Carbohydrates. If you think about it, it's a mix of the three words.
Answer:
52.45g
Explanation:
The computation of the mass of pure acetic acid in 125mL of this solution is shown below:
The percentage of mass would be equivalent to the g of solute in each 100g of water
As we know that
density = mass ÷ volume
So,
Volume = mass ÷ density
V = 100g / 1.049 (g / ml)
V = 95.328 mL
Now In every 95,328 ml of C_2H_4O_2 there are 40g of C_2H_4O_2
i.e.
each 125ml of C_2H_4O_2 there are 52.45g
SO,
x = 40g. 125ml ÷ 95.328
x = 52.45g
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: calories?
Explanation: but isn’t it 1 calorie = 4.184 joules of energy
Presumptive tests, also known as preliminary tests or field tests, allow drugs to be quickly classified into a particular chemical group, but do not unequivocally identify the presence of a specific chemical compound.
