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2 years ago

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20cm³ of sulphuric acid reacts completely with excess zinc to form 0.1 mole of hydrogen gas. Calculate the molarity of the acid

used.

1 answer:

VMariaS [17]2 years ago

7 0

Answer:

20cm³ of sulphuric acid reacts completely with excess zinc to form 0.1 mole of hydrogen gas. Calculate the molarity of the acid used.?

Molarity= mol/volume

Molarity= 0.1/20

Molarity= 0.005mol/L

Explanation:

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1 year ago

At a certain temperature, Kc = 0.0500 and ∆H = +39.0 kJ for the reaction below, 2 MgCl2(s) + O2(g) → 2MgO(s) + Cl2(g) Calculate

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Explanation:

Since, it is shown that the reaction has been reversed. Therefore, value of $K_{c}$ will become $\frac{1}{K_{c}}$ .

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Also, the number of moles of each reactant has been halved. So, $K_{c''}$ for the reaction $MgO(s) + \frac{1}{2}Cl2(g) → MgCl_{2}(s) + \frac{1}{2} O2(g)$ will also get halved.

Therefore, $K_{c''}$ = $K_{c'}$ = $(20)^{0.5}$

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As the value of $\Delta H$ is given as +39.0 kJ. So, it means that the reaction is endothermic in nature. So, energy of reactants will be more than the products. Hence, according to Le Chatelier's principle reaction will move in the forward direction.

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The given cell reactions is:

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Oxidation half reaction (anode): $Zn\rightarrow Zn^{2+}+2e^-$

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

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

where,

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R = gas constant = 8.314 J/mol.K

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

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

$E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}$

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<u>Answer:</u>

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The given expression V1T2 = V2T1 is the formula for the Charles' Law.

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