Answer:
It is mentioned that the student is mixing chemicals A and B and observes the time taken for the color to change. However, in the experiment, it is noticed that the student has repeated the procedure five times and each time he or she is modifying the concentration of chemical B. Thus, it is clear that the concentration of chemical B is the independent variable in the experiment. An independent variable is illustrated as the variable, which is controlled or modified in the experiment.
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
<span>If I done the math correctly it is 3729J because you multiply 16.5 g by the 2260 J/g and get 3729 J</span>
Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g
Answer:
B
Explanation:
The tendency of a liquid to convert to vapour increases smoothly with increasing temperature. Vapour pressure shows the tendency of a liquid to convert to vapour. Increase In vapour pressure shows an increased tendency to convert to vapour. The higher the temperature, the higher the vapour pressure.
