The substance that is a diatomic molecule is B. O2 Oxygen.
I'd say it's A because while the energy cannot be lost, it can not be 100% efficient because if it was we wouldn't have to worry about energy.
We can define atomic mass the total of number of protons and number of neutrons in an atom or isotope.
<span>So when an isotope of yttrium has 39 protons and 59 neutrons, its atomic mass is equal to;
number of protons + number of neutrons = 39 + 59 = 98</span>
E
θ
Cell
=
+
2.115
l
V
Cathode
Mg
2
+
/
Mg
Anode
Ni
2
+
/
Ni
Explanation:
Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]
Mg
2
+
(
a
q
)
+
2
l
e
−
→
Mg
(
s
)
−
E
θ
=
−
2.372
l
V
Ni
2
+
(
a
q
)
+
2
l
e
−
→
Ni
(
s
)
−
E
θ
=
−
0.257
l
V
The standard reduction potential
E
θ
resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (
298
l
K
,
1.00
l
kPa
) is defined as
0
l
V
for reference. [2]
A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.
Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher
E
θ
and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower
E
θ
will experience oxidation and act the anode.
E
θ
(
Ni
2
+
/
Ni
)
>
E
θ
(
Mg
2
+
/
Mg
)
Therefore in this galvanic cell, the
Ni
2
+
/
Ni
half-cell will experience reduction and act as the cathode and the
Mg
2
+
/
Mg
the anode.
The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:
E
θ
cell
=
E
θ
(
Cathode
)
−
E
θ
(
Anode
)
E
θ
cell
=
−
0.257
−
(
−
2.372
)
E
θ
cell
=
+
2.115
Indicating that connecting the two cells will generate a potential difference of
+
2.115
l
V
across the two cells.
Answer:
Rate constant of the reaction is 
.
Explanation:
A + B + C → D + E
Let the balanced reaction be ;
aA + bB + cC → dD + eE
Expression of rate law of the reaction will be written as:
![R=k[A]^a[B]^b[C]^c](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5Ea%5BB%5D%5Eb%5BC%5D%5Ec)
Rate(R) of the reaction in trail 1 ,when :
![[A]=0.30 M,[B]=0.30 M,[C]=0.30 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.30%20M%2C%5BB%5D%3D0.30%20M%2C%5BC%5D%3D0.30%20M)
![9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c](https://tex.z-dn.net/?f=9.0%5Ctimes%2010%5E%7B-5%7D%20M%2Fs%3Dk%5B0.30%20M%5D%5Ea%5B0.30%20M%5D%5Eb%5B0.30%20M%5D%5Ec)
...[1]
Rate(R) of the reaction in trail 2 ,when :
![[A]=0.30 M,[B]=0.30 M,[C]=0.90 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.30%20M%2C%5BB%5D%3D0.30%20M%2C%5BC%5D%3D0.90%20M)
![2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c](https://tex.z-dn.net/?f=2.7%5Ctimes%2010%5E%7B-4%7D%20M%2Fs%3Dk%5B0.30%20M%5D%5Ea%5B0.30%20M%5D%5Eb%5B0.90%20M%5D%5Ec)
...[2]
Rate(R) of the reaction in trail 3 ,when :
![[A]=0.60 M,[B]=0.30 M,[C]=0.30 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.60%20M%2C%5BB%5D%3D0.30%20M%2C%5BC%5D%3D0.30%20M)
![3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c](https://tex.z-dn.net/?f=3.6%5Ctimes%2010%5E%7B-4%7D%20M%2Fs%3Dk%5B0.60%20M%5D%5Ea%5B0.30%20M%5D%5Eb%5B0.30%20M%5D%5Ec)
...[3]
Rate(R) of the reaction in trail 4 ,when :
![[A]=0.60 M,[B]=0.60 M,[C]=0.30 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.60%20M%2C%5BB%5D%3D0.60%20M%2C%5BC%5D%3D0.30%20M)
![3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c](https://tex.z-dn.net/?f=3.6%5Ctimes%2010%5E%7B-4%7D%20M%2Fs%3Dk%5B0.60%20M%5D%5Ea%5B0.60%20M%5D%5Eb%5B0.30%20M%5D%5Ec)
...[4]
By [1] ÷ [2], we get value of c ;
c = 1
By [3] ÷ [4], we get value of b ;
b = 0
By [2] ÷ [3], we get value of a ;
a = 2
Rate law of reaction is :
![R=k[A]^2[B]^0[C]^1](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E2%5BB%5D%5E0%5BC%5D%5E1)
Rate constant of the reaction = k
![9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1](https://tex.z-dn.net/?f=9.0%5Ctimes%2010%5E%7B-5%7D%20M%2Fs%3Dk%5B0.30%20M%5D%5E2%5B0.30%20M%5D%5E0%5B0.30%20M%5D%5E1)
![k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B9.0%5Ctimes%2010%5E%7B-5%7D%20M%2Fs%7D%7B%5B0.30%20M%5D%5E2%5B0.30%20M%5D%5E0%5B0.30%20M%5D%5E1%7D)
