Lajdhskskdhskhdksnsjsndjs
false my dude......;;;;;;;;;;;;;;;[[[[[[
Answer:
1) The volume occupied by an atom is composed of mainly empty space
2) Atoms have a very small, relatively dense, central nucleus that is positively charged
3) The region around the nucleus of an atom are orbited by negatively charged electrons in a the same fashion planets orbit around the Sun.
Explanation:
The selection of gold for the gold foil experiment was due to its ability to be rolled into extremely thin sheets such that it was expected for alpha particle to perforate or pass through the foil.
This problem is requiring the empirical formula for CaCO₃, which is its molecular formula, and turns out to be equal, this is A. CaCO3 according to the following:
<h3>Empirical formulas:</h3><h3 />
In chemistry, molecular formulas show both the actual type and number of atoms in a chemical compound, based on the elements across the periodic table and the subscripts standing for the number of atoms in the compound.
However, the empirical formula is a reduced expression of the molecular one, which shows the minimum number of atoms in a compound after simplifying to the smallest whole numbers.
In such a way, since the given compound is CaCO₃ and both Ca and C have a one as their subscript, it is not possible to simplify any further and therefore the empirical formula equals the molecular one this time, making the answer to be A. CaCO3.
Learn more about empirical formulas: brainly.com/question/1247523
The reaction involved in this problem is called the combustion reaction where a hydrocarbon reacts with oxygen to product carbon dioxide and water. The reaction of C2H5OH would be as follows:
C2H5OH + 3O2 = 2CO2 + 3H2O
To determine the number of molecules of CO2 that is formed, we need to determine the number of moles produced from the initial amount of C2H5OH and the relation from the reaction. Then we multiply avogadros number which is equal to 6.022x10^23 molecules per mole.
2.00 g C2H5OH ( 1 mol C2H5OH / 46.08 g C2H5OH ) ( 2 mol CO2 / 1 mol C2H5OH ) = 0.0868 mol CO2
0.0868 mol CO2 ( 6.022x10^23 molecules / mol ) = 5.23x10^22 molecules CO2