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
See the image below.
An excited electron is in a <em>high-energy state</em>.
When it drops to the lower-level ground state, it must get rid of this excess energy by <em>emitting it</em> as a quantum of light.
Answer:
dS= 1.79*169.504
j/k = 303.41 j/k
Explanation:
Fe3O4(s) + 4H2(g) --> 3Fe (s)+ 4H2O(g)
dS(Fe3O4) =146.4 j/k
dS(H2) =130.684
dS(Fe) =27.78
dS(H2O) =188.825
dSrxn = dS[product]-dS[reactants]
= 3*dS(Fe)+ 4*dS(H2O)-[1*dS(Fe3O4)+ 4dS(H2)]
= [3*27.78 +4*188.825-146.4 -4*130.684] j/k = 169.504 j/k
This is the dS for 1mole Fe3O4
for 1.79 mols Fe3O4
dS= 1.79*169.504 j/k = 303.41 j/k
Its molecular formula : C₆H₁₂O₆
<h3>Further explanation </h3>
The empirical formula(EF) is the smallest comparison of atoms of compound forming elements.
A molecular formula(MF) is a formula that shows the number of atomic elements that make up a compound.
(EF) n = MF
its empirical formula is CH₂O
CH₂O : 12+2+16=30
![\tt [30]n=180\rightarrow n=6\\\\(CH_2O]_6=C_6H_{12}O_6](https://tex.z-dn.net/?f=%5Ctt%20%5B30%5Dn%3D180%5Crightarrow%20n%3D6%5C%5C%5C%5C%28CH_2O%5D_6%3DC_6H_%7B12%7DO_6)
