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Yogita Ingle 6 years, 3 months ago
Step 1: Preparation of sulphur dioxide:
SO2 is prepared by burning sulphur in the presence of excess air so that the product combines with oxygen which is helpful for the next stage.
S(s) + O2 (g) → SO2(g)
Step 2: Preparation of sulphur trioxide:
Sulphur trioxide is formed when sulphur dioxide reacts with oxygen in a ratio of 1:1 at a temperature of 400°C – 450°C and a pressure of 1-2 atm in the presence of N2O5 as a catalyst. This reaction is reversible in nature.
2SO2(g) + O2(g) ⇌ 2SO3(g)
Step 3: Preparation of concentrated sulphuric acid:
The sulphur trioxide formed is first made to react with concentrated sulphuric acid. Sulphur trioxide cannot be dissolved in water directly as it leads to the formation of fog. The product obtained after this reaction is known as oleum. The oleum obtained is then dissolved in water to obtain concentrated sulphuric acid.
H2SO4 + SO3(g) → H2S2O7(l)
H2S2O7(l) + H2O(l) → 2H2SO4
Posted by Shruti Kamal 6 years, 3 months ago
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Yogita Ingle 6 years, 3 months ago
CO2 is linear because carbon is sp-hybridised. CO2 is linear because carbon is sp-hybridised.
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Khuskaran Sidhu 6 years, 3 months ago
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..... ... 6 years, 3 months ago
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Yogita Ingle 6 years, 3 months ago
Cannizzaro reaction: Aldehydes having no α-hydrogen atoms undergo self oxidation and reduction (disproportionation) reactions on treatment with a concentrated alkali. In such reactions, one molecule of aldehyde gets oxidised to form an acid and the other molecule of aldehyde gets reduced to form an alcohol.
For example, two molecules of formaldehyde, in the presence of concentrated NaOH, produce methanol and sodium formate.
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Yogita Ingle 6 years, 3 months ago
or C = Л/RT
Л = 7.2 atm
R = 0.0821 L atm K-1 mol-1
T = 37°C = 37 + 273 = 310 K.
Molar concentration (C) = (7.2 atm)/(0.082L Latm K-1mol-1 ) x (310 K)
= 0.283 mol-1
= 0.283 M. Read more on Sarthaks.com - https://www.sarthaks.com/240928/molar-concentration-solute-particles-human-blood-osmotic-pressure-normal-temperature</div>
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Yogita Ingle 6 years, 3 months ago
Oxidation States of lanthanoids
- The elements belonging to lanthanide series exhibit an oxidation state of +3. For example, Praseodymium (Pr) shows +3 oxidation state.
- Some elements exhibit +2 oxidation states in their complexes in solutions. For example, Samarium (Sm), Europium (Eu), Thulium (Tm) and Ytterbium (Yb) show +2 oxidation state.
- Some elements exhibit +4 oxidation states due to high stability of empty, half-filled or fully filled f-subshells.
- Praseodymium (Pr), Neodymium (Nd), Terbium (Tb) and Dysprosium (Dy) exhibits +4 oxidation state in their oxides.
- Cerium (Ce) shows both +3 as well as +4 state.
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Khuskaran Sidhu 6 years, 3 months ago
2Thank You