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2 years ago

Ignore the one I chose hehe! but please help?!

Chemistry

2 answers:

FrozenT [24]2 years ago

6 0

Answer:

umm i believe that the one you chose is correct

Explanation:

gulaghasi [49]2 years ago

3 0

Answer:

It's A you are correct

Explanation:

Comounds contain two or more elements

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The energy level where an electron is located, is shown by the ______ on the periodic table

Norma-Jean [14]

Answer:

the first energy level is closest to nuclear the second energy level is a little farther away than the first

8 0

3 years ago

What is the ph of a solution of 0.400 m k2hpo4, potassium hydrogen phosphate?

dusya [7]

When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :

when Pka = - ㏒ Ka

6.86 = -㏒ Ka

∴Ka = 1.38 x 10^-7

by using ICE table:

H2PO4- → H+ + HPO4
initial 0.4 m 0 0

change -X +X +X

Equ (0.4-X) X X

when Ka = [H+][HPO4] / [H2PO4-]

by substitution:

1.38 X 10^-7 = X^2 / (0.4-X) by solving for X

∴X = 2.3x 10^-4

∴[H+] = X = 2.3 x 10^-4

∴PH = -㏒[H+]

= -㏒ (2.3 x 10^-4)
∴PH = 3.6

3 0

3 years ago

If a lab experiment is not completed, you should:___________.

stellarik [79]

The answer to this question is a

7 0

3 years ago

Read 2 more answers

CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial concentrations o

Katarina [22]

Answer:

9.4

Explanation:

The equation for the reaction can be represented as:

$CO_{(g)}$ + $2H_2O_{(g)}$ ⇄ $CH_3OH_{(g)}$

The ICE table can be represented as:

$CO_{(g)}$ + $2H_2_{(g)}$ ⇄ $CH_3OH_{(g)}$

Initial 0.27 0.49 0.0

Change -x -2x x

Equilibrium 0.27 - x 0.49 -2x x

We can now say that the concentration of $CH_3OH_{(g)}$ at equilibrium is x;

Let's not forget that at equilibrium $CH_3OH_{(g)}$ = 0.11 M

So:

x = [ $CH_3OH_{(g)}$ ] = 0.11 M

[ $CO_{(g)}$ ] = 0.27 - x

[ $CO_{(g)}$ ] = 0.27 - 0.11

[ $CO_{(g)}$ ] = 0.16 M

[ $2H_2_{(g)}$ ] = (0.49 - 2x)

[ $2H_2_{(g)}$ ] = (0.49 - 2(0.11))

[ $2H_2_{(g)}$ ] = 0.49 - 0.22

[ $2H_2_{(g)}$ ] = 0.27 M

$K_C = \frac{[CH_3OH]}{[CO][H_2]^2}$

$K_C = \frac{(0.11)}{(0.16)[(0.27)^2}$

$K_C = 9.4307$

$K_C$ = 9.4

∴ The equilibrium constant at that temperature = 9.4

8 0

2 years ago

What is the oxidation state of S in so ??

Marina86 [1]

Answer:

Explanation:

If a compound $SO$ existed, we would identify the oxidation state of sulfur using the following logic:

oxygen is more electronegative than sulfur, so it's more electron-withdrawing and it should have a negative oxidation state producing a positive oxidation state for sulfur;
oxygen typically has an oxidation state of -2;
we may then apply the fact that SO is expected to be a molecule with a net charge of 0;
if the net charge is 0 and the oxidation state of oxygen is -2, we may set the oxidation state of S to x;
write the equation for the net charge of 0 by adding all individual charges of the two atoms: $x + (-2) = 0$ ;
hence, x = 2.

That said, in this hypothetical compound S would have an oxidation state of +2.

6 0

3 years ago