3 years ago

WILL MARK YOU BRAINLIEST!! how are a elements abundance in nature and percent composition related

Chemistry

1 answer:

Drupady [299]3 years ago

8 0

Because they both have to do with and chemistry science

You might be interested in

126785033% Divided by 348675948849 = <br> what does it equal

meriva

Answer:

0.00000363618

could be wrong.

double check me someone or just trust me

(don't blame me if you get it wrong)

5 0

3 years ago

Calculated from measurements of distance and time defined

prisoha [69]

Answer:

Any of the six chemical elements that markup group1

of the periodic table.

Explanation:

8 0

3 years ago

You and your roommates have already decided to spend a week volunteering in the relief effort over the winter break, but in orde

Vlada [557]

Answer:

I dont know the answer for that question it's hard question isn't it

5 0

3 years ago

Read 2 more answers

[01.03]Einstein’s Theory of Relativity was built on the research and ideas of other scientists. The Theory of Relativity propose

11Alexandr11 [23.1K]

Can you attach the statements?

5 0

3 years ago

Read 2 more answers

The PH of a 0.1 M MCl (M is an unknown cation) was found to be 4.7. Write the net ionic equation for the hydrolysis of M and its

Serggg [28]

Answer:

6.25 X10^{-9} = Ka

$Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}$

Explanation:

The ionic equation for the hydrolysis of the cation of the given salt will be:

$M^{+} + H_{2}O ---> MOH + H^{+}$

The expression for Ka will be:

Ka = $\frac{[H^{+}][MOH]}{[M^{+}]}$

As given that the concentration of the salt is 0.1 M and pH of solution = 4.7, we can determine the concentration of Hydrogen ions from the pH

pH = -log [H⁺]

[H⁺] = antilog(-pH) = antilog (-4.7) = 2 X 10⁻⁵ M = [MOH]

Let us calculate Ka from this,

Ka = ${2.5X10^{-5}X2.5X10^{-5}}{0.1} = 6.25 X10^{-9}$

The relation between Ka an Kb is

KaXKb =10⁻¹⁴

$Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}$

6 0

3 years ago