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3 years ago

A teacher makes the following statement.

Chemistry

2 answers:

Natali [406]3 years ago

6 0

Answer:

The correct answer is option a.

Explanation:

According the question, boiling points of different liquids are discussed which means that if we take all these liquids in single mixture we can separate them from each other on the basis of their difference in boiling point. And the technique which employs this principle is distillation process.

Hence, we can say teacher is most likely talking about distillation of a mixture of oils.

Distillation is a process which is used to separate mixture of different liquids from each other. In this process, during the course of boiling of mixture liquid compound with lower boiling points vaporizes first leaving behind the liquid with higher boiling point behind.

And vapors of boiled liquid are condensed down in a separate container for the collection.

DedPeter [7]3 years ago

5 0

I believe the correct answer from the choices listed above is option A. The topic that the teacher is talking about would be distillation of a mixture. Gasoline is processed by distillation. Hope this answers the question. Have a nice day.

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A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn

nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_{[Ni^{2+}/Ni]}=-0.23V$

$E^0_{[Zn^{2+}/Zn]}=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^{2+}+2e^-$ $E^0_{[Zn^{2+}/Zn]}=-0.76V$

Reaction at cathode (reduction) : $Ni^{2+}+2e^-\rightarrow Ni$ $E^0_{[Ni^{2+}/Ni]}=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}$

$E_{cell}=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M.

New concentration of $Ni^{2+}$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}$

$E_{cell}=0.52V$

(c) Now we have to calculate the concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}$

Now put all the given values in the above equation, we get:

$0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}$

$x=1.49M$

The concentration of $Ni^{2+}$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0

4 years ago

Please answer these for the brainliest

k0ka [10]

Magnetic attraction and evaporation are the answers

8 0

3 years ago

Read 2 more answers

Pat paid $85.50 for renting a scooter for 6 hours. What was the rate per hour for renting the scooter? (Input only numeric value

wlad13 [49]

I believe the answer would be 14.25 an hour because $85.50/ 6 would be 14.25 but i could be wrong.

3 0

3 years ago

Convert 6.35 grams of aluminum sulfate to moles

vampirchik [111]

Answer:

There are 0.0186 moles of formula units in 6.35 grams of aluminum sulfate $\rm Al_2(SO_4)_3$ .

Explanation:

What's the empirical formula of aluminum sulfate?

Sulfate is an anion with a charge of -2 per ion. When sulfate ions are bonded to metals, the compound is likely ionic.

Aluminum is a group III metal. Its ions tend to carry a charge of +3 per ion.

The empirical formula of an ionic compound shall balance the charge on ions with as few ions as possible.

The least common multiple of 2 and 3 is 6. That is:

Three sulfate ions $\rm {SO_4}^{2-}$ will give a charge of -6.
Two aluminum ions $\rm Al^{3+}$ will give a charge of +6.

Pairing three $\rm {SO_4}^{2-}$ ions with two $\rm Al^{3+}$ will balance the charge. Hence the empirical formula: $\rm Al_2(SO_4)_3$ .

What's the mass of one mole of aluminum sulfate? In other words, what's the formula mass of $\rm Al_2(SO_4)_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Al: 26.982;
S: 32.06;
O: 15.999.

There are

two Al,
three S, and
twelve O

in one formula unit of $\rm Al_2(SO_4)_3$ .

Hence the formula mass of $\rm Al_2(SO_4)_3$ :

$\underbrace{2\times 26.982}_{\rm Al} + \underbrace{3\times 32.06}_{\rm S} + \underbrace{12\times 15.999}_{\rm O} = \rm 342.132\;g\cdot mol^{-1}$ .

How many moles of formula units in 6.35 grams of $\rm Al_2(SO_4)_3$ ?

$\displaystyle n = \frac{m}{M} = \rm \frac{6.35\;g}{342.132\;g\cdot mol^{-1}} = 0.0186\;mol$ .

8 0

3 years ago

A block of iron has a mass of 826 g. What is the volume of the block of iron whose density at 25°C is 7.9 ?

omeli [17]

Answer:

104.56cm³

Explanation:

Mass=826g

Density =7.9g/cm³

Volume =Mass /Density

Volume =826/7.9

Volume =104.56cm³

4 0

3 years ago