Hello there.
An atom that has the same amount of atoms of each element on both sides of the reaction obeys what law
<span>law of conservation of mass</span>
Answer:
is the correct answer to the given question.
Explanation:
Given k=6.40 x 10-3 min-1.
According to the first order reaction .
The concentration of time can be written as
![[\ A\ ]\ = \ [\ A_{0}\ ] * e \ ^\ {-kt}](https://tex.z-dn.net/?f=%5B%5C%20A%5C%20%5D%5C%20%3D%20%5C%20%5B%5C%20%20A_%7B0%7D%5C%20%5D%20%20%2A%20e%20%5C%20%5E%5C%20%20%7B-kt%7D)
Here
= Initial concentration.
So ![[\ A\ ]_{0}= 0.0314 M](https://tex.z-dn.net/?f=%5B%5C%20A%5C%20%5D_%7B0%7D%3D%200.0314%20M)
Putting this value into the above equation.

=0.211 M
This can be written as

C. carbon is the correct answer
In my opinion, the answer is the second option.
43.56 grams of are produced if 16g of CH4 reacts with 64g of O2.
Explanation:
Balance equation for the reaction:
CH4 + 2O2⇒ CO2 +2H2O
Data given : mass of CH4 =16 grams atomic mass = 16.04 grams/mole
mass of water 36 gram atomic mass = 18 grams/moles
mass of CO2=? atomic mass = 44.01 grams/mole
number of moles =
equation 1
number of moles in CH4
n = 
= 0.99 moles
Since combustion is done in presence of oxygen hence it is an excess reagent and methane is limiting reagent so production of CO2 depends on it.
From the equation
1 mole of CH4 gave 1 mole of CO2
O.99 moles of CH4 will give x moles of CO2
= 
x = 0.99 moles of carbon dioxide
grams of CO2 = number of moles x atomic mass
= 0.99 x 44.01
= 43.56 grams of CO2 is produced.