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Nina [5.8K]
2 years ago
15

a solution of copper sulfate slowly evaporates beautiful blue crystals made Cu(11) and sulfate ions from such that water molecul

es are trapped inside the crystals. the overall formula of the compound is CuSO4•5H2O. what is the percent water in this compound?​
Chemistry
1 answer:
adelina 88 [10]2 years ago
6 0

Answer:

36.08%

Explanation:

We are given the overall formula of the compound as: CuSO4•5H2O

Now, atomic mass of the elements are;

Cu = 63.55 g/mol

S = 32.07 g/mol

O = 16 g/mol

H20 = 18.02 g/mol

Now, let's calculate the total mass of the compound:

(63.55 g/mol) + (32.07 g/mol) + 4(16 g/mol) + 5(18.02 g /mol) = 249.72 g/mol

From the above, water in the compound is 5(H20)

Thus, total water atomic mass = 5 × 18.02 = 90.1 g/mol

Thus, percentage of water = (atomic mass of water/total mass of compound) × 100%

Percentage of water = (90.1/249.72) × 100% = 36.08%

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How to reproduce with a dolphin?<br> and how long will it take for the dolphin baby to come out?
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3 years ago
Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu
pochemuha

Answer : The volume of CO_2 will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2

First we have to calculate the  mass of HCO_3^- in tablet.

\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g

Now we have to calculate the moles of HCO_3^-.

Molar mass of HCO_3^- = 1 + 12 + 3(16) = 61 g/mole

\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles

Now we have to calculate the moles of CO_2.

From the balanced chemical reaction, we conclude that

As, 1 mole of HCO_3^- react to give 1 mole of CO_2

So, 0.0202 mole of HCO_3^- react to give 0.0202 mole of CO_2

The moles of CO_2 = 0.0202 mole

Now we have to calculate the volume of CO_2 by using ideal gas equation.

PV=nRT

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 37^oC=273+37=310K

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)

V=0.51411L=514.11ml

Therefore, the volume of CO_2 will be, 514.11 ml

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Answer:

0.366 moles to the nearest thousandth.

Explanation:

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3 years ago
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