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2 years ago

How many atoms are found in the products of the following chemical equation?

Chemistry

1 answer:

coldgirl [10]2 years ago

5 0

There are 13 atoms in the product

<h3>Further explanation</h3>

Given

Reaction

H2SO4 + 2KOH --> K2SO4 + 2H2O

Required

The number of atoms

Solution

In a chemical equation, there are reactants on the left and products on the right

Reactants : H2SO4 + 2KOH

Products : K2SO4 + 2H2O

The number of atoms is determined by their reaction coefficient and the subscript of the atoms in the compound

K2SO4 (coefficient = 1) :

K = 2 atoms

S = 1 atom

O = 4 atoms

Total atoms = 7 atoms

2H2O(coefficient = 2) :

H = 2 x 2 = 4 atoms

O = 2 x 1 = 2 atoms

Total atoms = 6 atoms

Total = 13 atoms

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Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

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Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

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:

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

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

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