Chemistry

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New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

All the solution have higher ion production with respect to 0.1 M C2H5OH i.e. 0.1 M Ba3 (PO4)2, 0.1 M Na2SO4, 0.1 M KCl and 0.1 M Li3PO4. Hence all have lowered freezing point than 0.1 M C2H5OH (which is non-ionisable in aqueous medium).

Ans. = 4

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Δ H f 0 = ( + Δ H s u b 0 ) + ( + Δ H I E 0 ) + ( + 1 2 Δ H B D E 0 ) + ( Δ H E A 0 ) + ( Δ H L E 0 )

Δ H L E 0 = 8 9 . 2 + 4 1 9 + 1 2 1 . 5 3 4 8 . 6 + 4 3 6 . 7 k J m o l 1

Lattice energy = 717.8

New answer posted

11 months ago

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R
Raj Pandey

Contributor-Level 9

H 2 S ( a c i d ) + H 2 O B a s e ? H 3 O + + H S

H 2 O ( a c i d ) + N H 3 ( B a s e ) ? N H 4 O H

With H2S water acts like base and with NH3 it acts like acid.

New answer posted

11 months ago

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R
Raj Pandey

Contributor-Level 9

In electro-refining, impure metal (blister copper) is used as an anode while precious metal like Au, Pt gets deposited as anode mud.

New answer posted

11 months ago

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R
Raj Pandey

Contributor-Level 9

In electro-refining, impure metal (blister copper) is used as an anode while precious metal like Au, Pt gets deposited as anode mud.

New answer posted

11 months ago

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R
Raj Pandey

Contributor-Level 9

Elemente- gain enthalpy

F-333

Cl-349

Te-190

Po-174

New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

Reactivity order of halogens and inter halogens are F2 > ClF > Cl2 and ClF forms HOCl and HF after the reaction with water.

New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

Statement-I is true, because metal reduced by decreasing Δ G in Elligham diagram but statement-II is false, Ellingham diagram is Δ G vs temperature graph has no relation with Δ S .

New answer posted

11 months ago

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A
alok kumar singh

Contributor-Level 10

23 * 3 gm of Na contained by 164 gm of Na3PO4

So, 3.45 gm of Na contained by ( 1 6 4 * 3 . 4 5 6 9 ) gm of Na3PO4

Therefore mole of Na3PO4= 1 6 4 * 3 . 4 5 6 9 * 1 6 4 = 0.05 mol

Molarity = 0 . 0 5 * 1 0 0 0 1 0 0 M = 0.5 M

= 50.0 * 10-2 M

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