Answer:
B
Explanation:
the group number is=valence electrons. element 1 is in group 1 element 18 is in group 8. 1<8
Answer:
Part A
The volume of the gaseous product is
Part B
The volume of the the engine’s gaseous exhaust is
Explanation:
Part A
From the question we are told that
The temperature is ![T = 350^oC = 350 +273 =623K](https://tex.z-dn.net/?f=T%20%3D%20350%5EoC%20%3D%20350%20%2B273%20%3D623K)
The pressure is ![P = 735 \ torr = \frac{735}{760} = 0.967\ atm](https://tex.z-dn.net/?f=P%20%3D%20735%20%5C%20torr%20%3D%20%5Cfrac%7B735%7D%7B760%7D%20%3D%20%200.967%5C%20atm)
The of ![C_8 H_{18} = 100.0g](https://tex.z-dn.net/?f=C_8%20H_%7B18%7D%20%3D%20100.0g)
The chemical equation for this combustion is
![2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}](https://tex.z-dn.net/?f=2%20C_8%20H_%7B18%7D_%7B%28l%29%7D%20%2B%2025O_2_%7B%28l%29%7D%20----%3E%2016CO_2_%7B%28g%29%7D%20%2B%2018%20H_2%20O_%7B%28g%29%7D)
The number of moles of
that reacted is mathematically represented as
![n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7Bmass%20%5C%20of%20%5C%20%20C_8H_%7B18%7D%20%20%7D%7BMolar%20%5C%20%20mass%20%5C%20of%20%20C_8H_%7B18%7D%20%7D)
The molar mass of
is constant value which is
So ![n = \frac{100 }{114.23} }](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B100%20%20%7D%7B114.23%7D%20%7D)
![n = 0.8754 \ moles](https://tex.z-dn.net/?f=n%20%3D%200.8754%20%5C%20moles)
The gaseous product in the reaction is
and water vapour
Now from the reaction
2 moles of
will react with 25 moles of
to give (16 + 18) moles of
and ![H_2 O_{(g)}](https://tex.z-dn.net/?f=H_2%20O_%7B%28g%29%7D)
So
1 mole of
will react with 12.5 moles of
to give 17 moles of
and ![H_2 O_{(g)}](https://tex.z-dn.net/?f=H_2%20O_%7B%28g%29%7D)
This implies that
0.8754 moles of
will react with (12.5 * 0.8754 ) moles of
to give (17 * 0.8754) of
and ![H_2 O_{(g)}](https://tex.z-dn.net/?f=H_2%20O_%7B%28g%29%7D)
So the no of moles of gaseous product is
![N_g = 17 * 0.8754](https://tex.z-dn.net/?f=N_g%20%3D%2017%20%2A%200.8754)
![N_g = 14.88 \ moles](https://tex.z-dn.net/?f=N_g%20%3D%2014.88%20%5C%20moles)
From the ideal gas law
![PV = N_gRT](https://tex.z-dn.net/?f=PV%20%3D%20N_gRT)
making V the subject
![V = \frac{N_gRT}{P}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7BN_gRT%7D%7BP%7D)
Where R is the gas constant with a value ![R = 0.08206 \ L\cdot atm /K \cdot mole](https://tex.z-dn.net/?f=R%20%3D%200.08206%20%5C%20%20L%5Ccdot%20atm%20%2FK%20%5Ccdot%20mole)
Substituting values
Part B
From the reaction the number of moles of oxygen that reacted is
![N_o = 0.8754 * 12.5](https://tex.z-dn.net/?f=N_o%20%3D%200.8754%20%2A%2012.5)
![N_o = 10.94 \ moles](https://tex.z-dn.net/?f=N_o%20%3D%2010.94%20%5C%20moles)
The volume is
![V_o = \frac{10.94 * 0.08206 *623}{0.967}](https://tex.z-dn.net/?f=V_o%20%20%3D%20%5Cfrac%7B10.94%20%2A%200.08206%20%2A623%7D%7B0.967%7D)
![V_o = 579 \ L](https://tex.z-dn.net/?f=V_o%20%20%3D%20579%20%5C%20L)
No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as
![V_e = V_o * \frac{0.79}{0.21}](https://tex.z-dn.net/?f=V_e%20%3D%20V_o%20%2A%20%5Cfrac%7B0.79%7D%7B0.21%7D)
Substituting values
I think the one that cause the equilibrium to shift would be :
3. adding a noble gas
Adding the noble gas will add more concentration to the KNO3, which will create different amount of equilibrium
hope this helps
Q1. TI (210/81Thallium)
Q2.
The answers are opposite from each other