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2 years ago

What happens when The vapour obtained by dropping conc. H2SO4 in a mixture of KI and MnO2 is treated with hypo solution

Chemistry

1 answer:

Shalnov [3]2 years ago

6 0

Iodine is decolorized.

The first reaction stated in the question occurs as follows;

2 KI (aq) + 2 H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (s) + 2 H2O (l)

The reaction here is the formation of iodine from MnO2 and KI in the presence of dropwise H2SO4.

Hypo is the common name of sodium thio-sulphate or sodium hypo-sulfite.

The equation of the titration reaction is;

2Na2S2O3 + I2→ Na2S4O6 + 2NaI

When this reaction takes place, iodine is decolorized due to its reduction to I^-.

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What is the length of a one-dimensional box for an electron (9.109 x 10-31 kg) with an n=1 energy of 479 kJ/mol? Give the answer

OverLord2011 [107]

Answer : The length of a one-dimensional box for an electron is $2.819\times 10^{31}\AA$

Explanation :

The energy level of quantum particle in a one-dimensional box is given as:

$E_n=\frac{n^2h^2}{8mL^2}$

where,

$E_n$ = 479 kJ/mol = 479000 J/mol

n = energy level = 1

h = Planck's constant = $6.626\times 10^{-34}Js$

m = mass of electron = $9.109\times 10^{-31}kg$

L = length of a one-dimensional box = ?

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

$479000J/mol=\frac{(1)^2\times (6.626\times 10^{-34}Js)^2}{8\times (9.109\times 10^{-31}kg)\times L^2}$

$L=2.819\times 10^{21}m$

conversion used : $1m=10^{10}\AA$

$L=2.819\times 10^{31}\AA$

Therefore, the length of a one-dimensional box for an electron is $2.819\times 10^{31}\AA$

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2 years ago

What happens when The vapour obtained by dropping conc. H2SO4 in a mixture of KI and MnO2 is treated with hypo solution​

What happens when The vapour obtained by dropping conc. H2SO4 in a mixture of KI and MnO2 is treated with hypo solution