Answer:
waste gas he should use algae
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
1. I think it is true?
2. Low melting points
3. True
4. Atomic number, I think it’s periods?
5. Groups?
Sorry, I might not get all of them right :(
Hope this helps you in any way!!
The patient needs 1000 ml of 5% (w/v) glucose solution
i.e 1000 ml x 5 g/ 100 ml
where the stock solution is 55% (w/v) = 55 g / 100 ml
So, 1000 ml x 5 g / 100 ml = V (ml) x 55 g / 100 ml
V = 1000 x (5 / 100) / (55 / 100) = 5000 / 55 = 90.9 ml
∴ the patient needs 90.9 ml of 55% (w/v) glucose solution
