This problem is about conversion and dimensional analysis. Important information to know:
1 atm = 760 torr = 101.325 kPa
For atm to torr conversion:
0.875 atm * (760 torr / 1 atm) = 665 torr
For atm to kPa conversion:
0.875 atm * (101.325 kPa / atm) = 88.7 kPa
Thus the answer is b) 665 torr and 88.7 kPa
Answer:
1.03
Explanation:
You would take 0.00103 and move the decimal like this; 0001.03, we wouldn't have the zeroes in front of the one, as it can throw us off. Therefore, your answer would be 1.03.
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The mass of ore required is
21 700 t.
r = 750 cm
V =

=

= 1.767 × 10⁹ cm³
The density of lead is 11.34 g/cm³.
So mass of lead sphere = 1.767 × 10⁹ cm³ ×

= 2.004 ×10¹⁰ g
2.004 ×10¹⁰ g ×

= 2.004 × 10⁷ kg
2.004 × 10⁷ kg ×

= 2.004 × 10⁴ t
92.5% efficiency means 92.5 t Pb per 100 t of ore.
Mass of ore = 2.004 × 10⁴ t Pb ×

= 2.17 × 10⁴ t ore = 21 700 t ore