Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
Acetylene has a chemical formula which can be written as:
C2H2
We can see that there are two positive ions, H+. Now what
deprotonation means is that the H+ is removed from acetylene to form acetylene
ion and water. In this case, I believe that the answer would be:
<span>LiOCH3</span>
7.2 mol H × (1 mol H_2SO_4/2 mol H) = 3.6 mol H_2SO_4
Answer:
a) 0.13 g = 0.13X0.001 kg = 1.3 X 10⁻⁴ kg
b) 232 Gg = 232 X 1000000 kg = 2.32 X 10⁸ kg
c) 5.23 pm = 5.23 X 10⁻¹² m
d) 86.3 mg = 86.3 X 10⁻⁶Kg = 8.63 X 10⁻⁵Kg
e) 37.6 cm = 37.6 X 10⁻² m = 3.76 X 10⁻¹m
f) 54 = 5.4 X 10¹
g) 1 Ts = 10¹² seconds
h ) 27 ps = 27 X 10⁻¹² s= 2.7 X 10⁻¹¹ s
i) 0.15 mK = 1.5 X10⁻⁴ K
Explanation:
A scientific notation is a representation of any number which is very large or very small. It is simplification of writing a number.
The SI units of
i) weight = Kg
ii) length = meter
iii) time = second
iv) temperature = K
a) 0.13 g = 0.13X0.001 kg = 1.3 X 10⁻⁴ kg
b) 232Gg
1 Gg = 1000000 kg
232 Gg = 232 X 1000000 kg = 2.32 X 10⁸ kg
c) 5.23 pm
1 pm = 10⁻¹² m
5.23 pm = 5.23 X 10⁻¹² m
d) 86.3 mg
1mg = 10⁻⁶Kg
86.3 mg = 86.3 X 10⁻⁶Kg = 8.63 X 10⁻⁵Kg
e) 37.6cm
1 cm = 10⁻² m
37.6 cm = 37.6 X 10⁻² m = 3.76 X 10⁻¹m
f) 54 (no units)
54 = 5.4 X 10¹
g) 1 Ts = 10¹² seconds
h ) 27 ps
1 ps = 10⁻¹² s
27 ps = 27 X 10⁻¹² s= 2.7 X 10⁻¹¹ s
i) 0.15 mK
1mK = 10⁻³ K
0.15 mK = 1.5 X10⁻⁴ K
Answer:
139/56 Ba
Explanation:
In writing nuclear equations, there must be a balance of the mass and charge on both sides of the reaction equation. That is, the sum of masses on the lefthand side of the nuclear reaction equation must equal the sum of masses on the righthand side of the nuclear reaction equation. Similarly, the charges on the lefthand side of the nuclear reaction equation must equal the sum of the charges on the right hand side of the nuclear reaction equation.
Hence, the isotope of barium formed in the equation mentioned in the question is 139/56 Ba.
