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2 years ago

What is the poH of a 4.8 x 10-10 M H+ solution?

Chemistry

1 answer:

stiks02 [169]2 years ago

6 0

Answer: pOH = 4.68

Explanation:

pOH = 14 - pH

pH = - Log [H+]

= - Log [4.8 x 10^-10]

= -(-9.32)

pH =+9.32

Therefore, pOH= 14 - 9.32

= 4.68

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True/False: All of the elements combine in a variety of ways to make up all living and all nonliving things

likoan [24]

Answer:

true

Explanation:

5 0

2 years ago

Consider 100.0-g samples of two different compounds consisting only of carbon and oxygen. one compound contains 27.2 g of carbon

Sliva [168]

This is a tricky question. All that matters are ratios of percentages, not percentages themselves. So no one should directly compare 27.2 with 42.9. We must and shall compare the ratios (27.2 to 72.8) and (42.9 to 57.1).

Take them both down to 1 to and see what happens.

Working out the formulas knowing atomic masses is a bit beside the point; this is how people first DISCOVERED the idea of atomic mass.

Carbon Oxygen

27.2g 72.8g (100-27.2)

Moles 27.2/12 72.8/16

2.27 4.55

Ratio 1 2

Do the same with the other

4 0

3 years ago

Read 2 more answers

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

3 0

3 years ago

Explain where the water that collects on the outside of a cold drink on a hot day comes from.

krok68 [10]

Answer:

It is because water molecules in the air condensed on to the container of the drink.

Explanation:

The way this works is the water molecules outside are hot and in the gas state, so when they come into contact with the cold side of the container they lose energy due to heat transfer between the molecules and the container, becoming a liquid on the side of the drink.

4 0

2 years ago

Plz help ugh!! will upvote

erma4kov [3.2K]

All are CORRECT except (d)

6 0

2 years ago