At 298K, particular redox reaction that involves an exchange of 2 electrons has
ID: 998390 • Letter: A
Question
At 298K, particular redox reaction that involves an exchange of 2 electrons has a E of +0.100 Volts. What is the equilibrum constant of this reaction? Enter your answer as either a regular number or in scientific notation to 3 significant figures using E to represent "*10^". For example, Avogadro's number would be written as 6.02E23
The alpha decay of what isotope of what element produces lead-206?
mercury-202
bismuth-208
radon-210
polonium-210
radon-222
thallium-204
What happens to the mass number and the atomic number of an element when it undergoes ?- decay?
The mass number does not change and the atomic number increases by 1.
The mass number does not change and the atomic number decreases by 1.
The mass number does not change and the atomic number decreases by 2.
The mass number decreases by 4 and the atomic number decreases by 2.
The mass number increases by 2 and the atomic number increases by 1.
mercury-202
bismuth-208
radon-210
polonium-210
radon-222
thallium-204
What happens to the mass number and the atomic number of an element when it undergoes ?- decay?
The mass number does not change and the atomic number increases by 1.
The mass number does not change and the atomic number decreases by 1.
The mass number does not change and the atomic number decreases by 2.
The mass number decreases by 4 and the atomic number decreases by 2.
The mass number increases by 2 and the atomic number increases by 1.
Neither the mass number nor the atomic number changes.
At 298K, particular redox reaction that involves an exchange of 2 electrons has a E of +0.100 Volts. What is the equilibrum constant of this reaction? Enter your answer as either a regular number or in scientific notation to 3 significant figures using E to represent10 For example, Avogadro's number would be written as 6.02E23 QUE S TION 11 The alpha decay of what isotope of what element produces lead-206? O mercury-202 O bismuth-208 O radon-210 O polonium-210 O radon-222 radon-222 O thallium-204 QUESTION 12 What happens to the mass number and the atomic number of an element when it undergoes decay? The mass number does not change and the atomic number increases by 1 O The mass number does not change and the atomic number decreases by 1 The mass number does not change and the atomic number decreases by 2 O The mass number decreases by 4 and the atomic number decreases by 2 O The mass number increases by 2 and the atomic number increases by 1Explanation / Answer
E0 cell = RT/nF ln K , at 298 K, = 0.0592/n log K where K is the equilibrium constant.
Then, log K = E0 cell x n/ 0.0592 = 0.1 V x 2 / 0.0592 = 3.37
K= 2389.89 = 2.39E3
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