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mr Goodwill [35]
3 years ago
6

(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

e place and before equilibrium is established), was 5.56. On the following graph, plot the points representing the initial concentrations of all three gases. Label each point with the formula of the gas.

Chemistry
1 answer:
Taya2010 [7]3 years ago
7 0

Answer:

Here's what I get  

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

                   H₂ +    I₂    ⇌ 2HI

I/mol·L⁻¹:    0.30   0.15         x

2. Calculate the concentration of HI

Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} =  5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}

3. Plot the initial points

The graph below shows the initial concentrations plotted on the vertical axis.

 

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Answer:

The molarity is 8.52M

Explanation:

molality = moles of solute/kg of solvent = 5.2mol/kg

moles of solute = 5.2mole

mass of solute = moles of solute × MW = 5.2 × 56.106 = 291.8g

mass of solvent = 1kg = 1000g

mass of solution = mass of solute + mass of solvent = 291.8g + 1000g = 1291.8g

Density of KOH = 2120g/L

Volume of solution = mass/density = 1291.8/2120 = 0.61L

Molarity = number of moles of solute/volume of solution = 5.2/0.61 = 8.52M

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A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:
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<u>Answer:</u> The cell voltage of the given chemical reaction is 2.74 V

<u>Explanation:</u>

For the given chemical reaction:

Sn^{2+}(aq.)+Ba(s)\rightarrow Ba^{2+}(aq.)+Sn(s)

Here, gold is getting reduced because it is gaining electrons and nickel is getting oxidized because it is loosing electrons.

<u>Oxidation half reaction:</u>  Ba(s)\rightarrow Ba^{2+}(aq.)+2e^-;E^o_{Ba^{2+}/Ba}=-2.90V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s);E^o_{Sn^{2+}/Sn}=-0.14V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=-0.14-(-2.90)=2.76V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]}{[Ba^{2+}]}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +2.76 V

n = number of electrons exchanged = 2

[Ba^{2+}]=5.15M

[Sn^{2+}]=1.59M

Putting values in above equation, we get:

E_{cell}=2.76-\frac{0.059}{2}\times \log(\frac{1.59}{5.15})\\\\E_{cell}=2.74V

Hence, the cell voltage of the given chemical reaction is 2.74 V

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