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3 years ago

What metal is easily oxidized in rocks?

Chemistry

1 answer:

Tom [10]3 years ago

4 0

Sodium, magnesium, iron. :)

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Can someone please try to help me

ahrayia [7]

Answer:

I would say B all of these

4 0

2 years ago

How many moles of water are in 1.23 x 10^18 water molecules?

TiliK225 [7]

1.23x10^24 atoms/6.022x10^23 atom/mol = 2.04 mol H20

3 0

3 years ago

Read 2 more answers

What is a meteorite?

yawa3891 [41]

Answer:

I think its C

Explanation:

6 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

3 years ago

Help lols plsssksksksksskd

artcher [175]

Answer:

I'd say "a" because not everything with plants and animals is perfectly organized together nothing ever is.

3 0

3 years ago