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2 years ago

Are carboxylic acids of more than 10 carbons more soluble in polar or nonpolar solvents? Explain.

Chemistry

1 answer:

lina2011 [118]2 years ago

8 0

Answer: Non polar solvents

Explanation:

Since with increasing the size of alkyl group hydrophobic nature increases and solubility in polar solvents decreases .

Hence Carboxylic acids with more than 10 carbon atoms, solubility is more in non polar solvents.

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All light is the same whether it comes from the sun a light bulb or anywhere

ziro4ka [17]

Answer:

no because the suns light is way more bright than lightbulbs.

Explanation:

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3 years ago

Read 2 more answers

What type of isomerism is shown by alkanes

Luda [366]

Answer:

Alkanes with more than 3 carbons can show constitutional isomerism. They can be either linear or branched structures. This is categorized as chain isomerism. Butane is the smallest alkane to show such isomerism with 2 isomers.Alkanes with more than 3 carbons can show constitutional isomerism. They can be either linear or branched structures. This is categorized as chain isomerism. Butane is the smallest alkane to show such isomerism with 2 isomers.

Explanation:

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2 years ago

Acetic acid (HC2H3O2, Ka=1.8x10^-5) is a weak acid. Calculate the pH of an aqueous solution of .25M acetic acid.

tensa zangetsu [6.8K]

Answer: The pH of an aqueous solution of .25M acetic acid is 2.7

Explanation:

$HC_2H_3O_2\rightarrow H^+C_2H_3O_2^-$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.25 M and $\alpha$ = ?

$K_a=1.8\times 10^{-5}$

Putting in the values we get:

$1.8\times 10^{-5}=\frac{(0.25\times \alpha)^2}{(0.25-0.25\times \alpha)}$

$(\alpha)=0.0084$

$[H^+]=c\times \alpha$

$[H^+]=0.25\times 0.0084=0.0021$

Also $pH=-log[H^+]$

$pH=-log[0.0021]=2.7$

Thus pH is 2.7

5 0

3 years ago

Calculate the specific heat of a substance if a 35g sample absorbs 48 j as the temperature is raised from 293 k to 313 k

bulgar [2K]

Answer:

c = 0.07 j/g.k

Explanation:

Given data:

Mass of sample = 35 g

Heat absorbed = 48 j

Initial temperature = 293 K

Final temperature = 313 K

Specific heat of substance = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

ΔT = 313 k - 293 K

ΔT = 20 k

Now we will put the values in formula.

48 j = 35 g × c× 20 k

48 j = 700 g.k ×c

c = 48 j/700 g.k

c = 0.07 j/g.k

8 0

2 years ago

How to write a balanced chemical equation from empirical formula?

netineya [11]

Let's use the example: H2O ---> H2 + O2
We find how many elements of a product are on one side and how many elements on the other side.
Reactant: H=2 O=1
Product: H=2 O=2
We need to make the same amount of hydrogen and oxegyn atoms on each side, regardless of how high the numbers are, and we do this by adding coefficients to the compounds.

Reactant: H=4 O=2
Product : H=4 O=2
2 H2O---> 2 H2 + O2

8 0

2 years ago