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salantis [7]
3 years ago
6

A hydrated form of copper sulfate (CuSO4·x H2O) is heated to drive off all the water. If we start with 8.79 g of hydrated salt a

nd have 6.57 g of anhydrous CuSO4 after heating, find the number of water molecules associated with each CuSO4 formula unit.
Chemistry
1 answer:
exis [7]3 years ago
4 0

Answer:

3 water molecules

Explanation:

The molar ratio between CuSO₄ and H₂O needs to be calculated.

The mass of water that was removed by heating is calculated, then converted to moles. The molecular weight of water is 18.02g/mol.

(8.97 g - 6.57 g) = 2.22 g H₂O

(2.22 g)/(18.02g/mol) = 0.1232 mol H₂O

The moles of anhydrous copper sulfate are calculated. The molecular weight is 159.609 g/mol

(6.57 g)/(159.609 g/mol) = 0.04116 mol CuSO₄

Now the molar ratio between CuSO₄ and H₂O can be calculated:

0.1232 mol H₂O ÷ 0.04116 mol CuSO₄ = 3 H₂O / CuSO₄

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25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is
vlabodo [156]

Answer:

a)   119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

\bigtriangleup T_f=K_fm, \  \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\

#We use the molality above to calculate the molar mass:

m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol

Hence, the molar mass of the compound is 119 g/mol

5 0
3 years ago
What is the percent by mass of oxygen in carbon dioxide
san4es73 [151]
C = 12
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32/44*100 = 72.73%
6 0
3 years ago
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KengaRu [80]

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Most common insulation materials work by slowing conductive heat flow and--to a lesser extent--convective heat flow. Radiant barriers and reflective insulation systems work by reducing radiant heat gain. To be effective, the reflective surface must face an air space.

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Modify the given fatty acid so that it represents the 18‑carbon fatty acid designated 18:2(Δ9,12). Draw any double bonds in the
Mekhanik [1.2K]

Answer:

See explanation

Explanation:

In this case, we have to remember the meaning of the nomenclature "18:2Δ9,12".  Where 18 is the <u>number of carbon atom</u>s, 2 is the <u>number of double bonds,</u> and the numbers successive to Δ "delta" the position of the double bonds <u>starting</u> to count from the carboxylic -COOH end of the molecule.

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Lets see figure 1

I hope it helps!

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