Chemistry

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New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

P 1 T 1 = P 2 T 2 3 5 3 0 0 = 4 0 T 2

T 2 = 4 0 * 3 0 0 3 5 = 3 4 2 . 8 5 K

T 2 ( ° C ) = 6 9 . 7 0 7 7 0 ° C

New answer posted

10 months ago

0 Follower 19 Views

A
alok kumar singh

Contributor-Level 10

A 2 B 3 ? 2 A + 3 + 3 B 2

1-α        2α        3α

i = 1 + 4 α = 1 + 4 * 0 . 6 = 1 + 2 . 4 = 3 . 4

Δ T b = i k b m = 3 . 4 * 0 . 5 2 * 1 = 1 . 7 6 8 1 . 7 7 K

T b = 3 7 4 . 7 7 K 3 7 5 K .         

New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

  ? U = ? 7 4 2 . 2 4 ? k J / m o l e ? H 2 9 8 = ?

NH2CN (s) + 3 2 O2 (g) ® N2 (g) + O2 (g) + H2O (l)

 = ? 7 4 2 . 2 4 + 1 2 * 8 . 3 1 4 1 0 0 0 * 2 9 8  

? 7 4 2 . 2 4 + 1 . 2 3 8 = ? 7 4 1 . 0 0 2 ? ? 7 4 1 k J / m o l                  

So, magnitude of  ? H = 741 kJ/mol.

 

New answer posted

10 months ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

NaOH + Na2CO3

(1) When Hph is added

m.e NaOH + 1 2 m . e N a 2 C O 3 = m . e . H C l = 1 7 . 5 * 1 1 0 = 1 . 7 5  (m.e = milli equivalents)

(2) When MeOH is added after Hph

1 2 m e N a 2 C O 3 = m . e H C l = 1 . 5 * 1 1 0 = 0 . 1 5

m . e N a O H = 1 . 7 5 0 . 1 5 = 1 . 6 m . e N a 2 C O 3 = 0 . 1 5 * 2 = 0 . 3  

W N a 2 C O 3 E N a 2 C O 3 * 1 0 0 0 = 0 . 3   

W e i g h t % o f N a 2 C O 3 = ( 0 . 3 * 5 3 1 0 0 0 0 . 4 ) * 1 0 0 = 0 . 3 * 5 3 1 0 * 0 . 4 = 1 5 . 9 4 = 3 . 9 7 5 % 4 % .               

New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

( + 6 ) ( + 2 ) ( + 3 ) ( + 6 ) C r O 4 2 + S 2 O 3 2 C r ( O H ) 4 + S O 4 2 ?  

O.N.C = 4

n factor of S2O3-- = 8

n factor of CrO4-- = 3

m . e C r O 4 2 = m . e . S 2 O 3 2        

V * ( 0 . 1 5 4 * 3 ) = 4 0 * ( 0 . 2 5 * 8 ) v = 8 0 0 . 4 6 2       

= 173.16 ml 173 ml

New answer posted

10 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

l o g K = l o g A E a 2 . 3 0 3 R T

S l o p e = E a 2 . 3 0 3 R = 1 0 , 0 0 0 K  

E a 2 . 3 0 3 R = 1 0 4            

l o g ( 1 0 5 ) = l o g A 1 0 4 * 1 5 0 0   (at 500 K temperature)

T = 1 0 4 1 9 K = 1 0 0 1 9 * 1 0 2 K = 5 . 2 6 3 1 * 1 0 2 K = 5 2 6 . 3 1 K 5 2 6 K

New answer posted

10 months ago

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P
Payal Gupta

Contributor-Level 10

(A) Ethane                   one (CC)σ bond

(B) Ethene                         one (CC)σ and one (CC)π bond

(C) C2                                                       two (CC)π bonds

(D) Ethyne HCCH                       two (CC)π bonds and one (CC)σ bond

New answer posted

10 months ago

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P
Payal Gupta

Contributor-Level 10

En=RH (Z2n2)J

For + (n=1) ,

En=x=RH (2212)=4RH

RH=x4

For 3+ (n=2) ,

En=RH (z2n2)J

=x4* (4*42*2)=xJ

New answer posted

10 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

(A) Isothermal process  Temperature is constant throughout the process

(B) Isochoric process  Volume is constant throughout the process

(C) Isobaric process  Pressure is constant throughout the process

(D) Adiabatic process  No exchange of heat  (q) between system and surrounding

New answer posted

10 months ago

0 Follower 20 Views

P
Payal Gupta

Contributor-Level 10

(1) 4 mol of =4A atoms

(2) 4 u of =4u4u=1 atom

(3) 4 g of Helium =44 mole =1 mole =NA He atom

(4) 2.2710982 of He at STP =2.27122.710982 mole

=0.1

=0.1A

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