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3 years ago

9

Based on solubility, boiling point, and spectroscopic results, a chemist identified an unknown compound as an alcohol. Yet when

she tried to confirm this conclusion using a chromic acid test, she got a negative result. All of her results were correct. How is that possible? Briefly explain.

1 answer:

mojhsa [17]3 years ago

4 0

Answer:

The compound was a 3° alcohol.

Explanation:

The chromic acid test is positive for 1° and 2° alcohols, but negative for 3° alcohols.

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My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

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

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

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

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

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

7 0

2 years ago

Which of the following statements about the combustion of glucose with oxygen to form water and carbon dioxide (C6H12O6 + 6 O2 →

KATRIN_1 [288]

Answer:

what are your statements?

3 0

2 years ago

you produce 115 grams of water in a combustion reaction. how much methane do you need in grams for this to react?

lora16 [44]

Answer:about 2 grams of methane

Explanation:

idk just got this right

6 0

2 years ago

The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?

vazorg [7]

C + O2= CO2
$n = \frac{m}{mw}$
$n = \frac{5.4}{12} \\ n = 0.45 \: mol \: of \: carbon$
$n = \frac{13.6}{12 + 16 \times 2} \\ n = \frac{13.6}{44} \\ n = 0.31 \: mol \: of \: carbon \: dioxide$
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %

4 0

2 years ago

How many moles of H2O2 are needed to react with 1.07 moles of N2H4?

I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄ + 2H₂O₂ → N₂ + 4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

N₂H₄ : H₂O₂

1 : 2

1.07 : 2×1.07 = 2.14 mol

6 0

2 years ago