Answer:
Excess Reagent = oxygen
Explanation:
Limiting reagent: The substance that is totally consumed when the reaction is completed.
Excess reagent: The substance left after the limiting reagent is consumed completely
The balanced chemical equation for formation of water is as follow:
This means when 2 moles of hydrogen reacts with 1 mole of oxygen, 2 moles of water is produced.
Hence the ratio in which hydrogen and oxygen gas reacts is 2:1
Now if 2 mole hydrogen require 1 mole of oxygen ,then 4 mole hydrogen need 2 mole of oxygen.
or
Here 5 mole of oxygen is reacting but only 2 mole is required .
Oxygen is in excess.
Answer:
I definitely think he mostly observed that it was clear in color.
Explanation:
• Aerobic respiration, photosynthesis
It is aerobic because it requires oxygen to be carried out. Anaerobic on the other hand does not require oxygen.
It’s two products would be the reactants for photosynthesis because in photosynthesis the reverse reaction takes place as glucose is produced along with water from carbon dioxide (CO2) and water (H2O) in the presence of chlorophyll and sunlight.
The answer is 100%
let me know if you want an explanation
Answer:
Explanation:
<u>1) Rate law, at a given temperature:</u>
- Since all the data are obtained at the same temperature, the equilibrium constant is the same.
- Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:
r = K [A]ᵃ [B]ᵇ
<u>2) Use the data from the table</u>
- Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s
Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0
- Use the first and second set of data to find the exponent a:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s
Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]
2ᵃ = 2 ⇒ a = 1
<u>3) Write the rate law</u>
This means, that the rate is independent of reactant B and is of first order respect reactant A.
<u>4) Use any set of data to find K</u>
With the first set of data
- r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹
Result: the rate constant is K = 0.167 s⁻¹
