I believe it is b. you only want to change one thing at a time so you know which one thing caused the effect
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
Answer:
Selenium
Explanation:
selenium has 4 energy levels (think a 4 story building) while oxygen only has 2 (think 2 story building). which one is taller?
Answer:
Explained below
Explanation:
Metalloids are elements that possess properties between that of metals and non - metals or that have properties that a re a combination of those of metals and non - metals. Whereas a metal is one that conducts electricity and heat very well.
Now, the property they both share is that they both possess valence orbitals which are highly de-localized over macroscopic volumes, thereby allowing both of them to be conductors of electricity although it has to be said that Metalloids don't conduct electricity as much as metals.
Answer:
The heat of combustion for the unknown hydrocarbon is -29.87 kJ/mol
Explanation:
Heat capacity of the bomb calorimeter = C = 1.229 kJ/°C
Change in temperature of the bomb calorimeter = ΔT = 2.19°C
Heat absorbed by bomb calorimeter = Q


Moles of hydrocarbon burned in calorimeter = 0.0901 mol
Heat released on combustion = Q' = -Q = -2,692 kJ
The heat of combustion for the unknown hydrocarbon :
