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2 years ago

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Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with a ba

se in the form of an antacid. Magnesium hydroxide is a common active ingredient in antacids. As a government chemist testing commercial antacids, you use 0.107 M HCl to simulate the acid concentration in the stomach. How many milliliters of this "stomach acid" will react with a tablet containing 0.252 g of magnesium hydroxide? Enter to 2 decimal places.

1 answer:

skelet666 [1.2K]2 years ago

5 0

Answer:

.

Explanation:

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An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

$Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)$

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

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

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

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

where,

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

$E_{cell}$ = 0.507 V

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

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

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

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

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

where,

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

$E_{cell}$ = 0.507 V

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

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

6 0

3 years ago

How much heat is absorbed in the complete reaction of 3.00 grams of SiO2 with excess carbon in the reaction SiO2(g) + 3C(s) → Si

defon

Answer:

31.24 kJ

Explanation:

SiO₂(g) + 3C(s) → SiC(s) + 2CO(g) ΔH° = 624.7 kJ/mol

First we <u>convert 3.00 grams of SiO₂ to moles</u>, using its <em>molar mass</em>:

3.00 g SiO₂ ÷ 60.08 g/mol = 0.05 mol

Now we <u>calculate the heat absorbed</u>, using the <em>given ΔH°</em>:

If the complete reaction of 1 mol of SiO₂ absorbs 624.7 kJ, then with 0.05 mol:

0.05 mol * 624.7 kJ/mol = 31.24 kJ of heat would be absorbed.

6 0

2 years ago

What is the third alkali metal

andrey2020 [161]

Answer:

potassium

The third alkali metal is K (potassium). The atomic number of K (potassium) is 19. Thus, the atomic number of third alkali metal is 19

Explanation:

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2 years ago

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Differentbetween sound and vibration

Fiesta28 [93]

Answer:

Both are similar concepts.

Sound is the vibration of air particles (compression and expansion) the can reach your ears. But you can have vibration being propagated in liquids and solids as well.

Some sounds are generated in structures, so the vibration of a structure is converted to sound in air — for instance, a loudspeaker.

Explanation:

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3 years ago

Does sound travel faster in a warm room or a cold room? Explain your answer.

RUDIKE [14]

Answer:

Explanation:

Sound travels outwards from the source in all directions. So there you have it sound does travel faster in warm air BUT it may appear to travel farther in cold air. This is how that works……if the air close to the ground is colder than the air above it then sound waves travelling upwards will be bent downwards.

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3 years ago

Read 2 more answers