 11.491g

Mass of argon = 7.613g

Unknown:

Mole fraction of neon = ?

Solution:

The mole fraction of an element is the number of moles of that element to the total number of moles in the gas mixture.

We need to calculate the number of moles of each element first;

 Number of moles  = \frac{mass}{molar mass}

Molar mass of Helium = 4g/mol

Molar mass of Neon = 20g/mol

Molar mass of Argon = 40g/mol

Number of moles of He = \frac{0.628}{4} = 0.16moles

Number of moles of Ne = \frac{11.491}{20} = 0.58moles

Number of moles of Ar = \frac{7.613}{40} = 0.19moles

Total number of moles = 0.16moles + 0.58moles + 0.19moles = 0.93moles

Mole fraction Neon = \frac{0.19}{0.93}  = 0.2

4 0
3 years ago
If a gas is initially at a pressure of nine ATM and a volume at 21 L at a temperature of 253K and the pressure is raise to 15 AT
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Answer:

15.0L

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v=15.0

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