Answer:
The answer is yes.
Explanation:
When driving any type of motorized vehicle, fuel is importatn. Fuel is what it runs on, so without it it wouldn't run. With trucks a lot more is needed because trucks are bigger, and tend to carry more than a little car does.
Its an element Im pretty sure
Answer: 18.65L
Explanation:
Given that,
Original volume of oxygen (V1) = 30.0L
Original temperature of oxygen (T1) = 200°C
[Convert temperature in Celsius to Kelvin by adding 273.
So, (200°C + 273 = 473K)]
New volume of oxygen V2 = ?
New temperature of oxygen T2 = 1°C
(1°C + 273 = 274K)
Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law
V1/T1 = V2/T2
30.0L/473K = V2/294K
To get the value of V2, cross multiply
30.0L x 294K = 473K x V2
8820L•K = 473K•V2
Divide both sides by 473K
8820L•K / 473K = 473K•V2/473K
18.65L = V2
Thus, the new volume of oxygen is 18.65 liters.
The equilibrium constant for the reaction is 0.00662
Explanation:
The balanced chemical equation is :
2NO2(g)⇌2NO(g)+O2(g
At t=t 1-2x ⇔ 2x + x moles
The ideal gas law equation will be used here
PV=nRT
here n= 
= 
= density
P = 
density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm
putting the values in reaction
0.75 = 
M = 34.61
to calculate the Kc
Kc=![\frac{ [NO] [O2]}{NO2}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BNO%5D%20%5BO2%5D%7D%7BNO2%7D)
x M NO2 + 
M NO+ 
M O2
Putting the values as molecular weight of NO2, NO,O2
34.61= 
x= 0.33
Kc= 
putting the values in the above equation
Kc = 0.00662
