Answer:
The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.
The answer is 4.
Gases have low densities, because of the increased space between hight-energy particles.
Answer: ![C_2H_2O_2](https://tex.z-dn.net/?f=C_2H_2O_2)
Explanation:
Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.
Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.
The empirical formula is ![CHO](https://tex.z-dn.net/?f=CHO)
The empirical weight of
= 1(12)+1(1)+1(16)= 29 g.
The molecular weight = 60 g/mole
Now we have to calculate the molecular formula:
![n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{60}{29}=2](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%5Ctext%7BMolecular%20weight%7D%7D%7B%5Ctext%7BEquivalent%20weight%7D%7D%3D%5Cfrac%7B60%7D%7B29%7D%3D2)
The molecular formula will be=![2\times CHO=C_2H_2O_2](https://tex.z-dn.net/?f=2%5Ctimes%20CHO%3DC_2H_2O_2)
Thus molecular formula will be ![C_2H_2O_2](https://tex.z-dn.net/?f=C_2H_2O_2)
Answer:
Explanation:
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