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3 years ago

True or false carboxylic acids are weak acids

1 answer:

lisabon 2012 [21]3 years ago

4 0

If we add 1 mole of acetic acid (carboxylic acid), then we will not obtain 1 mole of hydrogen ions or 1 mole of ions in the solution. This shows that carboxylic acid does not dissociate completely in water. Hence, carboxylic acid is a weaker acid than the inorganic acids. Hence, the given statement is true.

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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

When you cut your knee what part of the blood helps the blood stop flowing?

IgorC [24]

Answer:

the red blood cells

Explanation:

white blood cells fight off infection

4 0

4 years ago

Read 2 more answers

Type the correct answer in each box.

krok68 [10]

Answer:

2KI → 2K + I₂

Explanation:

(not my definition, but form a trustworthy source)

The definition of a decomposition reaction is:

a chemical reaction in which one reactant breaks down into two or more products.

In this case, potassium iodide will decompose into potassium and iodine.

2KI → 2K + 2I

2 Potassium and 2 Iodine on both sides, so equation is balanced.

Or, it can be written as 2KI → 2K + I₂

8 0

2 years ago

Which pair of compounds has the same empirical formula? c3h8 and c3h c2h2 and c2h c4h10 and c6h c4h10 and c2h5.

BartSMP [9]

Answer:

The correct answer is the final pair: C4H10 and C2H5

Explanation:

Took the test and it was right. :)

8 0

3 years ago

Which word goes we’re ?

Lady_Fox [76]

1. Adjustment knob
2.arm
3. Diaphram
4. Base
5. Light source
6. Tube
7. Stage clips
8. Objectives
9. revolving nose piece
10. eye piece
11. Stage

4 0

3 years ago