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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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How are changes between phases consistent with the law of conservation of matter
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4 years ago
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What is the molar solubility of marble (i.e., [ca2 ] in a saturated solution in normal rainwater, for which ph=5.60? express you
Brut [27]
Missing in your question :

Ksp of(CaCO3)= 4.5 x 10 -9

Ka1 for (H2CO3) =  4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

1) equation 1 for Ksp = 4.5 x 10^-9 

CaCO3(s)→ Ca +2(aq)    +  CO3-2(aq)  

2) equation 2 for Ka1 = 4.7 x 10^-7

 H2CO3 + H2O → HCO3- + H3O+

3) equation 3 for Ka2 = 5.6 x 10^-11

 HCO3-(aq) + H2O(l) → CO3-2 (aq)  + H3O+(aq)

so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s)  +  H+(aq)  → Ca2+(aq)   + HCO3-(aq)

note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :

CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l)  Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq)  +  CO3-2(aq)   Ksp = 4.5 x 10^-9

∴ the overall equation will be as we have mentioned before:
when H3O+ = H+

CaCO3(s) + H+(aq)  ↔ Ca2+ (aq) + HCO3-(aq)   K= 80.55

from the overall equation:

∴K = [Ca2+][HCO3-] / [H+]

when we have [Ca2+] = [HCO3-] so we can assume both = X

∴K = X^2 / [H+]

when we have the PH = 5.6 so we can get [H+]

PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6

so, by substitution on K expression:

∴ 80.55 = X^2 / (2.5 x10^-6)

∴X = 0.0142

∴[Ca2+] = X = 0.0142 
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