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Alborosie
3 years ago
13

Janice is given a mixture of alcohol and water. The teacher tells her that she can use temperature to separate these compounds.

What property is being demonstrated by this experiment.
A) conduction
B) density
C) solubility
D) boiling
Chemistry
2 answers:
Korolek [52]3 years ago
6 0
Since Janice was given a mixture of alcohol and water, her teacher suggested that she use temperature to separate the two substances. The property demonstrated by the experiment is D. boiling. The boiling point refers to the temperature at which the liquid phase of the substance will turn into vapor. Water is known to boil at 100°C at atmospheric pressure while alcohols are generally known to have a boiling point lower than that of water. In this experiment, knowing that the two substances had a significant difference in boiling temperature was crucial to be able to separate them into their pure substances.
4vir4ik [10]3 years ago
5 0

Answer: D) boiling

Explanation:

In order to separate a mixture of alcohol and water the mixture can be heated until the alcohol boils. The boiling point of alcohol is lower than water. The alcohol will boil off leaving the water.

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One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867
Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

So  

Mass = Density × Volume

= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene

1C_6 H_5 CH_3  + 9 O_2  > 7 CO_2  + 4 H_2 O

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of CO_2 gas and 4 moles of H_2 O Vapour

So the mole ratio is 1 : 11

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $

(C)

1mole contains 6.022\times10^{23} molecules

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $

6 0
3 years ago
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A double bond between carbon atoms is longer than a triple bond between carbon atoms.

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3 years ago
What has 39 neutrons which element
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No of Neutrons =Mass No-No of protons

  • Proton=Neutron+1

\\ \sf\longmapsto x+1=39

\\ \sf\longmapsto x=39-1=38

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The element is Stroncium(Sr)

6 0
2 years ago
Ethanol (CH3CH2OH), the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuri
Fed [463]

Answer:

a) 88.48%

b) 0.05625 mol

Explanation:

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CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g)                        Reaction 2

a) CH₃CH₂OH = 46.0684 g/mol

   CH₃CH₂OCH₂CH₃ = 74.12 g/mol

1 mol CH₃CH₂OH ______  46.0684 g

x                            ______   50.0 g

x = 1.085 mol  CH₃CH₂OH

1 mol  CH₃CH₂OCH₂CH₃ ______  74.12 g g

y                           ______   35.9 g

y = 0.48 mol   CH₃CH₂OCH₂CH₃

100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃

w                _____  0.48 mol CH₃CH₂OCH₂CH₃

w = 88.48%

b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.

4 0
3 years ago
What does the position of an element in the periodic table tell you?
loris [4]
They are in order according to their atomic number, so the position should tell you the atomic number. 
3 0
3 years ago
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