Hi
The volume is 58.1
The moler mass is 10.8
Molar mass details
10.81B (1*10.81)
Hope this helps
Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration of
that should used originally is 
Explanation:
From the question we are told that
The necessary elementary step is

The time taken for sixth of 0.5 M of reactant to react 
The time available is 
The desired concentration to remain
Let Z be the reactant , Y be the first product and X the second product
Generally the elementary rate law is mathematically as

Where k is the rate constant ,
is the concentration of Z
From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )
For second order reaction

Where
is the initial concentration of Z which a value of 
From the question we are told that it take 9 hours for the concentration of the reactant to become


So


=> 
For 





Answer:

Explanation:
Hello,
In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

Best regards.
Explanation:
Ions are always formed when metals and non-metals interact because metals are electropositive. They willing release electrons to non-metals that are electronegative.
This activity results in charge separation. The transfer of electrons from one specie to another is what results in an ionic bond and the precedence of charged particles.
Between non-metals, the electrons are jointly shared. Therefore, there is no charge separation.
The ideal gas under STP is 22.4 L/mol. While the gas has a rule of P1V1/T1=P2V2/T2. So the volume under 101 kPa and 273 K is 0.2*22.4=4.48 L.