Answer:
The rate of consumption of
is 2.0 mol/L.s
Explanation:
Applying law of mass action to this reaction-
![-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B4%7D%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D-%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7B%5CDelta%20%5BO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BN_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
where
represents rate of consumption of
,
represents rate of consumption of
,
represents rate of formation of
and
represents rate of formation of
.
Here rate of formation of
is 3.0 mol/(L.s)
From the above equation we can write-
![-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B4%7D%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
Here ![\frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D%3D3.0%20mol%2F%28L.s%29%29)
So, ![-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B4%7D%7B6%7D%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
Hence,
The mass of copper deposited is 9.589 g .
Given ,
A volatile cell using Cu/Cu2+ and Sn/Sn2+ half cells is set up at standard conditions ,and each compartment has a volume of 345 ml .
current = 0.17 A
time = 48 hrs =48 (60) (60) =172800 sec
we know , I = Q/t
thus , Q ,charge= It = 0.17 (17280) =29376 C
Then ,atomic mass of copper is 63 g
And valency of Cu is 2.
Thus , the equivalent mass of copper is 63/2 = 31.5
We know , 96500 coulombs of electricity produce copper = 31.5 g
29376 C of electricity produce copper =9.589 g
Hence , 9.589 g of Copper is deposited .
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Fire, explosions, acid, fumes-take your pick here.
See the sketch attached.
<h3>Explanation</h3>
The Lewis structure of a molecule describes
- the number of bonds it has,
- the source of electrons in each bond, and
- the position of any lone pairs of electrons.
Atoms are most stable when they have eight or no electrons in their valence shell (or two, in case of hydrogen.)
- Each oxygen atom contains six valence electrons. It demands <em>two</em> extra electrons to be chemically stable.
- Each sulfur atom contains six valence electrons. It demands <em>two </em> extra electrons to be chemically stable.
- Each hydrogen atom demands <em>one</em> extra electron to be stable.
H₂O contains two hydrogen atoms and one oxygen atom. It would take an extra 2 + 2 × 1 = 4 electrons for all its three atoms are stable. Atoms in an H₂O would achieve that need by sharing electrons. It would form a total of 4 / 2 = 2 O-H bonds.
Each O-H bond contains one electron from oxygen and one from hydrogen. Hydrogen has no electron left. Oxygen has six electrons. Two of them have went to the two O-H bonds. The remaining four become 4 / 2 = 2 lone pairs. The lone pairs repel the O-H bonds. By convention, they are placed on top of the two H atoms.
Similarly, atoms in a SO₂ molecule demands an extra 2 × 2 + 2 = 6 electrons for its three atoms to become chemically stable. It would form 6 / 2 = 3 chemical bonds. Loops are unlikely in molecules without carbon. As a result, one of the two O atoms would form two bonds with the S atom while the other form only one.
Atoms are unstable with an odd number of valence electrons. The S atom in SO₂ would have become unstable if it contribute one electron to each of the three bond. It would end up with 3 × 2 + 3 = 9 valence electrons. One possible solution is that it contributes two electrons in one particular bond. One of the three bonds would be a coordinate covalent bond, with both electrons in that bond from the S atom. In some textbooks this type of bonds are also known as dative bonds.
Dots and crosses denotes the origin of electrons in a bond. Use the same symbol for electrons from the same atom. Electrons from the oxygen atoms O are shown in blue in the sketch. They don't have to be colored.
Answer:
Extinction
Explanation:
When a species becomes extinct, every last specimen has died out