Moles of Zn: 26 / 65 = 0.4
Moles of S: 12.8 / 32 = 0.4
Molar ratio of Zn : S = 1 : 1
Empircal formula: ZnS
The answer is C
Decomposed organic matter ,mud,silt ,and sand
The last one will be It's better to use the average data instead of the single trail because the average you don't have to multiply add subtract and all that nonsense but if you use the single trail you will have to do all the adding multiplying etc.
We know that the element Z = 119 would be placed right below the Fr, in the column of the alcaline metals.
We also know that the trend in the electronegativity is to decrease when you go up-down ia group.
The known electronegativities of the elements of this group are:
Li: 0.98
Na: 0.93
K: 0.82
Rb: 0.82
Cs: 0.79
Fr: 0.70
Then the hypotetical element Z = 119 would probably have an electronegativity slightly below 0.70, for sure in the range 0.60 - 0.70.