Consider the following data concerning the equation: H2O2 + 3I– + 2H+ → I3– + 2H2O [H2O2] [I–] [H+] rate I 0.100 M 5.00 × 10–4 M
1.00 × 10–2 M 0.137 M/sec II. 0.100 M 1.00 × 10–3 M 1.00 × 10–2 M 0.268 M/sec III. 0.200 M 1.00 × 10–3 M 1.00 × 10–2 M 0.542 M/sec IV. 0.400 M 1.00 × 10–3 M 2.00 × 10–2 M 1.084 M/sec Two mechanisms are proposed: I. II. Which mechanism and which step as the rate determining step would best fit the data?
1 answer:
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Explanation:
The volumetric flow rate of water will be as follows.
q =
= 0.0378
Diameter =
= 0.2032 m
Relation between area and diameter is as follows.
A =
=
= 0.785 x 0.2032 x 0.2032
= 0.0324
Also, q = A × V
or, V =
=
= 1.166 m/s
As, viscosity of water = 1 cP = Pa-s
Density of water = 1000
Therefore, we will calculate Reynolds number as follows.
Reynolds number =
=
= 236931.2
Hence, the flow will be turbulent in nature.
Thus, we can conclude that the Reynolds number is 236931.2 and flow is turbulent.
The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea =
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)
Molarity of the urea solution ;
Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL
(1 g = 0.001 kg)
Molality of the urea solution ;
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.