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2 years ago

Why is gunshot residue important for a forensic scientist? What are the potential complications with this type of evidence?

Chemistry

1 answer:

sergij07 [2.7K]2 years ago

3 0

Answer:

Gunshot residue is important to forensic scientists because it can help measure the firing distance of a weapon, or if someone has fired a weapon lately based on what's on their hands. The potential complications of this evidence include it being uncertain if the residue is from another firing. This type of evidence can also be uncertain in a courtroom when firing distance is drawn into question.

Explanation:

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Maria wants to determine which type of disinfectant kills the most bacteria.

murzikaleks [220]

Answer:

well she can test both of the soap by putting one on and plate and another on the other plate and which ever is cleaner is your answer

6 0

3 years ago

Read 2 more answers

Determine the electron-group arrangement, molecular shape, and ideal bond angle(s) for each of the following:

Bad White [126]

The electron group arrangement of NO²⁻is trigonal planar. The molecular shape is bent, and the bond angle is 120°.

<h3>What is the molecular shape of a compound?</h3>

The molecular geometry of the compound shows the position of nuclei and the electron of the compound. It shows how the joining of electrons and nuclei makes the shape of the compound.

Like here, the shape of nitrite is bent with lone pair which is shown by Lewis's structure The bond angle will be the distance between the nuclei of the neighbor atoms.

Thus, the electron geometry arrangement of nitrite is trigonal planer with a bent shape and the bond angle will be 120°.

To learn more about molecular geometry, refer to the link:

brainly.com/question/7558603

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1 year ago

In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to

labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of $NH_3$ = 100 g

Molar mass of $NH_3$ = 27 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate moles of $NH_3$ .

$\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles$

The given balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the given reaction, we conclude that

2 moles of $NH_3$ produced from 1 mole of $N_2$

3.7 moles of $NH_3$ produced from $\frac{1mole}{2mole}\times 3.7mole=1.85moles$ of $N_2$

Now we have to calculate the mass of $N_2$ .

Mass of $N_2$ = Moles of $N_2$ × Molar mass of $N_2$

Mass of $N_2$ = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

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2 years ago

No matter where you are on Earth, the center of the Earth is always _______.

FinnZ [79.3K]

Answer:

D) straight down

Hope this helps!

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2 years ago

Read 2 more answers

If 97.3 L of NO2 forms measured at 35 C and 632 mm Hg. What is the percent yield?

enyata [817]

<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.

b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.

</span>and % will be 60%.

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2 years ago