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Bond [772]
3 years ago
7

To determine the mass of CO2 that forms from a given mass of CaCO3, which of the following must be known? Check all that apply.

Chemistry
2 answers:
Pavlova-9 [17]3 years ago
7 0

Answer:

1 3 and 5

Explanation:

ZanzabumX [31]3 years ago
3 0

To determine the mass of CO₂, the following must be known :

  • the molar mass of CaCO₃
  • the mole ratio of CaCO₃ to CO₂
  • the molar mass of CO₂
<h3>Further explanation</h3>

Reaction

Decomposition of CaCO₃

CaCO₃ ⇒ CaO + CO₂

Given the mass of  CaCO₃, so to determine the mass of CO₂ :

1. Find the mol of  CaCO₃ from the molar mass of  CaCO₃

\tt n_{CaCO_3}=\dfrac{mass~CaCO_3}{MW~CaCO_3}

2. Find the mole ratio of  CaCO₃ : CO₂(from equation = 1 : 1)

\tt n_{CaCO_3}\div n_{CO_2}=1\div 1

3. Find the mass of CO₂ from the molar mass of CO₂

\tt mass_{CO_2}=n_{CO_2}\times MW_{CO_2}

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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:

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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

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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)}

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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

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