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3 years ago

What’s the IUPAC name

Chemistry

1 answer:

hichkok12 [17]3 years ago

5 0

Answer:

Methyl pentanoate.

Explanation:

Hello there!

In this case, according to the given information, we can see the correct structure will be:

CH3CH2CH2CH2COCH3

Which matches with the structure of an ester due to the -COO- functional group. In such a case, the first part of the name is in function of the right side of the ester, in this case, methyl, followed by the left side, pentanoate, as it has five carbon atoms and is an ester (similar to an inorganic salt, but organic) and therefore, the name will be methyl pentanoate.

Regards!

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How can you separate the components soil and water

Vaselesa [24]

Answer:

Put the soil in one area and keep the water heated so the water can evaporate.

Explanation:

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3 years ago

If 3.25 mol of Ar occupies 100. L at a particular temp and pressure, what volume does

gtnhenbr [62]

Answer:

435.38 L

Explanation:

From the question,

Applying ideal gas equation,

PV = nRT................... Equation 1

Where P = Pressure, V = Volume, n = number of moles, R = molar gas constant, T = Temperature.

Since Temperature and Pressure were constant,

V ∝ n

V/n = V'/n'.................... Equation 2

Where V and V' = the initial and final Volume of Ar respectively, n and n' = the initial and final moles of Ar respectively

make V' the subject of the equation

V' = Vn'/n............... Equation 3

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Substitute into equation 3

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3 years ago

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3 years ago

List three properties of metal that nonmetals typically DO NOT have

finlep [7]

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3 years ago

Read 2 more answers

Calculate the pH of a 1.00 L buffer of 0.97 M CH3COONa / 1.02 M CH3COOH before and after the addition of the following species.

ivanzaharov [21]

Answer:

a) 4.73

b) 4.78

c) 4.66 (further addition) or 4.60 (starting from the original buffer solution)

Explanation:

Step 1: Data given

volume of the buffer = 1.00 L

Buffer = 0.97 M CH3COONa / 1.02 M CH3COOH

pKa CH3COOH = 4.75

Step 2: pH = pKa + log [CH3COONa]/[CH3COOH]

pH = 4.75 + log (0.97/1.02)

pH = 4.73

(b) pH after addition of 0.065 mol NaOH

Adding 0.065 mol NaOH will reduce the acid by that amount leaving 1.02 - 0.065 = 0.955 moles HA in 1 L so [HA] = 0.955; the neutralized acid produces A- in the same amount, increasing [A-] to 0.97 +0.065 = 1.035

pH = pKa + log[CH3COONa]/[CH3COOH]

pH = 4.75 + log(1.035/0.955)

pH = 4.78

c) pH after further addition of 0.144 mol HCl

The reverse will happen after the addition of HCl:

[HA] = 0.955 + 0.144 = 1.099

[A-] = 1.035 - 0.144 = 0.891

pH = 4.75 + log(0.891/1.099)

pH = 4.66

If we add 0.144 mol of HCl to the original buffer we will get:

[HA] = 1.02 + 0.144 = 1.164

[A-] = 0.97 - 0.144 = 0.826

pH = 4.75 + log(0.826/1.164)

pH = 4.60

3 0

3 years ago