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dlinn [17]
3 years ago
7

A forensic scientist reports that a dead woman had high concentrations of a pain-fighting drug in her blood. What type of analys

is was done? A) the scientist made an association through individualized evidence. B) The scientist made a comparison. C) The scientist made an identification. D) The scientist supplied direct evidence
Chemistry
2 answers:
zepelin [54]3 years ago
7 0

Answer: C). The scientist made an identification.

Explanation:

Forensic science deals with the searching, identification, collection and analysis of the evidences found at the crime scene or those which are related to the crime.

In an analysis of evidences the identification, quantification, and comparison of the evidences obtained from the crime scene with those which are requested from the suspects or victims takes place.

In the given situation, the scientist will identify the concentration of drug in the blood after the process of analysis of the blood.

AlladinOne [14]3 years ago
3 0
C) the scientist made an identification by identifying the amount of drug in her blood and realising it was high
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When sodium and oxygen react to form an ionic compound, the mass of the compound ____ the sum of the masses of the individual el
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What is a combination of two or more substances that are all solids
Sophie [7]

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Solids cant react to each other because they are the same room temperature

7 0
3 years ago
If an ideal gas has a pressure of 5.49 atm, a temperature of 88.78 °c, and has a volume of 22.03 l, how many moles of gas are in
kirza4 [7]
N = (PV)/RT
(T = 88.78 + 273 = 361.78K)
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7 0
4 years ago
At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106
lana [24]

Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>          

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M        

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant (K_{c}) for the given chemical reaction, is given by the equation:

K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}

<u><em>At the original equilibrium state:</em></u>

K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}

K_{c} = \frac {1.6641}{0.002332} = 713.59

<u><em>Therefore, at the new equilibrium state:</em></u>

K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}

\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}

\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88

\Rightarrow [HI] = \sqrt {12.88} = 3.589 M

<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>

6 0
3 years ago
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