Chemistry

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New answer posted

8 months ago

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A
alok kumar singh

Contributor-Level 10

Po2 = PTotal -   P N 2 5 = ( x / 3 2 x / 3 2 + 2 0 0 2 0 ) * 2 5 x = 8 0 g m

= 25 – 20 = 5bar

Po2 =   n O 2 n O 2 + n N e * P T o t a l

5 = ( x / 3 2 x / 3 2 + 2 0 0 2 0 ) * 2 5 x = 8 0 g m
 

New answer posted

8 months ago

0 Follower 15 Views

V
Vishal Baghel

Contributor-Level 10

No. of moles =  5 9 2 ( t o l u e n e )

Moles of benzaldehyde = 5 9 2 * 9 2 1 0 0 = 5 * 1 0 2  

Mass of benzaldehyde = 5 * 10-2 *106 = 5.3 gm

= 530 * 10-2 gm

New answer posted

8 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Fehling's solution is a complex of   C u 2 + , C u 2 + = 3 d 9

No. of unpaired e- = 1

MM =  1 ( 1 + 2 ) = 3 = 1 . 7 3 B M

New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

SO3 -> Sp2 Planar

BF3 -> Sp2 Planar

N O 3 -> Sp2 Planar

SF4 -> Sp3d non-planar

H2O2 -> Sp3 Non-planar

PCl3 -> Sp3 Non –planar

[Al (OH)4]- -> Sp3 Non-planar

XeF4 -> Sp3d2 planar

XeO3 -> Sp3 Non-planar

P H 4 -> Sp3 Non-planar

New answer posted

8 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

  2 K M n O 4 + 3 H 2 O 2 B a s i c m e d i u m 2 M n O 2 + 3 O 2 + 2 H 2 O + 2 K O H

New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

Bauxite = AlOx (OH)3-2x (where O < x < 1)

Siderite = FeCO3

Cuprite = Cu2O

Calamine = ZnCO3

Haemetite = Fe2O3

Kaolinite = Al2 (OH)4Si2O5

Malachite = CuCO3 Cu (OH)2

Magneite = Fe3O4

Sphalerite = ZnS

Limonite = Fe2O3.3H2O

New answer posted

8 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

K = 0 . 6 9 3 t 1 / 2 = 0 . 6 9 3 7 0 * 6 0 = 6 9 3 0 7 * 6 * 1 0 6

= 165 * 10-6 s-1

New answer posted

8 months ago

0 Follower 34 Views

V
Vishal Baghel

Contributor-Level 10

P 0 P S P 0 = n s o l u t e n s o l v e n t

P 0 P 0 2 P 0 = n s o l u t e n s o l v e n t

n s o l u t e = n s o l v e n t 2 = 1 0 0 1 8 * 2 = 2 . 7 8 m o l

New answer posted

8 months ago

0 Follower 14 Views

V
Vishal Baghel

Contributor-Level 10

Let x gm is burnt

Moles = x/280

Heat released by x/280 mol = 2.5 * 0.45 kJ

Heat released by 1 mol =  2 . 5 * 0 . 4 5 * 2 8 0 x = k J

Δ H = Δ U + Δ n g R T Δ H = Δ U            

9 = 2 . 5 * 2 8 0 * 0 . 4 5 x            

x = 35 gm

New answer posted

8 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

For real gas under high pressure,

Z = 1 + P b R T b = R T P = 0 . 0 8 3 * 2 9 8 9 9        

= 0.25 * 10-2 L mol-1

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