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Tom [10]
3 years ago
10

The rock in a lead ore deposit contains 84% PbS by mass. How many kilograms of the rock must be processed to obtain 2.3 kg of Pb

?
kg
Chemistry
2 answers:
Alex Ar [27]3 years ago
7 0

Answer: 3.162 kg


Explanation:


1) Mass percent = (mass of a part / total mass) × 100


2) 84% PbS by mass.


Percent PbS = ( mass of pure PbS / mass of the rock) × 100 = 84%


⇒ mass of the rock = mass of pure PbS × 100 / 84 =


⇒ mass of the rock = mass of pure PbS / 0.84 ← this is fhe formula that you will uise at the end.


2) Determine the mass of pure PbS that contains 2.3 kg of Pb


i) Unit formula: PbS


ii) molar mass of PbS: 207.2 g/mol + 32.065 g/mol = 239.265 g/mol


iii) proportion:


207.2 g of Pb / 239.265 g of PbS = 2,300 g of Pb / x


⇒ x = 2,300 × 239.265 g of PbS / 207.2 = 2,655.93 g of PbS = 2.656 Kg of PbS


4) Use the formula to determine how many kilograms of the rock contain 2.565 kg of pure PbS


mass of the rock = mass of pure PbS / 0.84 = 2.565 / 0.84 = 3.162 kg

lesya [120]3 years ago
6 0
Well take 2.3 kg and divide it by .84 to get 2.738 kg then divide the percent of S in the compounds by adding the atomic masses and dividing the atomic mass of S by the atomic mass of the molecule to get 13.40% S by mass
Then you know that Pb is 86.6% by mass so divide 2.738 by .866 to get your answer. Your answer is 3.162 kg of rock is needed
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Answer: Step 1: Calculate qsur (the surrounding is

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qsur = ? J

m = 75.0 g water

c = 4.184 J/g

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ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

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c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

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Step 4: Calculate ΔH ΔH = q sys .

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ΔH= + 2949.72 J

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ΔH= +17179.81 J/mol or +1.72 x 104

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i hope this helps

5 0
3 years ago
Read 2 more answers
For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car
Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • Ca(NO₃)₂: 1 mole
  • CaCO₃: 1 mole
  • NaNO₃: 2 mole

Being the molar mass of the compounds:

  • Na₂CO₃: 106 g/mole
  • Ca(NO₃)₂: 164 g/mole  
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  • NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
  • Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
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You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of  CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of  CaCO₃?

mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

6 0
3 years ago
What is the mass, in grams, of 1.4 mol W
algol [13]
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3 years ago
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belka [17]

Answer:

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We know that volume=4/3×πr³

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k=R/Na

R=k×Na

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R is real gas constant

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PV=NkT

P=NkT/V

Since there is only one molecule of oxygen so N=1

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8 0
3 years ago
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GalinKa [24]
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