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New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

For isotonic solution

? i n j e c t o r = ? B l o o d

? C * 0 . 0 8 2 * 3 0 0 = 7 . 4 7

? C = 0 . 3 ? ? m o l e / l

= 0 . 3 * 1 8 0 g m / l = 5 4 g m / l

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

  K = A . e E a / R T = ( 6 . 5 * 1 0 2 ) e 2 6 0 0 0 K / T

E a 8 . 3 1 4 = 2 6 0 0 0

 Ea = 216.164kJ/mol 216

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Volume of H2 adsorbed = nRTP=2*0.083*3002*1=24.9lit=24900ml

Therefore volume of gas adsorbed per gram of the adsorbent = 249002.5=9960

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

π = i * c * R * T = 2 * 0 . 8 3 * 3 0 0 6 0 * 1 0 3 = 0 . 0 0 4 1 5 b a r

= 415 Pa

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Mass = d * v = 1.02 * 1.2 = 1.224gm

Moles of acetic acid = 0.0204 moles in 2L

So molality = 0.0102 mol/kg

Δ T f = i * K f * m

i = 1 + a for acetic acid

0.0198  = (1 + a) * 1.85 * 0.0102

α = 0.04928 = 5%

New answer posted

2 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

ΔTb=iKbm1

ΔTbΔTf=Kb*1Kf*236=12=Kbkf*12

KbKf=1x=1

New answer posted

2 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

(c) 2CH3CH2Clether2NaC2H5C2H5+2NaCl (WurtzReaction)

(d) 2C6H5Clether2NaC6H5C6H5+2NaCl (WittingReaction)

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 PT=PA+PB=0.8? ? atm

PA=PB=0.4atm

YA=0.5YB=0.5

XA=0.2

PAo=2atm.

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

PT=XAPA0-PB0+PB0

ATQ

550=14PA0-PB0+PB0

2200=PA0-PB0+4PB0

560=15PA0-PB0+PB0

2200=PA0-PB0+5PB0

PA0+3PB0=2200

PA0±4PB0=2800PB0=600

PA0=400mmHg

 

New answer posted

2 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

P A 0 = 5 0 t o r r

P B 0 = 1 0 0 t o r r

Mole fraction of A in liquid phase, xA = 0.3

Mole fraction of B in liquid phase, xB = 0.7

Now; PA=PA0xA

PB=PB0xB

= 100 * 0.7 = 70 torr

Mole fraction of B in vapour phase,  yB=PBPA+PB=7085

7085=x17

X = 14

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