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3 years ago

Can someone please help!!! Its for chemistry and i looked everywhere even my text book and got no answer

Chemistry

1 answer:

Tatiana [17]3 years ago

8 0

Which question the top part, middle or the 7 covalents?

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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

3 years ago

What can you infer about copper's properties?

tiny-mole [99]

Answer:

good electrical conductor

lustre in appearance

7 0

3 years ago

What is the molality of a solution containing 10.0 G of NA2 S04 and 1000.0 G of water

galina1969 [7]

Explanation:

1 literThe total of water is equal to 1000.0 g of water

we need to find the molality of a solution containing 10.0 g of dissolved in Na₂S0₄1000.0 g of water

1. For that find the molar mass

Na: 2 x 22.99= 45.98

S: 32.07

O: 4 x 16= 64

The total molar mass is 142.05

We have to find the number of moles, y

To find the number of moles divide 10.0g by 142.05 g/mol.

So the number of moles is 0.0704 moles.

For the molarity, you need the number of moles divided by the volume. So, 0.0704 mol/1 L.

The molarity would end up being 0.0704 M

The molality of a solution containing 10.0 g of Na2SO4 dissolved in 1000.0 g of water is 0.0704 Mliter

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2 years ago

Which of the following is a galvanic cell?

Gemiola [76]

Answer:

i think Aluminum (Al) oxidized, zinc(Zn) reduced

5 0

3 years ago

Manganese, Mn, forms two ions, one with a 2+ charge and one with a 3+ charge. What is the formula for manganese (II) sulfide?

maria [59]

It would be MnSO4

The (II) lets you know it’s the form with a 2+ charge and Sulfate has a 2- charge

These will cancel out making it plain MnSO4

If it was manganese (iii) sulfide the answer would be Mn2(SO4)3

4 0

3 years ago