First, we determine the mass of each element from the data collected. We can get the mass of molybdenum Mo from the difference between the mass of crucible and molybdenum and the mass of crucible:
Mass of molybdenum = 39.52 – 38.26 = 1.26 g Mo
We can calculate for the mass of molybdenum oxide from the difference between the mass of crucible and molybdenum oxide and the mass of crucible:
Mass of molybdenum oxide = 39.84 – 38.26 = 1.58g
We can now compute for the mass of oxygen O by subtracting the mass of molybdenum from the mass of molybdenum oxide:
Mass of oxygen in molybdenum oxide = 1.58 – 1.26 = 0.32g O
To convert mass to moles, we use the molar mass of each element.
1.26 g Mo * 1 mol Mo / 95.94 g Mo = 0.0131 mol Mo
0.32 g O * 1 mol O / 15.999 g O = 0.0200 mol O
0.0131 mol is the smallest number of moles. We divide each mole value by this number:
0.0131 mol Mo / 0.0131 = 1
0.0200 mol O / 0.0131 = 1.53
Multiplying these results by 2 to get the lowest whole number ratio,
0.0131 mol Mo / 0.0131 = 1 * 2 = 2
0.0200 mol O / 0.0131 = 1.5 * 2 = 3
Thus, we can write the empirical formula as Mo2O3.
I don't think so. No way that I know anyway. It it could be done then the need for more coal to be mined would have stopped hundreds of years ago. Once coal is burned, it forms water and carbon dioxide (essentially) with some sulfur oxides.
How do you put that back together again. It's a little like humpty dumpty.
Answer:
Mark has a speed of 6 MPS (Miles Per Hour)
Explanation:
It took mark 2 hours to ride his bike to his grandma's house, which was 12 miles away. I divided 2 by 12 and got 6. Remember this: mph = miles away ÷ time.
The atom<span> then has more protons than electrons and so it will be positively charged, a positive </span>ion<span>. Example: A </span>magnesium atom<span> may lose two electrons and </span>become<span> a Mg2+ </span>ion<span>. Non-metal </span>atoms<span> may gain electrons and </span>become<span> negatively charged. ... (It loses two electrons.)</span>
Wind isn’t an external force that acts in the rock cycle.
