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3 years ago

10

If some the hexane had evaporated from the stearic acid solution, what effect (if any) would it have on the experimental value o

f Avogadro’s number? Why?

1 answer:

Vanyuwa [196]3 years ago

4 0

Answer:

If some of the stearic-acid hexane solution had evaporated before using it, this would increase Avogadro's number, or the number of atoms per mole. This is because with less hexane in the solution (because only the hexane evaporates), the grams of stearic acid used would be smaller.

Explanation:

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Suppose a phosphorus atom forms a bond with a fluorine atom. What type of bond must between these two elements?

Liula [17]

Answer:

Covalent bond

Explanation:

they will make a covalent bond with each other because both are non- metals and have a low electronegativity difference.

6 0

3 years ago

In the liquid and solid states, molecules are held together by attractions called intermolecular forces. there are several types

BaLLatris [955]

<span>Since these molecules are all non-polar, the only intermolecular force of attraction will be London dispersion forces. Since these increase by the size of the molecule, the boiling points will decrease in the same order: Parafin > Heptadecane > hexane > 2,2-dimethylbutane > propane For these two, hexane > 2,2-dimethylbutane, dispersion forces are greater in a molecule which is longer and unbranched compared to one which is branched and more compact.</span>

3 0

3 years ago

What are the primary chemical components present in a phosphate buffer at ph 7.4? H3po4 and po43–?

fenix001 [56]

Answer :

The correct answer for primary component of phosphate buffer at pH = 7.4 is H₂PO₄⁻ and HPO₄²⁻ .

<u>Buffer solution :</u>

It is a solution of mixture of weak acid and its conjugate base OR weak base and its conjugate acid . It resist any change in solution when small amount of strong acid or base is added .

<u>Capacity of a good buffer : </u>

A good buffer is identified when pH = pKa .

From Hasselbalch - Henderson equation which is as follows :

$pH = pka + log \frac{[A^-]}{[HA]}$

If [A⁻] = [HA] ,

pH = pka + log 1

pH = pKa

This determines that if concentration of weak acid and its conjugate base are changed in small quantity , the capacity of buffer to maintain a constant pH is greatest at pka . If the amount of [A⁻] or [HA] is changed in large amount , the log value deviates more than +/- 1M and hence pH .

Hence Buffer has best capacity at pH = pka .

<u>Phosphate Buffer : </u>

Phosphate may have three types of acid-base pairs at different pka ( shown in image ).

Since the question is asking the pH = 7.4

At pH = 7.4 , the best phosphate buffer will have pka near to 7.4 .

If image is checked the acid - base pair " H₂PO₄⁻ and HPO₄²⁻ has pka 7.2 which is near to pH = 7.4 .

Hence we can say , the primary chemical component of phosphate buffer at pH = 7.4 is H₂PO₄⁻ and HPO₄²⁻ .

8 0

3 years ago

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

3 years ago

What is the pH of a solution within a solution with pH = 4.50? [H +] = 3.25×10-6 M?

Lisa [10]

Answer: 5.48

Explanation:

pH is the negative logarithm of hydrogen ion concentration in a solution.

Mathematically, pH = - log(H+)

where H+ represent the concentration of hydrogen ion

So, to get the pH of the solution with [H +] = 3.25×10-6 M:

Apply, pH = -log(H+)

pH = - log (3.25×10-6 M)

pH = - ( -5.48)

(Note that the minus signs will cancel out each other)

Therefore pH = 5.48

Now we know that the pH of the solution with hydrogen ion concentration of 3.25×10-6 M is 5.48 (i.e slightly acidic)

Thus, we can finally say 5.48 is the pH of the solution within a solution with pH = 4.50

6 0

3 years ago