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castortr0y [4]
3 years ago
7

Jamie rolls a 6-sided die 30 times and determines that the experimental probability of rolling a 2 is Start Fraction 1 over 15 E

nd Fraction. The theoretical probability of rolling a 2 is Start Fraction 1 over 6 End Fraction. What could Jamie do to make his experimental results more closely match the theoretical probability?
Chemistry
1 answer:
anyanavicka [17]3 years ago
8 0

Answer:

Conduct more trials

Explanation:

Theoretical Probability can be defined as what someone is expecting to happen

Experimental Probability on the other hand, is defined as what actually happens.

Probability is usually calculated in the same way for experimental probability and that of theoretical probability. You divide the total number of possible ways in which a particular outcome can happen, by the total number of outcomes itself.

In Experimental probability, the more times a probability is tried, it gets closer and even more closer to theoretical probability.

So, for the question, Jamie should improve the number of tries more, so as to get his experimental probability results to be closer to the theoretical probability result.

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Explanation:

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There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
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Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ...[1]

Step 2 :

6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ..[2]

Adding 6 × [1] and [2]:

6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)

6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)

we get :

6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?

\Delta H'=6\times \Delta H_1+\Delta H_2

\Delta H'=6\times 414.0 kJ+132.0kJ

\Delta H'=2,626 kJ

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

\frac{ 2,626 kJ}{5 } = 523.2 kJ

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