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3 years ago

Which equilibrium expression represents the ionization of water?

Chemistry

2 answers:

Tems11 [23]3 years ago

8 0

Keq = [H3O+][OH-] .............................

pav-90 [236]3 years ago

5 0

H₂O + H₂O ⇄ H₃O⁺ + OH⁻

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How many grams of F are in 12.56 g of SF6? h.

natali 33 [55]

Answer:

9.80 g

Explanation:

The molecular mass of the atoms mentioned in the question is as follows -

S = 32 g / mol

F = 19 g / mol

The molecular mass of the compound , SF₆ = 32 + ( 6 * 19 ) = 146 g / mol

The mass of 6 F = 6 * 19 = 114 g /mol .

The percentage of F in the compound =

mass of 6 F / total mass of the compound * 100

Hence ,

The percentage of F in the compound = 114 g /mol / 146 g / mol * 100

78.08 %

Hence , from the question ,

In 12.56 g of the compound ,

The grams of F = 0.7808 * 12.56 = 9.80 g

4 0

3 years ago

Drag each label to the correct location on the image identify the parts of the energy diagram

diamong [38]

If you would’ve attached a picture I’m sure it would’ve been a lot easier.

4 0

2 years ago

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Im sooooooooooooooooooooooooo cufused on science

Elena-2011 [213]

Answer:

What do you need please to understand?

6 0

3 years ago

Could the cuttlefish tell the difference between the blue and yellow in the lab?

jeka94

Answer:

<em>yes</em>

Explanation:

the cuttle fish tell the difference between blue and yellow

5 0

1 year ago

Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs

saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

Percent of $Ag^+$ remains is, 0.0018 %

Explanation :

$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

$K_{sp}$ for AgBr is $7.7\times 10^{-13}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

$CuBr\rightleftharpoons Cu^++Br^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][Br^-]$

$4.2\times 10^{-8}=0.073\times [Br^-]$

$[Br^-]=5.75\times 10^{-7}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgBr\rightleftharpoons Ag^++Br^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][Br^-]$

$7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M$

$[Ag^+]=1.34\times 10^{-6}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{1.34\times 10^{-6}}{0.073}\times 100$

Percent of $Ag^+$ remains = 0.0018 %

3 0

3 years ago