What they have in common is that they both have the same number of atoms. Hope this helps. :)
When you are collecting DNA, you could be looking for a few different things. A few examples could be skin cells, strands of hair, or possibly even a fingernail. Anything that comes from a person, including blood or saliva can be potential DNA that could help investigators to link a person back to a crime.
Investigators do not need a warrant for analyzing crime scenes due to the fact of the dangers of the fire. You must work quickly because accelerants tend to evaporate within days, sometimes hours. It is also important to note that finding the origin of the fire is very important, to make sure it won't be reignited. Debris is usually cleaned away quickly to ensure health and safety issues.
The point of origin of a fire is the lowest point, since fire burns upwards.
High explosive: Ignite almost instantly, like dynamite and TNT. Two different types are primary and secondary.
<em>Primary: easily ignited, very sensitive to heat and friction. often used to ignite other explosives. </em>
<em>Secondary: much less sensitive to heat and friction, might be ignited using other explosive materials. TNT and dynamite are both secondary. </em>
Low explosive: decompose slowly and include black and smokeless powder. They are the most common type of explosives, and are readily available.
Answer:
1.58x10⁻⁵
2.51x10⁻⁸
0.0126
63.10
Explanation:
Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:
pH = pKa + log[In-]/[HIn]
pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,
i) pH = 4.9
4.9 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = - 4.8
[In-]/[HIn] = 
[In-]/[HIn] = 1.58x10⁻⁵
ii) pH = 2.1
2.1 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -7.6
[In-]/[HIn] = 
[In-]/[HIn] = 2.51x10⁻⁸
iii) pH = 7.8
7.8 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -1.9
[In-]/[HIn] = 
[In-]/[HIn] = 0.0126
iv) pH = 11.5
11.5 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = 1.8
[In-]/[HIn] = 
[In-]/[HIn] = 63.10
