Sulphate ion is used but fornyour kind information there ain't no place
Critical Thinking Questions
1. Why do you think forensic scientists are so careful that the tests they do are sensitive, reproducible, and specific? What might happen if they were less careful about this?
They have to be careful to ensure as much accuracy as possible.
2.Which type of evidence do you think is most useful in an investigation? Why?
Physical evidence would probably be most important because it is the best way to connect someone directly with that crime.
3.Why do you think that forensic scientists continue to look for class characteristics given their limitations?
Class characteristics are good in court because it provides details of different aspects of the crime.
Answer:
A
Explanation:
All of the other options are ways that could help the environment from the chemicals, but if they just continue to use the chemicals then it would hurt the environment
Answer:
C. By being honest in recording the results of their own experiments
Explanation:
Honesty in recording results is ethical behaviour and the basis of scientific research.
A. is wrong. If you repeat the same experiment, you should always get the same result.
B. is wrong. Copying someone else's results is cheating and certainly unethical behaviour.
D. is wrong. Making up data is dishonest and unethical behaviour.
E. is wrong. Following your own instincts about safety procedures is unethical because your instincts may be wrong. following them may result in injury to yourself or to others.
Answer:
0.2193 μm
Explanation:
The reaction showing the Photodissociation of ozone (O3) is given below as:
O₃ + hv --------------------------> O₂ + O⁺
H° (142.9) (0) (438kJ/mol).
The first thing to do here is to determine the change in the enthalpy of the total reaction, this can be done by subtracting the change in the enthalpy of the reactant from the change in enthalpy in the product. Hence, we have:
ΔH° = [438 kJ/mol + 247.5 kJ/mol] - (142.9) = 542.6 KJ/mol.
This value, that is 542.6 KJ/mol will then be used in the determination of the value for the maximum wavelength that could cause this photodissociation.
Therefore, the maximum wavelength could cause this photodissociation ≤ h × c/ E = [ 1.199 × 10⁻⁴]/ 542.6 = 2.193 × 10⁻⁷ = 0.2193 μm
