There are approximately 7.98182*10^27 S atoms in 425.0kg of Sulfur
How many sulfur (S) atoms are there in this chemical formula 3Al₂(SO₄)₃? *
72.0 grams sulfur (1 mole S/32.07 grams)(6.022 X 10^23/1 mole S) = 1.35 X 10^24 atoms of sulfur
32 g S * (1 mole S/32 g S) * (6.022*1023 atoms/1 mole S) = 6.022*1023 atomsThere are 6.022*1023 atoms in 32 grams of sulfur.
multiply 5 with 6.02x10x23 and the answer will be 3.01x10
I rounded up to 1.47 X 10^24 atoms of Sulfur in 78.4 grams Sulfur78.4 grams X (1 mole S/32.07 g S) X (6.02 X 10^23 atoms S/ 1 mole) = 1.472 X 10^24 atomTaking significant digits into account (3 sig. figs.), I get 1.47 X 10^24 atoms S
5 g of sulfur contain 0,94.10e23 atoms.
How many sulfur (S) atoms are there in this chemical formula 3Al₂(SO₄)₃? *
There is only one sulfur atom. The S is sulfur, and there is no number next to it.
6,35 moles of S contain 38,24059444195.10e23 sulfur atoms.
48.0 grams sulfur (1 mole S/32.07 grams)(6.022 X 10^23/1 mole S) = 9.01 X 10^23 atoms of sulfur
6.02 grams sulfur (1mole S/32.07 grams)(6.022 X 10^23/1 mole S) = 1.13 X 10^23 atoms sulfur
72.0 grams sulfur (1 mole S/32.07 grams)(6.022 X 10^23/1 mole S) = 1.35 X 10^24 atoms of sulfur
64 grams sulfur (1 mole S/32.07 grams)(6.022 X 1023/1 mole S) = 1.2 X 1024 atoms of sulfur ====================
"S"
8 g sulfur have 1,5027176187149.1023 atoms.
Full formal set up. 48.096 grams sulfur (1 mole S/32.07 grams)(6.022 X 1023/1 mole S)(1 mole S atoms/6.022 X 1023) = 1.4997 moles of sulfur atoms ---------------------------------------
1mol S = 6.022 E23 atoms S 3.7mol S x 6.022 E23 atoms S/1mol S = 2.2 E24 atoms S Since the superscript and subscripts are not working, I used E to represent "x 10", followed by the exponent like you might find on a calculator.