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Lubov Fominskaja [6]
2 years ago
12

Multiple-choice questions have a special grading rule determined by your instructor. Assume that your instructor has decided to

grade these questions in the following way: If you submit an incorrect answer to a multiple-choice question with nnn options, you will lose 1/(n−1)1/(n−1) of the credit for that question. Just like the similar multiple-choice penalty on most standardized tests, this rule is necessary to prevent random guessing.If a multiple-choice question has five answer choices and you submit one wrong answer before getting the question correct, how much credit will you lose for that part of the question?
a. 100%
b. 50%
c. 33%
d. 25%
e. 20%
Chemistry
1 answer:
leva [86]2 years ago
6 0

Answer:

The correct answer is option d.

Explanation:

On submission of an incorrect answer to a multiple-choice question with n options,  we will lose:

=\frac{ 1}{(n-1)} of credit

If we multiple-choice question has five answer choices. And we have submitted 1 wrong answer. the fraction of credit we loose :

=\frac{ 1}{(5-1)} of credit

=\frac{1}{4} of credit

=0.25 credit

The percentage of credit we loose:

=0.25\times 100 of credits

=25\% of credits

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Answer:

The mass of 3.491 × 10¹⁹ molecules of Cl₂  of Cl₂ is 4.11 × 10⁻³ grams

Explanation:

The number of particles in one mole of a substance id=s given by the Avogadro's number which is approximately 6.023 × 10²³ particles

Therefore, we have;

One mole of Cl₂ gas, which is a compound, contains 6.023 × 10²³ individual molecules of Cl₂

3.491 × 10¹⁹ molecules of Cl₂ is equivalent to (3.491 × 10¹⁹)/(6.023 × 10²³) = 5.796 × 10⁻⁵ moles of Cl₂

The mass of one mole of Cl₂ = 70.906 g/mol

The mass of 5.796 × 10⁻⁵ moles of Cl₂ = 70.906 × 5.796 × 10^(-5) = 4.11 × 10⁻³ grams

Therefore;

The mass of 3.491 × 10¹⁹ molecules of Cl₂  of Cl₂ = 4.11 × 10⁻³ grams.

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Snowcat [4.5K]

Answer:

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<h3>How to explain the reaction?</h3>

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Plasmid and the insert fragment are both present in the microfuge tube, and they both have compatible sticky ends. However, the ligase has been denatured and is no longer active because the prior student left it outside rather than freezing it; despite this, we had already put the ligase into the tube. Ligase aids in binding the plasmid and insert fragments together, but because it is denatured in this instance, it will no longer be able to do so. As a result, no transformation process will take place. And since ligase links DNA fragments together by catalyzing the development of connections between the nearby nucleotides, the two fragments will not be able to unite.

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