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kotykmax [81]
3 years ago
8

A student is given 1.525 g of pure CuO. To recover the Cu present in the compound, the dark powdery solid was dissolved in 15.0

mL of 6 M HCl, the solution diluted with 50.0 mL of water, and 0.50 g of Mg was added. Was this enough Mg to displace all the ions from the solution
Chemistry
1 answer:
ikadub [295]3 years ago
3 0

Answer:

The amount of Mg was enough

Explanation:

In this case, we have to start with the <u>reaction</u> between Mg and CuO, so:

CuO~+~Mg~->~MgO~+~Cu

If we check <u>the reaction is already balanced</u>. Now, we can do some stoichiometry to calculate the amount of Mg. The first step is the number of moles of CuO. To this we have to calculate the molar mass of CuO first, so:

Cu: 63.55 g/mol and O: 16 g/mol. So, (63.55+16)= 79.55 g/mol.

Now, we can calculate the moles:

1.525~g~CuO\frac{1~mol~CuO}{79.55~g~CuO}=0.0192~mol~CuO

The <u>molar ratio</u> between Mg and CuO is 1:1, so:

0.0192~mol~CuO=0.0192~mol~Mg.

Now we can <u>calculate the mass of M</u>g if we know the atomic mass of Mg (24.305 g/mol). So:

0.0192~mol~Mg\frac{24.305~g~Mg}{1~mol~Mg}=0.466~g~Mg

<u>With this in mind, the student added enough Mg to recover all the Cu.</u>

Note: The HCl doesn't take a role in the reaction. The function of HCl is to dissolve the CuO.

I hope it helps!

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Vinil7 [7]

Answer:

6+69

Explanation:

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3 years ago
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what is the concentration of hydroxide ions after 50.0 ml of 0.250 m naoh is added to 120 ml of 0.200 m na2so4? please show all
Galina-37 [17]

The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

What is meant by concentration?

Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.

Concentration of hydroxide ions can be calculated by,

M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.

where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.

Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

To learn more about concentration click on the given link brainly.com/question/17206790

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5 0
10 months ago
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insens350 [35]

Answer:

% purity of limestone = 96.53%

Explanation:

Question (4).

Weight of impure CaCO₃ = 25.9 g

Molecular weight of CaCO₃ = 40 + 12 + 3(16)

                                              = 100 g per mole

We know at S.T.P. number of moles of CO₂ = 1 and volume = 22.4 liters

From the given reaction, 1 mole of CaCO₃ reacts with 1 mole or 22.4 liters of

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∴ 1 liter of CO₂ will be produced by CaCO₃ = \frac{100}{22.4}

∴ 5.6 liters of CO₂ will be produced by CaCO₃ = \frac{100\times 5.6}{22.4}

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Therefore, % purity of CaCO₃ = \frac{\text{Weight calculated}}{{\text{Weight given}}}\times 100

                                                 = \frac{25}{25.9}\times 100

                                                 = 96.53 %

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From stoichiometry;

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And from molar mass analysis;

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