This will give substituted product which will be by SN2 mechanism
so here we will get product with inverted geometry
In SN2 mechanism the nucleophile attacks from back side and we always get product with inverted geometry
This is known as Walden inversion.
The molarity is count by dividing the mole of the solute within 1 liter of solvent. In this case, the KNO3 is 16.8g with 101.11 g/mol molar mass. Then we need to find the mol first. The calculation would be: 16.8g / (101.11g/mol)= 0.0166 mol.
Then the molarity would be: 0.0166mol/ 0.3l= 0.0498= 0.0553 M
Answer: The factor that lead to cyclopropane being less stable than the other cycloalkanes is the presence of a RING STRAIN.
Explanation:
In organic chemistry, the end carbon atoms of an open aliphatic chain can join together to form a closed system or ring to form cycloalkanes. Such compounds are known as cyclic compounds. Examples include cyclopropane, cyclobutane, cyclopentane and many among others.
Cyclopropane is less stable than other cycloalkanes mentioned above because of the presence of ring strain in its structural arrangement. The ring strain is the spatial orientation of atoms of the cycloalkane compounds which tend to give off a very high and non favourable energy. The release of heat energy which is stored in the bonds and molecules cause the ring to be UNSTABLE and REACTIVE.
The presence of the ring strain affects mainly the structures and the conformational function of the smaller cycloalkanes. cyclopropane, which is the smallest cycloalkane than the rest mentioned above, contains only 3 carbons with a small ring.
Answer: a) The concentration after 8.8min is 0.17 M
b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.
Explanation:
Expression for rate law for first order kinetics is given by:
![t=\frac{2.303}{k}\log\frac{a}{a-x}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2.303%7D%7Bk%7D%5Clog%5Cfrac%7Ba%7D%7Ba-x%7D)
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) concentration after 8.8 min:
![8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}](https://tex.z-dn.net/?f=8.8%5Ctimes%2060s%3D%5Cfrac%7B2.303%7D%7B6.7%5Ctimes%2010%5E%7B-4%7Ds%5E%7B-1%7D%7D%5Clog%5Cfrac%7B0.25%7D%7Ba-x%7D)
![\log\frac{0.25}{a-x}=0.15](https://tex.z-dn.net/?f=%5Clog%5Cfrac%7B0.25%7D%7Ba-x%7D%3D0.15)
![(a-x)=0.17M](https://tex.z-dn.net/?f=%28a-x%29%3D0.17M)
b) for concentration to decrease from 0.25M to 0.15M
![t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2.303%7D%7B6.7%5Ctimes%2010%5E%7B-4%7Ds%5E%7B-1%7D%7D%5Clog%5Cfrac%7B0.25%7D%7B0.15%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B2.303%7D%7B6.7%5Ctimes%2010%5E%7B-4%7Ds%5E%7B-1%7D%7D%5Ctimes%200.20)
![t=687s](https://tex.z-dn.net/?f=t%3D687s)
<h3>Answers:</h3>
1) 2 Units of Ozone
2) 3 Units of Ozone
3) 9 Units of Ozone
<h3>Solution:</h3>
1) From 6 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
6 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 2 Units of Ozone
2) From 9 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
9 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 3 Units of Ozone
3) From 27 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
27 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 9 Units of Ozone