Answer:
KO is the limiting reactant.
0.11 mol O₂ will be produced.
Explanation:
4 KO₂ + 2 H₂O ⇒ 4 KOH + 3 O₂
Find the limiting reagent by dividing the moles of the reactant by the coefficient in the equation.
(0.15 mol KO₂)/4 = 0.0375
(0.10 mol H₂O)/2 = 0.05
KO₂ is the limiting reagent.
The amount of product produced depends on the limiting reagent. To find how much is produced, take moles of limiting reagent and multiply it by the ratio of reagent to product. You can find the ratio by looking at the equation. For every 4 moles of KO₂, 3 moles of O₂ are produced.
0.15 mol KO₂ (3 mol O₂)/(4 mol KO₂) = 0.1125 mol O₂
0.11 mol O are produced.
Answer:
3
Explanation:
Sodium is not a gas, it is a Alkali Metal. I recommend studying the sections of the periodic table to help remember what each element should be.
Hope this helps.
Answer:
The units of the rate constant, k, depend on the overall reaction order. The units of k for a zero-order reaction are M/s, the units of k for a first-order reaction are 1/s, and the units of k for a second-order reaction are 1/(M·s).
Explanation:
Cold is the absence of heat! so heat moves into the object from the environment (room) rapidly~ and it slows down by time and water is formed on the surface of the object because of cold and hot air.
I hope this helps cx
Answer :
(a) The rate of 
formed is, 0.066 M/s
(b) The rate of 
formed is, 0.033 M/s
Explanation : Given,
![\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
= 0.066 M/s
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[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
The rate of formation of = ![\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
As we know that,
= 0.066 M/s
(a) Now we have to determine the rate of formed.
![\frac{1}{2}\frac{d[NO_2]}{dt}=\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[NO_2]}{dt}=\frac{d[NO]}{dt}=0.066M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D0.066M%2Fs)
The rate of formed is, 0.066 M/s
(b) Now we have to determine the rate of molecular oxygen reacting.
![-\frac{d[O_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.033M%2Fs)
The rate of formed is, 0.033 M/s
