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3 years ago

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What change occurs during the reaction:MnO4- --->Mn2+?Five electrons are lost.Three electrons are lost.Five electrons are gai

ned.Three electrons are gained.

1 answer:

tatiyna3 years ago

3 0

Answer:

Five electrons are gained.

Explanation:

Oxidation number or oxidation state of an atom in a chemical compound is the number of electrons lost or gained. It is also defined as the degree of oxidation of the atom in the compound.

This is a theoretical number which can help to decipher the oxidation and reduction in a redox reaction.

Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.

The given reaction is shown below as:

$MnO_4^-\rightarrow Mn^{2+}$

Manganese in $MnO_4^-$ has oxidation state of +7

Manganese in $Mn^{2+}$ has an oxidation state of +2

<u>It reduces from +7 to +2 . It means that 5 electrons are gained.</u>

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3 years ago

For the reaction, A(g) + B(g) => AB(g), the rate is 0.385 mol/L.s when the initial concentrations of both A and B are 2.00 mo

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Answer : The rate for a reaction will be $0.14Ms^{-1}$

Explanation :

The balanced equations will be:

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$\text{Rate}=k[A]^2[B]^1$

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$\text{Rate}=k[A]^2[B]$

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Now we have to calculate the initial rate for a reaction that starts with 1.48 M of reagent A and 1.32 M of reagents B.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(0.0481)\times (1.48)^2(1.32)^1$

$\text{Rate}=0.14Ms^{-1}$

Therefore, the rate for a reaction will be $0.14Ms^{-1}$

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4 years ago

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erastova [34]

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3 years ago

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Answer:

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4 years ago

A 18.08-g sample of the ionic compound , where is the anion of a weak acid, was dissolved in enough water to make 116.0 mL of so

Oksi-84 [34.3K]

Answer:

a) 129.14 g/mol

b) 8.87

Explanation:

Given that:

mass of the ionic compound [NaA] = 18.08 g

Volume of water = 116.0 mL = 0.116 L

Let the mole of the acid HCl = 0.140 M

Volume of the acid = 500.0 mL = 0.500 L

pH = 4.63

$V_{equivalence}_{acid}$ = 1.00 L

Equation for the reaction can be represented as:

$NaA_{(aq)} + HCl_{(aq)} -----> HA_{(aq)} + NaCl_{(aq)$

From above; 1 mole of an ionic compound reacts with 1 mole of an acid to reach equivalence point = 0.140 M × 1.00 L

= 0.140 mol

Thus, 0.140 mol of HCl neutralize 0.140 mol of ionic compound at equilibrium

Thus, the molar mass of the sample = $\frac{18.08g}{0.140 mole}$

= 129.14 g/mol

b) since pH = pKa

Then pKa of HA = 4.63

Ka = $10^{-4.63]$

= $2.3*10^{-5}$

$[A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}$

$= \frac{0.140 mol}{1.116 L}$

= 0.1255 M

$K_a$ of HA = $2.3*10^{-5}$

$K_b = \frac{1.0*10^{-14}}{2.3*10^{-5}}$

$= 4.35*10^{-10}$

$A_{(aq)}$ + $H_2O_{(l)}$ $\rightleftharpoons$ $HA_{(aq)}$ + $OH^-_{(aq)}$

Initial 0.1255 0 0

Change - x + x + x

Equilibrium 0.1255 - x x x

$K_b = \frac{[HA][OH^-]}{[A^-]}$

$4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}$

As $K_b$ is very small, (o.1255 - x) = 0.1255

$x = \sqrt{0.1255*4.35*10^{-10}}$

[OH⁻] = $x = 7.4 *10^{-6}$

But pOH = - log [OH⁻]

= - log [ $7.4*10^{-6}$ ]

= 5.13

pH = 14.00 = 5.13

pH = 8.87

6 0

3 years ago