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3 years ago

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A criminalist uses a fluorescent chemical called rhodamine, cyanoacrylate fumes, and a uv light source to see a fingerprint on a

gun. Which type of evidence is he detecting?
Gross evidence
latent evidence
direct evidence
trace evidence

1 answer:

Mumz [18]3 years ago

8 0

The answer is latent evidence. Evidence that can't be seen by the naked eye. These are hidden pieces of evidence.This evidence needs special tools, chemicals or materials to exposed them. Such items like cyanoacrylate fumes and UV light source are used to see the finger print on a gun.

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Explain why potassium atom would not bond with a calcium atom.

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3 years ago

Read 2 more answers

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

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Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

$Total~volume=50mL+10mL=60mL$

$60mL=0.06L$

Now we can calculate the concentration:

$M=\frac{0.00423mol}{0.06L}$

$M=0.0705$

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

$pOH=-Log(0.0705)$

$pOH=1.15$

Now we can <u>calculate the pH value</u>:

$14=~pH~+~pOH$

$pH=14-1.15=12.84$

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3 years ago

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Answer:

D. 5 frogs/sq meter

Explanation:

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3 years ago