The question is incomplete. The complete question is :
A 100.0 mL flask is filled with 0.065 moles of A and allowed to react to form B according to the reaction below. The following experimental data are obtained for the amount of A as the reaction proceeds. What is the average rate of appearance of B in units of M/s between t = 10 min. and t = 30 min.? Assume that the volume of the flask is constant. A(g) → B(g)
Time 0.0 10.0 20.0 30.0 40.0
Moles of A 0.065 0.051 0.042 0.036 0.031
Solution :
Consider the following reaction as follows :

The experiment data is given as follows :
Time (min) : 0.0 10.0 20.0 30.0 40.0
Moles of A : 0.065 0.051 0.042 0.036 0.031
According to the rate of reaction concept, the rate can be expressed as a consumption of the reactant and formation of the product as follows :
Average rate : ![$= -\frac{d[A]}{dt} = \frac{d[B]}{dt} $](https://tex.z-dn.net/?f=%24%3D%20-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3D%20%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%20%24)
Now we have to calculate the average rate between 10.0 to 30.0 min w.r.t. A as follows :
Rate 


Therefore, the rate = 
Answer:
when solid dry ice vaporizes (physical change), carbon dioxide molecules absorb energy. When liquid water becomes ice, energy is released. Remember that all chemical reactions involve a change in the bonds of the reactants.
Explanation:
I got you
The molecular formula for the combustion of butane in oxygen is:
2 C4H10 + 13 O2 ---> 8 CO2 + 10 H20
<span>You take the mass of carbon dioxide, 56.8g, divide by its molar mass, 44.01g/mol, to produce the moles of carbon dioxide. This is multiplied by the molar ratio of butane/CO2, (2/8) = 1/4, which gives the moles of butane required to produce the carbon dioxide. Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g</span>
One molecule of water contains two atoms of hydrogen and one atom of oxygen, the atomicity of water is three.