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4 years ago

6. The oxidation number of Mn in KMnO4 is A) +8. B) +7. C) +5. D) -7. E) -8.

Chemistry

2 answers:

QveST [7]4 years ago

7 0

The oxidation number of Mn in $KMnO_4$ is +7.

Explanation:

In a compound the sum of the oxidation states of all the atoms will be to equal the total charge on the ions.

$KMnO_4$ is neutral hence its total charge is zero

Oxidation number of potassium+ Oxidation number of Manganese+Oxidation number of Oxygen=0

We Know that

Oxidation number of potassium= +1

Oxidation number of Oxygen= -2

Substituting the values,

(+1)+ Oxidation number of Manganese+ 4(-2)=0

(+1)+ Oxidation number of Manganese+ (-8)=0

(-7)+ Oxidation number of Manganese=0

Oxidation number of Manganese= +7

Pani-rosa [81]4 years ago

6 0

Answer:

$kmno4 = 0\\ k(potassium) = + 1 \\ mn(manganese) = x\\ o(oxygen) = - 2 \\ \\ (+ 1) + x + ( - 2)4 = 0 \\ 1 + x - 8 = 0 \\ x = + 8 - 1 \\ x = + 7$

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Question 34 (1 point)

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Answer:

8.33 atm

Explanation:

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5/9 * 15 atm = 8.33 atm

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2 years ago

Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC

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<u>Answer:</u> The $\Delta G$ for the reaction is 54.6 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm$

Putting values in above equation, we get:

$K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 22.92

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.6 kJ/mol

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3 years ago

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3 years ago

7. What volume of a 0.10 mol/l HCl solution is needed to neutralize 10 ml of a 0.15

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Answer: The volume of a 0.10 mol/l HCl solution needed to neutralize 10 ml of a 0.15 mol/l LiOH solution is 15 ml

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

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