All categories

3 years ago

Draw the products of the acid-base reaction between stearic acid and OH-

Chemistry

1 answer:

jok3333 [9.3K]3 years ago

7 0

Answer:

Products are stearate anion and water.

Explanation:

Stearic acid is a 18-carbon chain molecule containing -COOH group. IUPAC name of stearic acid is octadecanoic acid.

Molecular formula of stearic acid is $C_{17}H_{35}CO_{2}H$ .

When $OH^{-}$ is added into stearic acid, $OH^{-}$ removes a proton ( $H^{+}$ ) from acidic -COOH group and forms stearate anion and water as products.

The balanced acid-base reaction is given as:

$C_{17}H_{35}CO_{2}H+OH^{-}\rightleftharpoons C_{17}H_{35}CO_{2}^{-}+H_{2}O$

Structure of products are given below.

You might be interested in

If 16 moles of al react with 3 moles of S8 how many moles of Al2 S3 will be formed

Gnom [1K]

Answer:

8 moles

Explanation:

Al reacts with $S_8$ to produce $Al_2S_3$ as

$Al+S_8\rightarrow Al_2S_3$

The balanced chemical equation is

$16Al+3S_8\rightarrow 8Al_2S_3$

In the reaction, 16 moles of $Al$ react with 3 moles of $S_8$ to produce 8 moles of $Al_2S_3$ .

8 0

2 years ago

Does acid in your stomach make you unhealthy?

Delvig [45]

No because it burns the food that u had and leaves it more space?????? I

8 0

3 years ago

Read 2 more answers

PLEASE HELP ME what would be the mass of 9.03 x 1021 molecules of hydrobromic acid (HBr)?

Nikitich [7]

We know 1 mole of any atom or molecules contains $6.033 \times 10^{23}$ atom or molecules.

1 mole of HBr i.e 81 gm/mol contains $6.033 \times 10^{23}$ atom or molecules.

So, mass of $9.03\times 10^{21}$ molecules is :

$m=\dfrac{81\times 9.03\times 10^{21} }{6.033 \times 10^{23}}\\\\m= 1.21\ gm$

Therefore, mass of $9.03\times 10^{21}$ molecules is 1.21 gm .

Hence, this is the required solution.

7 0

3 years ago

What are 3 things an Apple, Knife and Mirror have in common?

s2008m [1.1K]

1)They are all take up space.
2)They all have mass.
3)They are all solids.

Because of 1 and 2, they are all matter.

7 0

3 years ago

A balloon containing helium gas expands from 230

Anit [1.1K]

The answer for the following problem is mentioned below.

Therefore the final moles of the gas is 14.2 × $10^{-4}$ moles.

Explanation:

Given:

Initial volume ( $V_{1}$ ) = 230 ml

Final volume ( $V_{2}$ ) = 860 ml

Initial moles ( $n_{1}$ ) = 3.8 × $10^{-4}$ moles

To find:

Final moles ( $n_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

V ∝ n

$\frac{V_{1} }{V_{2} }$ = $\frac{n_{1} }{n_{2} }$

where,

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $n_{1}$ ) represents the initial moles of the gas

( $n_{2}$ ) represents the final moles of the gas

Substituting the above values;

$\frac{230}{860}$ = $\frac{3.8 * 10^-4}{n_{2} }$

$n_{2}$ = 14.2 × $10^{-4}$ moles

Therefore the final moles of the gas is 14.2 × $10^{-4}$ moles.

7 0

3 years ago