Given the percentage composition of HC as C → 81.82 % and H → 18.18 %
So the ratio of number if atoms of C and H in its molecule can will be:
C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8
So the Empirical Formula of hydrocarbon is:
C 3 H 8
As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol
Now let Molecular formula of the HC be ( C 3 H 8 ) n
Using molar mass of C and H the molar mass of the HC from its molecular formula is:
( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1
Hence the molecular formula of HC is C 3 H 8
Does that help?
Answer:
A. stored energy
Explanation:
potential energy is stored energy.
kinetic energy is the energy of motion
Answer:
7.71x10^23 molecules
Explanation:
Avogadro's # = 6.022x10^23
1.28 mol SiO2 x 6.022x10^23/ 1 mol SiO2 = 7.71x10^23 molecules
Answer:
Carbon and Oxygen, Argon and Helium.
Explanation:
noble gases have full outer shells of electrons, and so cannot share other atoms' electrons to form bonds. sodium and chlorine form an ionic bond.
Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.
Explanation : Given,
![\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
=21 torr/min
The balanced chemical reaction is,
The rate of disappearance of 
= ![-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
The rate of disappearance of 
= ![-\frac{d[Cl_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D)
The rate of formation of 
= ![\frac{1}{2}\frac{d[NOCl]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D)
As we know that,
=21 torr/min
So,
![-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2021torr%2Fmin%3D10.5torr%2Fmin)
And,
![\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D)
![\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%3D21torr%2Fmin)
Now we have to calculate the rate change.
Rate change = Reactant rate - Product rate
Rate change = (21 + 10.5) - 21 = 10.5 torr/min
Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.
