The given compound 3-chloro-2,2,5-trimethylhexane is an optically active compound .
Because this compound does not have plane of symmetry (POS) and center of symmetry (COS) i.e. does not have di-symmetry . And also forms non superimposable mirror image . the compound is optically active .
It has chiral center.
Here the chiral carbon has 4 distinct groups such as : chlorine , hydrogen , 2-methylpropyl , tertbutyl .
<h3>What is di-symmetry?</h3>
Di-symmetry is that which have no center of symmetry and plane of symmetry and alternate axis of symmetry .
<h3>Chiral center :</h3>
Have Sp3 hybridized center (4sigma bond ) .
4 distinct group is attached to the chiral atom. form non -superimposable mirror image .
<h3>What is optical isomerism ?</h3>
Same molecular formula and same structural formula . also have same physical and chemical properties .
They differ in their behavior towards plane polarized light (ppl) .
Learn more about chiral center here:
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C = 0.11 mol
V = 5.65 L
n = ???
n = C*V
n = 0.11 * 5.65
n = 0.622 mols
1 mol of CaCl2 = 40 + 2*35.5 = 111 grams
0.622 mol = x
x = 111 * 0.622
x = 69.0 grams CaCl2
Explanation:
The two half equations are;
3e + HNO3 → NO
S→ H2SO4 + 6e
When balancing half equations, we have to make sure the number of electrons gained is equal to the number of electrons lost.
<em>Which factor will you use for the top equation?</em>
We multiply by 2 to make the number of electrons = 6e
<em>Which factor will you use for the bottom equation?</em>
We multiply by 1 to make the number of electrons = 6e
Explanation:
The given data is as follows.
Thickness (dx) = 0.87 m, thermal conductivity (k) = 13 W/m-K
Surface area (A) = 5 
, 
According to Fourier's law,
Q = 
Hence, putting the given values into the above formula as follows.
Q =
= 
= 5902.298 W
Therefore, we can conclude that the rate of heat transfer is 5902.298 W.
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
