Answer:
C₁₃H₁₈O₂
Explanation:
Empirical formula is the simplest ratio in moles of the elements presents in the compound. Assuming a basis of 100g, we need to convert the mass of each element to moles and, find the ratio of each of these:
<em>Moles C:</em>
75.69g * (1mol / 12.01g) = 6.3 moles of C
<em>Moles H:</em>
8.80g * (1mol / 1.01g) = 8.7 moles of H
<em>Moles O:</em>
15.51g * (1mol / 16g) = 0.97 moles of O
The ratio dividing in the moles of O -Because is the element with the lower amount of moles-:
C = 6.3mol / 0.97mol = 6.5
H = 8.7mol / 0.97mol = 9
O = 0.97mol / 0.97mol = 1
As the ratio must be in whole-numbers, multiplying in 2:
C = 13;
H = 18
O = 2
And empirical formula is:
<h3>C₁₃H₁₈O₂</h3>
Answer:
so i never did this befor so i am sorry hope you get the answer
Explanation:
(I think you have a mistake in your question as the addition is 30mL, not 100mL)
when PH = - ㏒[H+]
and here we have HClO4 is the strong acid
So PH = - ㏒[HClO4]
moles of HClO4 = 0.1 L *0.18 m = 0.018 M
moles of LiOH = 0.03 L * 0.27 m = 0.0081 M
when the total volume = 0.1L + 0.03L = 0.13 L
∴ [HClO4] = (0.018-0.0081)/0.13 L
= 0.076 M
PH = -㏒ 0.076
= 1.12
The mixture of rock particle sand humus is called the soil.
If soil contains greater proportion of big particles it is called sandy soil. If the proportion of fine particles is relatively higher, then it is called clayey soil. If the amount of large and fine particles is about the same, then the soil is called loamy.
Answer:
It is <u> </u><u>propyl</u><u> </u><u>amine</u><u>.</u>
<u>as</u><u> </u><u>amine</u><u> </u><u>=</u><u>H-N-H</u>
<u>and</u><u> </u><u>it</u><u> </u><u>contain</u><u> </u><u>3</u><u>C</u><u> </u><u>=</u><u>prop</u><u> </u>