Answer:
Size of the nucleus of an atom is very small as compared to the size of the atom.
According to Rutherford gold foil experiment, nucleus is very small in size as compared to the size of the atom as a whole. Nucleus is very hard, dense and positively charged which consists of protons and neutrons.
Explanation:
The standard enthalpy change of a reaction is positive thus endothermic.
The standard enthalpy change of a reaction is negative thus exothermic.
Hope it helped!
Answer:
The dimerization of butadiene to 4-vinylcyclohexene folows second order kinetics and its rate law will be given by :
![R=k[C_4H_6]^2](https://tex.z-dn.net/?f=R%3Dk%5BC_4H_6%5D%5E2)
Explanation:
The rate of the reaction ;
![R=k[C_4H_6]^x](https://tex.z-dn.net/?f=R%3Dk%5BC_4H_6%5D%5Ex)
As given in the question , that graph of time verses ![\frac{1}{[C_4H_6]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BC_4H_6%5D%7D)
was linear but plots of ![[C_4H_6]](https://tex.z-dn.net/?f=%5BC_4H_6%5D)
or ![\ln[C_4H_6]](https://tex.z-dn.net/?f=%5Cln%5BC_4H_6%5D)
was curved.
Generally:
Graph of time verses ![[concentration]](https://tex.z-dn.net/?f=%5Bconcentration%5D)
for zero order reaction is linear with negative slope.
Graph of time verses ![\ln [concentration]](https://tex.z-dn.net/?f=%5Cln%20%5Bconcentration%5D)
for secon order reaction is linear with negative slope.
Graph of time verses ![\frac{1}{[concentration]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Bconcentration%5D%7D)
for secon order reaction is linear with positive slope.
So, the dimerization of butadiene to 4-vinylcyclohexene folows second order kinetics and its rate law will be given by :
<span> Molar mass (H2)=2*1.0=2.0 g/mol
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol
5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2
H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 1 mol
From problem 2.50 mol 1 .00mol
We can see that excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.
</span> H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 2 mol
From problem 1.00 mol 2.00mol
2.00 mol HF can be formed.
2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed
