Ask question

Ask question

All categories

3 years ago

5

the elements carbon hydrogen and oxygen are all part of the same blank on the periodic table........ i i think GROUP diangol row

period

1 answer:

Mrrafil [7]3 years ago

6 0

The elements Carbon, Hydrogen, and Oxygen are all part of Non-Metals.

Hope This Helps! :)

You might be interested in

Determine the number of valence electrons in each of the following neutral atoms.

Harlamova29_29 [7]

Answer:

Fluorine = 7 v.e.

Sulfur = 6 v.e.

Oxygen = 6 v.e.

Nitrogen = 5 v.e.

Carbon = 4 v.e.

Explanation:

Valence electrons can be easily determined by understanding groups (columns). Every element in. . .

column 3A (13) = 3 v.e.

column 4A (14) = 4 v.e.

column 5A (15) = 5 v.e.

column 6A (16) = 6 v.e.

column 7A (17) = 7 v.e.

6 0

3 years ago

In nature, the chlorine atom has two isotopic forms, 35Cl and 37Cl. The masses of each form are 34.96885amu (75.76%) and 36.9659

Anni [7]

Answer:

Explanation:

mass avg = ( abundance 1 x mass 1 = abundance 2 x mass 2 )

= ( 75.76% x 34.96885 + 24.24% x 36.9659)

= ( 0.7576 x34.96885 + 0.2424 x 36.9659 )

= ( 26.49241 + 8.960534 )

= 35.45294 amu

7 0

3 years ago

The ph of coffee is 5.6. The ph of grapefruit juice is 2.6. This means that the proton concentration in coffee is

Brrunno [24]

pH scale is used to measure the acidity or alkalinity on a scale of 0-14 where 0-6.9 is the acidic region , 7.1-14 is the basic region and 7 is for the neutral substance. We can calculate the concentration of proton from pH.

$pH=-log[H^+]$

where $[H^+]$ is the concentration of proton

As per the question ,the pH of coffee is 5.6 and we need to find the concentration of proton so putting the values in the above equation, we get

$5.6=-log[H^+]$

$-5.6=log[H^+]$

$antilog(-5.6)= [H^+]$

$[H^+]= 2.511\times10^-^6 M$

6 0

3 years ago

Write the word or phrase beginning with each letter of the word CHANGE

Mrrafil [7]

Creativity
Happiness
A-idk
New things
G-idk
Experience
Don't know what to do for G and A sorry!

4 0

4 years ago

Consider the dissolution of AB(s): AB(s)⇌A+(aq)+B−(aq) Le Châtelier's principle tells us that an increase in either [A+] or [B−]

Arlecino [84]

Answer:

A. 0.000128 M is the solubility of M(OH)2 in pure water.

B. $3.23\times 10^{-6} M$ is the solubility of $M(OH)_2$ in a 0.202 M solution of $M(NO_3)_2$ .

Explanation:

A

Solubility product of generic metal hydroxide = $K_{sp}=8.45\times 10^{-12}$

$M(OH)_2\rightleftharpoons M^{2+}+2OH^-$

S 2S

The expression of a solubility product is given by :

$K_{sp}=[M^{2+}][OH^-]^2$

$K_{sp}=S\times (2S)^2=4S^3$

$8.45\times 10^{-12}=4S^3$

Solving for S:

$S=0.000128 M$

0.000128 M is the solubility of M(OH)2 in pure water

B

Concentration of $M(NO_3)_2$ = 0.202 M

Solubility product of generic metal hydroxide = $K_{sp}=8.45\times 10^{-12}$

$M(OH)_2\rightleftharpoons M^{2+}+2OH^-$

S 2S

So, $[M^{2+}]=0.202 M+S$

The expression of a solubility product is given by :

$K_{sp}=[M^{2+}][OH^-]^2$

$8.45\times 10^{-12}=(0.202 M+S)(2S)^2$

Solving for S:

$S=3.23\times 10^{-6} M$

$3.23\times 10^{-6} M$ is the solubility of $M(OH)_2$ in a 0.202 M solution of $M(NO_3)_2$ .

8 0

4 years ago