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
Answer:
The value of the partial pressure of the oxygen 
= 690 torr
Explanation:
Total pressure of the mixture of gases = 736 torr
The partial pressure of water vapor = 46 torr
From the law of pressure we know that
Total pressure = The partial pressure of water vapor + The partial pressure of oxygen 
Put the values of pressures in above equation we get,
⇒ 736 = 46 +
⇒ = 736 - 46
⇒ = 690 torr
This is the value of the partial pressure of the oxygen.
<u>Given:</u>
Initial velocity (v1) = 0 m/s
Final velocity (v2) = 30 m/s
Acceleration (a) = 6.1 m/s2
<u>To determine:</u>
The time (t) taken to reach the final speed
<u>Explanation:</u>
Use the relation:
Acceleration (a) = [final velocity(v2) - initial velocity (v1)]/time (t)
t = (v2-v1)/a = 30-0/6.1 = 4.92 s
Ans: Time taken is around 4.9 s
Answer:
umm ok lol thx for the f r e e points
