This is what I got. Hope it helps :)
if a lead can be extrated with 92.5% efficiency,what is the mass of ore required to make a lead sphere with 750%cm radius
1 answer:
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
r = 750 cm
V =
The density of lead is 11.34 g/cm³.
So mass of lead sphere = 1.767 × 10⁹ cm³ ×
2.004 ×10¹⁰ g ×
2.004 × 10⁷ kg ×
92.5% efficiency means 92.5 t Pb per 100 t of ore.
Mass of ore = 2.004 × 10⁴ t Pb ×
You might be interested in
Answer:
C₃H₅O₂
4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O
Explanation:
The reaction can be expressed as:
CₓHₓOₓ + nO₂ → CO₂ + H₂O
Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:
All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:
Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:
0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O
In order to determine the empirical formula, we calculate the moles of each component:
- mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
- mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
- mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O
Then we divide those values by the lowest one:
0.0236 mol C ÷ 0.0141 = 1.67
0.0354 mol H ÷ 0.0141 = 2.51
0.0141 mol O ÷ 0.0141 = 1
If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.
- The reaction is:
4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O
Answer:
A beaker
Step-by-step explanation:
Specifically, I would use a 250 mL graduated beaker.
A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.
You don't need precisely 100 mL solution.
If the beaker is graduated, you can easily measure 100 mL of the stock solution.
Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).
