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3 years ago

A sample of paraffin wax undergoes the following change: C25H52(s) → C25H52(l)

Chemistry

1 answer:

nalin [4]3 years ago

5 0

Answer:

Explanation:

(a) Answer: Intermolecular forces

The reason for this answer is because the substance (paraffin wax) only changed it's state from solid to liquid and didn't undergo a breakage in it's covalent bond within it's carbon chain which would have produced another substance.

(b) Solid substances are generally more dense than there corresponding liquid substances because the more compact particles are (which occurs in solids), the more dense they become. They are thus more dense than liquids because liquids have there particles loosely packed and well spaced making them less dense than there corresponding solids. Hence, the solid paraffin wax was going to become less dense because it's particles moved from being tightly packed (as solids) to being loosely packed (as liquids). Density refers to mass per volume but can also be described as the level of compactness of a substance. Thus, since liquid is not as compact as solid, it can be said to be less dense than solids.

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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

A chemist has a sample of hydrated Li2SiF6 and it weighs 0.4813 grams. He heats it strongly to drive off the water of hydration,

Aleksandr-060686 [28]

Answer:

percentage of water = 18.76%

Explanation:

A chemist has a sample of hydrated Li2SiF6 and it weighs 0.4813 grams.

The overall weight of the compound is 0.4813 grams.

weight of hydrated sample = 0.4813 grams.

weight of anhydrous compound = 0.391 grams

percentage weight of water = mass of water/mass of the hydrated compound × 100

mass of water = mass of hydrated compound - mass of anhydrous compound

mass of water = 0.4813 - 0.391 = 0.0903 grams.

percentage of water = 0.0903/0.4813 × 100

percentage of water = 9.03/0.4813

percentage of water = 18.761687097

percentage of water = 18.76%

6 0

4 years ago

Read 2 more answers

How many moles are in 55 g or NaCk

seraphim [82]

Answer:

Explanation:

Ok, so I suppose you meant NaCl. But ok:

So we have the formula moles = grams/MM, where MM is molar mass. The molar mass of NaCl is 58.44 g/mol. That means that 55/58.44 is the number of moles which appears to be 0.94 moles.

4 0

3 years ago

If elasticity increases, does the speed of sound increase?

shtirl [24]

Sound increases my the sound waves

4 0

3 years ago

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:answer question number 10

Vesna [10]

Answer:

the last one