How many electrons can be held in the second orbital (closest to the nucleus)
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This problem is requiring the balanced chemical equation that takes place when copper hydroxide and potassium sulfate are produced when reacting potassium hydroxide with copper sulfate.
<h3>Balancing chemical equations:</h3>In chemistry, balancing chemical equations is based on the law of conservation of mass, which demands us to have equal number of atoms on both sides of the chemical equation. This can be accomplished by inserting coefficients in front of the chemical species.
For this particular case, we have potassium hydroxide with copper sulfate on the reactants side, however, copper can be copper (I) or copper (II) as it has 1+ and 2+ as its possible oxidation numbers. In addition, copper hydroxide and potassium sulfate as the products. Hence, we can assume this is all about copper (II) so we can write:
As we can see, potassium, hydrogen and oxygen have two atoms each on the products side, but just one on the reactants side; drawback we can overcome by putting a 2 in front of KOH so as to balance it:
Learn more about balancing chemical equations: brainly.com/question/8062886
So I haven’t got time to answer all of it for you but the id you look at the picture of the periodic table I’ve added the top number in the red boxes are the groups and the period is how many elements down from the top it is (remember that the hydrogen and helium make up period ONE) so remember to include them when counting the elements as you go down the table
