Answer:
Explanation:
C = 49.48
H = 5.19
O = 16.48
N = 28.85
ratio of moles
= 49.48 / 12 : 5.19 / 1 : 16.48 / 16 : 28.85 / 14
= 4.123 : 5.19 : 1.03 : 2.06
= 4 : 5 : 1 : 2
so the empirical formula = C₄ H₅O N₂
Let molecular formula = ( C₄ H₅ON₂ )ₙ ,
n ( 48 + 5 + 16 + 28 ) = 119.19
97 n = 194.19
n = 2 ( approx )
molecular formula = C₈ H₁₀O₂ N₄
2Cu(s) + O2(g) --> 2CuO(s)
<span>Nitrogen trifluoride - NF3.
1 mol NF3 contains 1 mol atoms of Nitrogen
2.49 mol NF3 contains 2.49 mol atoms of Nitrogen
1 mol ---- 6.02 *10²³ atoms
2.49 mol ----- 2.49*6.02*10²³ = 15.0*10²³ atoms of N</span>
Answer:
The correct answer:
8)- e)2, 2 dymetilpropane
9)- b) 2-chloropropane
10)- b) hydroxyl
See the explanation below, please.
Explanation:
In the cases of exercises 8 and 9:
Correspond to alkanes, having 3 carbons are named with the prefix prop and suffix anus. In 8 it has 2, 2 methyl groups in carbon and 9 in a 2-carbon chlorine group.
In the case of 10, it corresponds to a 3-carbon alcohol: suffix prop, and prefix ol: 2- propanol; in group 2 it has an OH group corresponding to alcohol (hydroxyl).
