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Agata [3.3K]
3 years ago
11

NO LINKS⚠️

Chemistry
1 answer:
Vesna [10]3 years ago
4 0
From Q = mcΔΤ, the specific heat capacity, c, of the metal that was cooled is c = Q/mΔT = (-769 J)/(46.4 g)(30.0 °C - 101.0 °C) = 0.233 J/g °C. From the table, it appears that this is the specific heat capacity of silver. So, the metal is most like silver.

Note: The value for Q was written as a negative value in the equation as heat energy was given off by the metal when the metal was cooled (from the metal’s point of view, it’s losing heat energy).
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Boiling point of a compound is determined by the strength of intermolecular forces of attraction between the molecules present in it. Stronger the intermolecular forces of attraction, higher will be the boiling point.

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How many milliliters of an aqueous solution of 0.170 M ammonium carbonate is needed to obtain 16.1 grams of the salt
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There will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.There will be needed mL of

Why?

In order to calculate how many milliliters are needed to obtain 16.1 grams of the salt given its concentration, we first need to find its chemical formula which is the following:

(NH_{4})2CO_{3}

Now that we know the chemical formula of the substance, we need to find its molecular mass. We can do it by the following way:

N_{2}=14g*2=28g\\\\2H_{4}=2*1g*4=8g\\\\C=12.01g*1=12.01g\\\\O_{3}=15.99g*3=47.97g

We have that the molecular mass of the substance will be:

MolecularMass=\frac{28g+8g+12.01g+47.97g}{mol}=95.98\frac{g}{mol}

Therefore, knowing the molecular mass of the substance, we need to calculate how many mols represents 16.1 grams of the same substance, we can do it by the following way:

mol_{(NH_{4})2CO_{3}=\frac{mass_{(NH_{4})2CO_{3}}}{molarmass_{(NH_{4})2CO_{3}}}

mol_{(NH_{4})2CO_{3}=\frac{16.1g}{95.98\frac{g}{mol}}=0.167mol

Finally, if we need to calculate how many milliliters are needed, we need to use the following formula:

M=\frac{moles_{solute}}{volume_{solution}}

M=\frac{moles_{solute}}{volume_{solution}}\\\\volume_{solution}=\frac{moles_{solute}}{M}

Now, substituting and calculating, we have:

volume_{solution}=\frac{0.167mol}{0.170\frac{mol}{L}}\\\\volume_{solution}=0.982L=0.982L*1000=982.35mL

Henc, there will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.

Have a nice day!

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