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Rama09 [41]
3 years ago
5

Caculate the number of moles in 1.70g of NH3

Chemistry
1 answer:
lutik1710 [3]3 years ago
5 0
Moles of NH3 = mass of NH3/molecular mass of NH3 = 1.7/17= O.1 MOLES. 
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For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell
frutty [35]

Answer:

\large\boxed{\text{0.512 V}}

Explanation:

We must use the Nernst equation

E = E^{\circ} - \dfrac{RT}{zF}lnQ

The equation for the cell reaction is is

2Cl⁻(0.384 mol·L⁻¹) + 2Co³⁺(0.324 mol·L⁻¹) ⇌ Cl₂(5.80 atm) + 2Co²⁺(0.158 mol/L)

Data:

E° = 0.483 V

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Calculation:  

T = 25 + 273.15 = 298.15 K

Q = \dfrac{\text{[Cl}^{-}]^{2}[\text{Co}^{3+}]^{2}}{p_{\text{Cl}_{2}}^{2}\text{[Co}^{3+}]^{2}} = \dfrac{0.384^{2} \times 0.324^{2}}{5.80 \times 0.158^{2}} =0.1069\\\\E = 0.483 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(0.1069)\\\\=0.483 -0.01285 \times (-2.236) = 0.483 + 0.02872 = \textbf{0.512 V}\\\text{The cell potential is } \large\boxed{\textbf{0.512 V}}

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