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3 years ago

Which of the following is a strong electrolyte

Chemistry

1 answer:

Ilya [14]3 years ago

6 0

A strong electrolyte is completely ionized. Ex: NaCl --h2o--> Na^+(aq) + Cl^-(aq)

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Object A has a molar heat of 31.2 J/mole∙°C and object B molar heat is 11.2 J/mole∙°C. Which object will heat up faster if they

riadik2000 [5.3K]

Answer:

Substance B

Explanation:

Molar heat of A = 31.2J/mole.°C

Molar heat of B = 11.2 J/mole∙°C.

The molar heat of a substance is the amount of heat that must be added to a mole of a substance to raise the temperature by 1°C.

Substance B will heat up faster compared to A.
It has a smaller molar heat compared to A.
This suggests that it will require lesser heat to raise its temperature by 1°C.

4 0

3 years ago

Please help! How many grams is 4.5 x 10^27

Eva8 [605]

Answer: The answer is 4.5e+27 grams, I hope I helped!

Explanation: N/A (I threw away my paper ^^")

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3 years ago

How much energy is released during the formation of 1 mol H2O

kari74 [83]

2.81 moles NO

I HOPE THAT RIGHT

7 0

3 years ago

Which of the following statement is not true?

konstantin123 [22]

I believe C is the answer <span />

4 0

3 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

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

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

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

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

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

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

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

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

4 years ago