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kifflom [539]
3 years ago
9

How do you balance redox equations in acidic solutions?

Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
7 0

Answer:

First, balance the half-reactions

Second, equalize the electrons

Third,add two reaction equations to get final answer

Explanation:

For example

H₂C₂0₄ + MnO⁻₄ ---------->CO₂+Mn²⁺

(i) Balancing the half reactions

H₂C₂O₄-------->2CO₂+2H⁺+2e⁻

5e⁻ +8H⁺+MnO₄⁻----------->Mn²⁺+4H₂O

(ii)

Equalizing the electrons

5H₂C₂O₄--------->10CO₂+10H⁺+10e⁻  ---here there is a factor of 5

10e⁻+16H⁺+2MnO₄⁻--------->2Mn²⁺+8H₂O -----here there is a factor of 2

(iii)

Add the two where electrons and some Hydrogen ions will cancel out

5H₂C₂O₄+6H⁺+2MnO₄⁻---->10CO₂+2Mn²⁺+8H₂O

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When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot
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Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the 'n^{th}' orbit of hydrogen atom.

Formula used :

E_n=-13.6\times \frac{Z^2}{n^2}ev

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number  of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV

Energy of n = 2 in an hydrogen atom:

E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0
3 years ago
What volume would 56.2 mL of gas at 820 mm of Hg occupy at 720 mm of Hg?
Andreyy89

Answer:

49.35  mL

Explanation:

Given: 56.2 mL of gas

To find: volume that 56.2 mL of gas at 820 mm of Hg would occupy at 720 mm of Hg

Solution:

At 820 mm of Hg, volume of gas is 56.2 mL

At 1 mm of Hg, volume of gas is \frac{56.2}{820}

At 720 mm of Hg, volume of gas is \frac{56.2}{820}(720)=49.35\,\,mL

3 0
3 years ago
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