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
the answer to your question is
M1V1=M2V2
V2=750-150=600 ml
0.75M*750 ml = M2*600
M2=0.75*750/600 ≈ 9.38 M
False
hope it helps comment if you have any questions hope you have an amazing day
Answer:
Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]
Explanation:
The equilibrium constant indicates the % of the yield reaction and can shows where the reaction is going to be equilibrated.
It works with molar concentrations on the equilibrium and it does not consider the solids compounds
Kc also can be modified by the time of the reaction.
This reaction is:
CS₂ (g) + 4 H₂O(g) ⇌ CH₄ (g) + 2H₂S (g)
Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]
