Answer:
Cyclopropane has a planar carbon back bone while propane does not
Explanation:
We have to recognize that in straight chain saturated organic compounds, carbon atoms have a tetrahedral geometry. Each carbon atom is bonded to four other atoms.
However, carbon atoms in cyclic compounds are also sp3 hybridized with each carbon bonded to only four other atoms but the ring system is highly strained.
Cyclopropane is a necessarily planar molecule with a bond angle that is far less than the expected tetrahedral bond angle due to strain in the molecule. Hence, the carbon atoms may have have a "planar backbone".
Answer:
Kauai
Explanation:
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Answer:
1.91×1021/6.023×10^23
Explanation:
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Answer:
18 g
Explanation:
We'll begin by converting 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:
Volume = 0.5 L
Molarity = 0.2 M
Mole of C₆H₁₂O₆ =?
Molarity = mole / Volume
0.2 = Mole of C₆H₁₂O₆ / 0.5
Cross multiply
Mole of C₆H₁₂O₆ = 0.2 × 0.5
Mole of C₆H₁₂O₆ = 0.1 mole
Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:
Mole of C₆H₁₂O₆ = 0.1 mole
Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)
= 72 + 12 + 96
= 180 g/mol
Mass of C₆H₁₂O₆ =?
Mass = mole × molar mass
Mass of C₆H₁₂O₆ = 0.1 × 180
Mass of C₆H₁₂O₆ = 18 g
Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.
