Answer:
2.42L
Explanation:
Given parameters:
V₁ = 1.8L
T₁ = 293K
P₁ = 101.3kPa
P₂ = 67.6kPa
T₂ = 263K
Unknown:
V₂ = ?
Solution:
To solve this problem, we are going to use the combined gas law to find the final volume of the gas. The combined gas law expression combines the equation of Boyle's law, Charles's law and Avogadro's law;

All the units are in the appropriate form. We just substitute and solve for the unknown;
101.3 x 1.8 / 293 = 67.6 x V₂ / 263
V₂ = 2.42L
I would say physical, because a physical change is affecting the form of a chemical substance, but not it's chemical makeup.
Answer:
(a) Rate at which
is formed is 0.050 M/s
(b) Rate at which
is consumed is 0.0250 M/s.
Explanation:
The given reaction is:-

The expression for rate can be written as:-
![-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
Given that:-
(Negative sign shows consumption)
![-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![-(-0.050\ M/s)=\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%28-0.050%5C%20M%2Fs%29%3D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\frac{d[NO_2]}{dt}=0.050\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D0.050%5C%20M%2Fs)
(a) Rate at which
is formed is 0.050 M/s
![-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-0.050%5C%20M%2Fs%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![\frac{d[O_2]}{dt}=0.0250\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D0.0250%5C%20M%2Fs)
(b) Rate at which
is consumed is 0.0250 M/s.
Answer:
The period 2 element would be more reactive because the attractive force of protons is stronger when electrons are attracted to a closer electron shell.
Explanation:
Nonmetals want to add more electrons to their valence electron shell. Elements in period 2 have their valence shell closer their nucleus, then they are more reactive than period 4 elements.