Step 1
The reaction is written and balanced:
4 Rb + O2 =>2 Rb2O
-----------
Step 2
Define % yield of product (Rb2O) = (Actual yield/Theoretical yield) x 100
The actual yield is provided by the exercise = 39.7 g
----------
Step 3
Determine the limiting reactant. The molar masses are needed to solve this:
For Rb) 85.4 g/mol
For O2) 32 g/mol
Procedure:
4 Rb + O2 =>2 Rb2O
4 x 85.4 g Rb ----- 32 g O2
82.4 g Rb ----- X = 7.72 g O2 are needed
For 82.4 g Rb, 7.72 g O2 is needed, but there is 11.6 g O2. Therefore, O2 is the excess agent. Rb is the limiting reactant.
--------
Step 4
Determine the theoretical yield from the limiting reactant:
The molar mass Rb2O) 187 g/mol
Procedure:
4 x 85.4 g Rb ------ 2 x 187 g Rb2O
82.4 g Rb ------ X = 90.2 g Rb2O = Theoretical yield
---------
Step 5
% yield = Actual y./Theoretical y. x 100 = (39.7 g/90.2 g) x 100 = 44 % approx.
Answer: % yield = 44 %
The answer would be how much greenhouse gas does your fuel produce compared with the current fuel sources.
hope this helps
Bohr suggested, that there are definitive shells of particular energy and angular momentum in which an electron can revolve. It was not in Rutherford's model
You want to divide by avagadros number (6.22 x 10^23). This will cancel the atoms unit and give moles, moles of Iridium. Now you want to calculate the atomic mass of Iridium which is in units of grams per mole. Multiply these two numbers and the moles will cancel giving you grams.
Setting up a dimension analysis type of thing helps tremendously. See what you have to cancel in order to get what you want. We canceled the atoms, then we canceled the moles, and then we got grams.
Explanation:
There are two aspects of a reaction:
Thermodynamics:
Combustion of the compound is reaction with oxygen.
The carbohydrates contain more content of oxygen as compared to fats.
Hence, carbohydrates have lot of oxygen content which are already partially oxidized as compared to fats. Hence, combustion of the carbohydrates is a faster process.
Kinetics:
The molecular reacts also shows that the combustion of the carbohydrates is a faster process.
