Answer:
Empirical formula of compound is C₄H₈O
Explanation:
Given data:
Mass of compound = 5.60 g
Mass of CO₂ = 13.7 g
Mass of H₂O = 5.60 g
Empirical formula of compound = ?
Solution:
Percentage of C:
13.7 g/5.60 g × 12/44× 100
2.45×0.273× 100 = 66.9%
Percentage of H:
5.60 g/ 5.60 g × 2.016/18 × 100
11.2%
Percentage of O:
(66.9% + 11.2%) - 100 = 21.9%
Grams atom of C , H, O
66.9/12 = 5.6
11.2 / 1.008 = 11.11
21.9 / 16 = 1.4
Atomic ratio:
C : H : O
5.6/1.4 : 11.11/1.4 : 1.4/1.4
4 : 8 : 1
Empirical formula:
C₄H₈O
The effects on the concentration of SO3 gas when the
following changes occur after initial equilibrium has been established in this
system (N.C. = no change) by adding a catalyst.
2SO2(g) + O2(g) 2SO3(g) + 46.8 kcal
So that even though there is an addition of catalyst, no
change in reactants or products has occurred because catalyst only provides a
faster pathway for the reaction to occur.
First, since l = n-1,
5,4,-5,1/2 and 2,1,0,1/2 are the only answer choices left.
Next, since ml = -l to l,
2,1,0,1/2
is the answer because in 5,4,-5,1/2, the ml value of -5 is not in the range of -4 to 4, as notes by the value 4 for l.
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