Answer:
Option C.
2 Mg (s) + O₂(g) → 2MgO (s)
Explanation:
Two moles of magnesium solid react with one mol of oxygen gas to
form two moles of magnesium-oxide solid
2 Mg (s) + O₂(g) → 2MgO (s)
That's the reaction for the magnessium oxide's formation.
Be careful cause we do not say molecules, they are moles.
The stoichiometry indicates the number of moles that react and the moles which are produced.
It is a redox reaction, because the magnessium is oxidized and the oxygen is reduced. Both elements, changed the oxidation states.
In a neutral atom they are both equal, and their even quantities makes the atom neutral...
Answer:
1.38 M
Explanation:
Need to use the Molarity equation M=n/L
23.5g/ 17.031g/mol NH3 = 1.38 moles
1.38 moles/ 1.0 L = 1.38 M
Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu
