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katovenus [111]
2 years ago
9

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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Why does such a small decrease in pH mean such a large increase in acidity?
den301095 [7]

Answer:

pH is an index of how many protons, or hydrogen ions (H+) are dissolved and free in a solution. The pH scale goes from 0 to 14. A fluid with a pH of 7 is neutral. Below 7, it is acidic; above 7, it is alkaline.

The more below or above 7 a solution is, the more acidic or alkaline it is. The scale is not linear—a drop from pH 8.2 to 8.1 indicates a 30 percent increase in acidity, or concentration of hydrogen ions; a drop from 8.1 to 7.9 indicates a 150 percent increase in acidity. Bottom line: Small-sounding changes in ocean pH are actually quite large and definitely in the direction of becoming less alkaline, which is the same as becoming more acidic.

If you think about it, we use descriptive words like this all the time. A person who stands 5’5” tall and weighs 300 pounds isn’t thin. If he loses 100 pounds, he still won’t be thin, but he will be thinner than he was before he went on the diet. (And we are more likely to comment that he’s looking trimmer than to say he’s not as fat as he used to be.)

7 0
2 years ago
an element has an isotope with a mass of 203.973 amu and and abundance of 1.40%. another isotope has a mass of 205.9745 amu with
ryzh [129]

Answer is: the average atomic mass 217.606 amu.

Ar₁= 203.973 amu; the average atomic mass of isotope.

Ar₂ = 205.9745 amu.

Ar₃ = 206.9745 amu.

Ar₄ = 207.9766 amu.

ω₁ = 1.40% = 0.014; mass percentage of isotope.

ω₂ = 24.10% = 0.241.

ω₃ = 22.10% = 0.221.

ω₄ = 57.40% = 0.574.

Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.  

Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.

Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.

Ar = 217.606 amu.

But abundance of isotopes is greater than 100%.

It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.

7 0
3 years ago
What is the formula of a compound formed between nitrogen (N) and potassium (K)?
Rina8888 [55]
<span>You need to consider the valences of the two elements. Potassium is +1; nitrogen is -3. To balance the molecule, you need 3 potassium to one nitrogen, or K3N</span>
5 0
3 years ago
Read 2 more answers
Hello<br>Thank you for not answering my question
julsineya [31]
**Visible confusion** no problem
4 0
2 years ago
Estradiol is a female sexual hormone that causes maturation and maintenance of the female reproductive system. Elemental analysi
Ipatiy [6.2K]

Answer:

The molecular formula of estradiol is: C_{18}H_{24}O_2.

Explanation:

Molar mass of of estradiol = M= 272.37 g/mol

Let the molecular formula of estradiol be C_xH_yO_z

Percentage of an element in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100

Percentage of carbon in estradiol  :

\frac{x\times 12 g/mol}{272.37 g/mol}\times 100=79.37\%

x = 18.0

Percentage of hydrogen in estradiol  :

\frac{y\times 1g/mol}{272.37 g/mol}\times 100=8.88\%

y = 24.2 ≈ 24

Percentage of oxygen in estradiol  :

\frac{z\times 16 g/mol}{272.37 g/mol}\times 100=11.75\%

z = 2

The molecular formula of estradiol is: C_xH_yO_z:C_{18}H_{24}O_2

7 0
2 years ago
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