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maria [59]
3 years ago
12

A box contains a mixture of small copper spheres and small lead spheres. The total volume of both metals is measured by the disp

lacement of water to be 423 cm^3 and the total mass is 4.35 kg.
What percentage of the spheres are copper?
Chemistry
1 answer:
kirill115 [55]3 years ago
3 0

Answer:44.01% of the spheres are copper

Explanation:

mass Cu + mass Pb = 4.35kg =  4.35 x 1000 = 4350g

But

Volume= mass / density

Let the volume of copper be C  

density of  Cu = 8.94 g/cm^3

Mass of copper = 8.94C

Let the volume of Lead be L  

density of  lead = 11.34g/cm^3

Mass of lead  = 11.34L

Since  

mass Cu + mass Pb = 4350kg

therefore

8.94C + 11.34L = 4350g----- EQN A

also

volume of copper + volume of lead =  423 cm^3

Volume of Lead ,L=423-C

putting the value of C = 423-L in Equation A becomes

8.94C + 11.34L = 4350g

8.94C + 11.34(423-C)= 4350g

8.94C+ 4796.82-11.34C=4350

4796.82-4350=11.34C- 8.94

446.82=2.4C

C= 186.175

Volume of copper = 186.175cm^3

Percentage copper = volume copper / total volume metal x 100/1

= 186.175 cm^3 / 423 cm^3 x 100/1

= 44.01%

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The correct answer is Solute

Explanation:

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3 years ago
You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of
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Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

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Moles=1.60 \times {300.0\times 10^{-3}}\ moles

<u>Moles of potassium iodide = 0.48 moles </u>

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

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Moles=1.20\times {285\times 10^{-3}}\ moles

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According to the given reaction:

2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

<u>Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.</u>

Also, consumed lead(II) nitrate = 0.24 moles  (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

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<u>Statement A is correct.</u>

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2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

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Molar mass of lead(II) iodide = 461.01 g/mol

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<u>Statement B is correct.</u>

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2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

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<u>Statement C is correct.</u>

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1 mole of lead(II) nitrate  produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate  produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

<u>So, Concentration = 0.684/0.585 M = 1.169 M</u>

<u>Statement D is incorrect.</u>

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