Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
If the lead can be extracted with 92.5 efficiency, the mass of ore is required to make a lead sphere with a 7.00 cm radius is 17.707 kg.
To find the mass of ore, the given values are,
The radius of the lead sphere is r = 7.0 cm,
The lead can be extracted with, η = 92.5% = 0.925 efficiency.
What is the Density?
Density is the substance's mass per unit of volume.
We know the density of lead as,
⇒ p = 11.4 g/cm³ .
The volume of the lead sphere is,
V = 4/3πr³
= 4/3π(7)³
V = 1436.7 cm³.
Mass of the lead present in the lead sphere is,
m = p . V
= 11.4 × 1436.7
= 16379 g
m = 16.379 kg
From the expression of efficiency, we calculate the mass of the ore,
η = Mass of lead obtained / Mass of the ore to be taken
0.925 = 16.379 kg / Mass of the ore to be taken
Mass of the ore to be taken = 17.707 kg
If the lead can be extracted with 92.5 efficiency, the mass of ore is required to make a lead sphere with a 7.00 cm radius is 17.707 kg.
Learn more about Lead extraction,
https://brainly.com/question/10982537
#SPJ4
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.