Explanation:
There are government khatris as well, which are maintained by the panchayat. Kuhls are a traditional irrigation system in Himachal Pradesh- surface channels diverting water from natural flowing streams (khuds). A typical community kuhl services six to 30 farmers, irrigating an area of about 20 ha.
3:00m cu of the current in electrical engineering and electrical toques
I will assume that the sign ? between the C and the CCH3 is a triple bond, and I will represent it by three vertical lines |||
So the reaction is:
<span>CH3CH2CH2CH2C ||| CCH3+2Br2 ---->
This is a typical reaction known as halogenation of alkines.
This is an addition reaction, i.e. the alkyne undergoes an addition of the Br2 (and it also happens with Cl2) to the triple bond to form a tetra halide.
.
Br Br
</span> | |
<span><span>CH3CH2CH2CH2C ||| CCH3+2Br2 ----> CH3 CH2 CH2 CH2 C - C</span> - CH3
| |
Br Br
</span>
Answer:
1.65 L
Explanation:
The equation for the reaction is given as:
A + B ⇄ C
where;
numbers of moles = 0.386 mol C (g)
Volume = 7.29 L
Molar concentration of C = 
= 0.053 M
A + B ⇄ C
Initial 0 0 0.530
Change +x +x - x
Equilibrium x x (0.0530 - x)
![K = \frac{[C]}{[A][B]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%7D%7B%5BA%5D%5BB%5D%7D)
where
K is given as ; 78.2 atm-1.
So, we have:
![78.2=\frac{[0.0530-x]}{[x][x]}](https://tex.z-dn.net/?f=78.2%3D%5Cfrac%7B%5B0.0530-x%5D%7D%7B%5Bx%5D%5Bx%5D%7D)


Using quadratic formula;

where; a = 78.2 ; b = 1 ; c= - 0.0530
=
or 
=
or 
= 0.0204 or -0.0332
Going by the positive value; we have:
x = 0.0204
[A] = 0.0204
[B] = 0.0204
[C] = 0.0530 - x
= 0.0530 - 0.0204
= 0.0326
Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326
= 0.0734
Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT
if we make V the subject of the formula; we have:

where;
P (pressure) = 1 atm
n (number of moles) = 0.0734 mole
R (rate constant) = 0.0821 L-atm/mol-K
T = 273.15 K (fixed constant temperature )
V (volume) = ???

V = 1.64604
V ≅ 1.65 L
Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2