Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams
One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams
Based on the above calculations, the sample is not CaCO3
Answer:
it is Calcium (Ca)
4th period, 2nd group, 2 valence electrons
Answer:
That is a compound. If it was an element it would either just be Na or Cl.
In the physical properties, it is mentioned that the element has 4 valence electrons. The elements in the periodic table are arranged such that, the number of valence electrons present in the neutral atoms belonging to a particular group is equal to the group number.
Thus, the unidentified element can be best classified as a nonmetal in period 4
Ans C)
I would say that descriptive investigations aren't repeatable because it means that you are only describing something - you ask certain question about something, but do not form a hypothesis at that point yet. So it would be a waste of time to simply ask the same questions over and over again with no hypothesis to prove, which is why these types of investigations cannot be repeated.
