Chemistry

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New answer posted

6 months ago

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A
alok kumar singh

Contributor-Level 10

H? SO? + 2NaOH → Na? SO? + 2H? O
Milli mole: 80 60 (limiting reagent)
H? + OH? → H? O
Milli mole: 160 60
Milli mole: 160-60 0 60
Total heat produced = (60/1000) x 57.1 x 1000 = 3426 J
Total heat absorbed = msΔT
ΔT = Totalheat / (m x s)
= 3426J / (1000 x 4.18) = 0.819K
= 81.9 x 10? ² K
Ans. = 82 (the nearest)

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

3 [Co (en)? Cl? ]Cl + 3AgNO? (excess) → 3AgCl↓ (ppt) + 3 [Co (en)? Cl? ] NO?
Secondary valency of complex = 6

New answer posted

6 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

E.C = σ1s² σ1s² σ2s² σ2s² σ2p? ² π2p? ² = π2p? ² π2p? ² = π2p? ²
Total electrons in BMO = 10

New answer posted

6 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

6 months ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

E_cell = E? _cell - (0.059/2)log { [Cu²? ]/ [Ag? ]²}
E? = E? _cell - (0.059/2)log [0.1/ (0.01)²] = 0.3095
E? _cell = 0.3095 + 0.0885 V
E? _cell = 0.398

Again, E? = E? _cell - (0.059/2)log [10? ²/ (10? ³)²]
E? = 0.398 - (0.059/2) x 4 V

E? = 0.28 V
E? = 28 x 10? ² V

New answer posted

6 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

Kp = (Po? )¹/² = 4
Po? = 16 atm

New answer posted

6 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

1.     The number of chiral carbons in open-chain aldohexose (such as glucose) is four, therefore, the number of stereoisomers? 2? 16.

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

ATP has phosphate group, PO? ³?
So x + 4 (-2) = - 3 or x = + 5

New answer posted

6 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

ΔH / ΔU = C? / C? = 75 / 25 = 3

New answer posted

6 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

A → 2B
t = 0: 1 mole 0
after 100 min: (1 - 0.1) mol 0.2 mol
K = (1/t)ln [a/ (a-n)]
K = (2.303/100)log (1/0.9) min? ¹
0.693/t? /? = (2.303/100) [log10 - log9] = (2.303/100) x 0.046
t? /? = 69.3 / (2.303 x 0.046) = 654.15 min
Ans. = 654 (the nearest integer)

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