2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
Answer: The oxidation state of a free element (uncombined element) is zero. For a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion. For example, Cl– has an oxidation state of -1. When present in most compounds, hydrogen has an oxidation state of +1 and oxygen an oxidation state of −2.
hope this helps........ Stay safe and have a Merry Christmas!!!!!!!!! :D Explanation:
Electron has the smallest mass
Answer:
The arguments both for and against allowing genetically altered meats into the United States is explained below in details.
Explanation:
Arguments for and against genetically altered meats:
- Supply the Society. By 2050, the world's inhabitants are supposed to increase from today's 7 billion to way beyond 9 billion.
- ENVIRONMENTAL RISK.
- STRONGER CROPS = LESS PESTICIDES.
- REMEMBER WHEN CIGARETTES WERE 'HARMLESS'?
- TAMPERING FOR TASTE. ...
- BIG BUSINESS EATS SMALL FARMERS.
- ENHANCED HEALTH.
- NOTHING TASTES BETTER THAN NATURE.
Answer:
There are 20.8 g of fluorine in 55.5 g of copper (II) fluoride
Explanation:
x % by mass of a species in a specimen means there are x g of the species in total 100 g of a specimen
37.42 % F by mass means 100 g of copper (II) fluoride contains 37.42 g of F.
So, 100 g of copper (II) fluoride contains 37.42 g of F
55.5 g of copper (II) fluoride contains
g of F or 20.8 g of F
Hence there are 20.8 g of fluorine in 55.5 g of copper (II) fluoride.