C.
This is basically the definition of percent composition which is how much mass a element contributes to a specific molecule.
Answer:
The answer to your question is:
a) 31.75 cm
b) 0.475 miles
c) 2.44 yards
d) 11496.04 inches
Explanation:
Convert
a) 12.5 in to cm
1 in ------------------- 2.54 cm
12.5 in ---------------- x
x = 12.5(2.54)/1 = 31.75/ = 31.75 cm
b) 2513 ft to miles
1 mile -------------- 5280 ft
x miles ------------ 2513 ft
x = 2513(1)/5280 = 0.475 miles
c) 2.23 m to yards
1 m ------------- 1,094 yards
2.23 m ---------- x
x= 2.23x1.094/1 = 2.44 yards
d) 292 m to inches
1 m ---------------- 39.37 inches
292 m ------------- x
x = 292 x 39.37/1 = 11496.04 inches
Answer:
Contains DNA, Contains Ribosomes, Lacks a nucleus
Explanation:
Answer:
positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).
The question is incomplete, the complete question is;
Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane? product. A) The terminal alkyne is more stable than the internal alkyne and is naturally the favored B) The terminal alkyne is not favored in this reaction. C) The resonance favors the formation of the terminal rather than internal alkyne. D) The strong base deprotonates the terminal alkyne and removes it from the equilibrium.
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Answer:
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Explanation:
In this reaction, sterric hindrance plays a very important role. We know that sodamide is a strong base, it tends to attack at the most accessible position.
The first deprotonation yields an alkene. The strong base attacks at the terminal position again and yields the terminal alkyne. Thus the structure of the dihalide makes the terminal hydrogen atoms most accessible to the base. Hence the answer.
