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konstantin123 [22]
3 years ago
5

If an ore is 32.4% ti by mass, what is the mass of ore that contains 200. g of ti?

Chemistry
1 answer:
Fed [463]3 years ago
6 0
Correct Answer: Mass of ore that contains 200 g of Ti is 617.28 g.

Reason:
Given: Ore contains 32.4% Ti by mass. 
It implies that, 100 g of ore ≡ 32.4 g of Ti
Therefore,              x g of ore   ≡ 200 g of Ti

Thus, x = \frac{100X200}{32.4} = 617.28 g

Hence, mass of ore that contains 200 g of Ti is 617.28 g.
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aleksandrvk [35]

Remark

The question with these kind of problems is "Which R do you use?" That's where dimensional analysis is so handy. You must look at the units of the givens and choose your R accordingly. You'll see how that works in a moment.

You need to list the givens along with their units and in this case the property you want to solve for. You need all that to determine the R value

Givens

n = 0.25 moles

T = 35°C = 35 + 273.15 = 308.15°K

V = 6.23 L

Pressure = P in kPa

Which R

The units of the R you want has to have units of moles, kPa, °K and liters

The R that  you want is 8.314

<em><u>Formula</u></em>

PV = nRT

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because the number of moles has only 2 sig digs.

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Second Question

You are going to have to clean up the numbers. I think I've got only 1 chance at this. The partial pressures of the 2 gases will add up to the total pressure. So the total pressure was 100 approx and the water vapor was 3.36 kPa. The difference is

Total = air + water vapor

100.18 = air + 3.36 about  Subtract 3.36 from both sides.

100.18 - 3.36 = 96.82 about. Pick the answer that is closest to that. I'll clean up the numbers if I can.

Answer  C

6 0
3 years ago
Read 2 more answers
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