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zhannawk [14.2K]
2 years ago
13

A well-known scientist testifies that she believes a piece of evidence is accepted by science. The judge rules that the evidence

is not, in his opinion, admissible. Which case applies to the decision?
a. Daubert v. Merrell Dow Pharmaceuticals
b. Daubert v. Frye
c. Kumho Tire Co. Ltd. v. Daubert
d. Frye v. United States
Chemistry
1 answer:
notsponge [240]2 years ago
5 0

Answer:

The correct answer is C:

Kumho Tire Co. Ltd V. Daubert

Explanation:

In this case, the appellant (Daubert) has sued Kumho Tire Co. Ltd (the defendant) when 3 months prior the right rear tyre of his car exploded while the car was in motion killing a passenger and injuring others severely.

The appellant employed the services of an expert who had worked for 10 years with Michelin to give his opinion as having studied the attributes of the tire technology.  

In a counter-argument, the defendant rather than dispute the opinion of the expert urged the court to exclude it as evidence because his methodology didn't conform to the requirements of the Law as given in Rule 702 of the Federal Rules of Evidence.

True to the defendants' position, the Law by the ruling in Daubert v Merrell Dow Pharmaceuticals required that for the opinion to count, it had to be scientifically correct. It was hence excluded as evidence in the case.

Cheers!

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Convert 5.0x10^24 molecules to liters.
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Number of moles = 5 x 10^24 / 6.02 x 10^23 = 8.305 moles. Volume= moles x 22.4 = 186.032 liters. Hope this helps!
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2 years ago
Consider a process in which the entropy of a system increases by 125 J K−1 and the entropy of the surroundings decreases by 125
expeople1 [14]

Answer : The process is not spontaneous.

Explanation :

As, we know that:

Change in entropy = Change in entropy of system + Change in entropy of surrounding

As we are given in question, the entropy of surroundings decrease by the same amount as the entropy of the system increases.

For the given reaction to be spontaneous, the total change in entropy should be positive.

Given :

Entropy change of system = +125J/K

Entropy change of surroundings = -125J/K

Total change in entropy = Entropy change of system + Entropy change of surroundings

Total change in entropy = 125 J/K + (-125 J/K)

Total change in entropy = 0

The process is at equilibrium because the entropy change is equal to zero. So, the process is not spontaneous.

4 0
2 years ago
Exactly 500 grams of ice are melted at a temperature of 32°f. (lice = 333 j/g.) calculate the change in entropy (in j/k). (give
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Entropy Change is calculated  by (Energy transferred) / (Temperature in kelvin) 
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Q = (mass)(latent heat of fusion) 
Q = m(hfusion) 
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T(K) = 32 + 273.15 = 305.15K 
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3 0
2 years ago
Eddie is going to do an experiment to find out which freezes more easily—distilled water or salt water. In what order should he
mel-nik [20]

1. Make a Prediction

2. Fill both beakers with water

3. Dissolve salt in one of the beakers

4. Place both in the freezer and observe

5. Write a report

(Always make the prediction first! That's a hypothesis!)

4 0
2 years ago
Read 2 more answers
When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at
STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

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\\ 1.22 * 10^{5} = -19953.6 * ln K

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So, the equilibrium constant is 0.0022.

4 0
2 years ago
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