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chubhunter [2.5K]
3 years ago
10

During an experiment, Juan rolled a six-sided number cube 18 times. The number two occurred four times. Juan claimed the experim

ental probability of rolling a two was approximately 1/9. Which of the following is true about Juan’s claim?
A. Juan’s claim is incorrect. The correct experimental probability is 2/9.
B. Juan’s claim is incorrect. The correct experimental probability is 1/3.
C. Juan’s claim is incorrect. The correct experimental probability is 4/9.
D. Juan’s claim is correct.
PLEASE HURRY!!!!!!!!!!
God bless, stay safe, and happy holidays! :)
Chemistry
1 answer:
Karolina [17]3 years ago
6 0

Answer:

Because u would have to find the undercorse of 010-1 witch makes the out of part by 6

Explanation:

Given :

Juan rolled a six-sided number cube 18 times.

The number two occurred four times.

To Find: Juan claimed the experimental probability of rolling a two was approximately 1/9. Why is Juan’s experimental probability incorrect?

Solution:

Total events = number of times cube rolled = 18

Favorable events = The number two occurred four times.  = 4

So, Experimental probability of rolling a two was approximately 1/9

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Lechateliers principle can be used here to determine the effect  of changes observed in the system.

Lechateliers principle states that if  any reaction at equilibrium  is subjected to change in concentration, temperature and pressure or even in reaction conditions  then the equilibrium of the reaction would shift in such a way so that it can oppose the change .

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The above reaction is following:

C(s)+H₂O(g)→CO(g)+H₂(g)

The enthalpy change  of this reaction is positive and hence the reaction is endothermic in nature.

So the given changes would lead to the following  results:

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b Since H₂O(g) is in gaseous state and a reactant and hence the addition of  H₂O(g) that is more reactant would lead to more formation of hydrogen gas according to lechatelier principle. The equilibrium would shift in such a way so that it can decrease the concentration of added H₂O(g) hence it would form H₂(g).

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\boxed {\boxed {\sf 0.4 \ mol \ HCl}}

Explanation:

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