Answer:
The empirical formula is CH3O2
Explanation:
From The question, number of moles of CO2= 4.89/44=0.111mol
Number of moles of H2O=3/18= 0.1667moles
Since 1mole of CO2 contains 1mole of carbon and 2 moles of O2,
Implies that there 0.111 moles of C and 0.222moles of O2
Similarly also 0.1667moles of water contains 2 moles of H2 hence = 0.333 moles of H2
Dividing by smallest number= 0.111mol
For C = 0.111/0.111= 1
For H = 0.333/ 0.111= 3
For O = 0.222/0.111= 2
Hence the empirical formula= CH3O2
Answer is: a. Mg + 2HCl → H₂ + MgCl₂.
Reduction hald reaction: 2H⁺ + 2e⁻→ H₂.<span>
Oxidation half reaction: Mg</span>⁰ → Mg²⁺ + 2e⁻.
In this chemical reaction, magnesium lost two electrons (oxidation, change oxidation number from 0 to +2) and hydrogen gain that two electrons (reduction, change oxidation number from +1 to 0).
Either carbon monoxide or NaOH
Period : the no. of electron shells
group: the last no. of electron configuration
