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Lubov Fominskaja [6]
3 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]3 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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Explanation:

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3 years ago
The zinc in a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (so th
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Answer:

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3 years ago
⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqu
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Answer:

pH=12.3\\\\pOH=1.7\\

[H^+]=5x10^{-13}M

[OH^-]=0.02M

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)

Thus, since the ionization occurs at an extent of 1/3, we can set  up the following relationship:

\frac{1}{3} =\frac{x}{[Mg(OH)_2]}

Thus, x for this problem is:

x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x=  0.01M

Now, according to an ICE table, we have that:

[OH^-]=2x=2*0.01M=0.02M

Therefore, we can calculate the H^+, pH and pOH now:

[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M

pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7

Best regards!

4 0
3 years ago
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[email protected]@dow

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3 years ago
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Are alcohols with more than one hydroxyl group attached to a carbon sequence.
Darya [45]

Answer:

Polyhydroxyl alcohols

Explanation:

Whenever we have several C-OH bonds, we have a polyhydroxyl alcohol. For example, if we have just one alcohol group, that is, an R-OH group, then the naming is simple, say, we have EtOH, it's ethanol.

The problem becomes more complicated when we have several hydroxyl groups present in the alcohol. Let's say we have an ethane molecule and we replace the hydrogen atoms of carbon 1 and 2 with hydroxyl groups. In that case, we have 1,2-ethanediol. Similarly, we can have triols etc.

That said, we have poly (several) hydroxyl groups and we can generalize this to having polyhydroxyl alcohols.

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