Chemistry: Principles and Reactions
8th Edition
ISBN: 9781305079373
Author: William L. Masterton, Cecile N. Hurley
Publisher: Cengage Learning
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Chapter 10, Problem 52QAP
Interpretation Introduction
Interpretation:
The molar mass of the enzyme needs to be determined, if the mass of enzyme is 7.89 g, volume of aqueous solution is 0.856 L and the osmotic pressure of the solution is 1.14 mm Hg at
Concept Introduction :
Osmotic pressure is a colligative property, and it its expression is as follows:
Where
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An 90 kg subject consumed a snack that contained 80 g of carbohydrate. Before eating the snack the subject’s blood glucose was 6.0 mmol.L-1. Calculate the concentration of glucose (mM) that would be reached in the blood after consuming the snack.Assume that all the carbohydrate in the snack was converted to glucose, rapidly absorbed and distributed throughout the extracellular water (ECW) within 30 minutes. Assume ECW is approximately 20% of body weight and the glucose is not metabolised further.
Consider the following equilibria:
2SO3(g) 2SO2(g) + O2(g) Kc1 = 2.3 × 10 – 7
2NO3(g) ⟷ 2NO2(g) + O2(g) Kc2 = 1.4 ×10 – 3
Calculate the equilibrium constant for the reaction:
SO2(g) + NO3(g) ⟷ SO3(g) + NO2(g)
A
1.3 × 10 – 2
B
1.6 × 10 – 4
C
78
D
6.1 × 103
3) Based on the information below, determine the value of the unknown equilibrium constant, Kc.
4 CO2 (g) + 2 H2O (g) « 4 O2 (g) + 2 CH2CO (g)Kc = 3.7 x 1017
CO2 (g) + 2 H2O (g) « CH4 (g) + 2 O2 (g)Kc = 8.3 x 10-15
CH4 (g) + CO2 (g) « CH2CO (g) + H2O (g)Kc = ?
Chapter 10 Solutions
Chemistry: Principles and Reactions
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