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3 years ago

Identify the neutralization reaction.

Chemistry

1 answer:

olchik [2.2K]3 years ago

3 0

Answer:

3Ba(OH)₂ + 2H₃PO₄ → Ba₃(PO₄)₂ + 6H₂O

Explanation:

Neutralization reaction:

When an acid and base react they form water and salt.The reaction is called neutralization reaction.

General equation:

HX + BOH → BX + H₂O

From given reaction equations only option C is correct because only this reaction gives salt and water.

Chemical equation:

3Ba(OH)₂ + 2H₃PO₄ → Ba₃(PO₄)₂ + 6H₂O

Barium hydroxide is acting as a base. H₃PO₄ is an acid. Both these reactants react to form salt which is barium phosphate and water.

Other options are incorrect because non of these reaction produced salt and water. So option A is correct.

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A solution is made by mixing equal masses of methanol, CH 4 O , and ethanol, C 2 H 6 O . Determine the mole fraction of each com

Fynjy0 [20]

Answer:

Mole fraction of $CH_4O$ = 0.58

Mole fraction of $C_2H_6O$ = 0.42

Explanation:

Let the mass of $CH_4O$ and $C_2H_6O$ = x g

Molar mass of $CH_4O$ = 33.035 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles_{CH_4O}= \frac{x\ g}{33.035\ g/mol}$

$Moles_{CH_4O}=\frac{x}{33.035}\ mol$

Molar mass of $C_2H_6O$ = 46.07 g/mol

Thus,

$Moles= \frac{x\ g}{46.07\ g/mol}$

$Moles_{C_2H_6O}=\frac{x}{46.07}\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ CH_4O=\frac {n_{CH_4O}}{n_{CH_4O}+n_{C_2H_6O}}$

$Mole\ fraction\ of\ CH_4O=\frac{\frac{x}{33.035}}{\frac{x}{33.035}+\frac{x}{46.07}}=0.58$

Mole fraction of $C_2H_6O$ = 1 - 0.58 = 0.42

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3 years ago

The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when

tia_tia [17]

Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

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

The half-cell reactions are:

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,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = $25^oC=273+25=298K$

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

$E^o_{cell}$ = standard electrode potential of the cell = +0.63 V

$E_{cell}$ = cell potential for the reaction = ?

$[Zn^{2+}]$ = 3.5 M

$[Pb^{2+}]$ = $2.0\times 10^{-4}M$

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

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

$E_{cell}=0.50V$

Therefore, the cell potential for this reaction is 0.50 V

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timama [110]

Answer:

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