The empirical formula of the compound is C4H1.
An empirical formula represents the whole number ratio of various atoms present in a compound.
If the mass percent of various elements present in a compound is known, its empirical formula can be determined
We are given that a compound contains 4 mole C and 1 mole H.
Hence, the ratio of C: H is 4:1, which is already in the form of the simplest whole number.
Therefore, the empirical formula of the given compound is C4H1 and its molecular mass is 12 x 4 + 1 = 49.
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N2 + 3H2 ---> 2NH3
nitrogen + hydrogen ---> ammonia
0.00702906176 moles
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Mercury (ii) oxide is made up of mercury and oxygen. The total mass of mercury (ii) oxide is 14.2 g, after decomposition 13.2 g of mercury were formed, therefore the mass of oxygen 1 g (14.2 g -13.2 g).
Percentage of oxygen = (1/14.2)×100 = 7.04%
Percentage of mercury = (13.2/14.2) × 100 = 92.96%
Therefore, percentage composition of the compound, oxygen is 7.04% and mercury is 92.96%.
Answer:
Mg would blow off. AI would be affective to copper but not to MG
Explanation:
