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3 years ago

What is the number of moles of hydrogen that is needed to produce 0.253 mol of water

Chemistry

1 answer:

kakasveta [241]3 years ago

7 0

the answer would be equal to 0.253 mol

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22 °C to K (approximate room temperature)

marta [7]

Answer:

295.15 i think

Explanation:

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3 years ago

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Vilka [71]

Answer:

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3 years ago

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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 %

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3 years ago

Familiar solutions can have a wide range of pH levels. According to the chart, which is the strongest acid, based on its pH leve

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3 years ago

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How many molecules are there in 79g of Fe2O3?

weqwewe [10]

There are approximately 160 grams in 1 mol of fe203 molecules therefore there would be 79/160 = 0.49375 mols of fe203 molecules in 79 grams therefore 5 atoms in total for each molecule of fe203 therefore 79/160 *5 =79/32=2.46875 mols in atoms

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3 years ago