Answer:
See explanation
Explanation:
The major difference between distance and displacement is that distance is a scalar quantity while displacement is a vector quantity. Scalar quantities have magnitude without direction while vector quantities have both magnitude and direction.
The distance covered by Benjamina in both cases just refer to the difference in her position at each time without reference to the direction in which she changed her position.
Her displacement will clearly mention the direction in which she moved both when she walked to the front door of her apartment and when she walked to a friend's apartment.
Delta H of solution = -Lattice Energy + Hydration
<span>Delta H of solution=- (-730)+(-793) </span>
<span>Delta H of solution= -63kJ/mol </span>
<span>Now we find moles of LiI: </span>
<span>10gLiI/133.85g=.075moles </span>
<span>multiply moles to the delta H of solution to cross cancel moles. .75moles x -64kJ/mol =4.7</span>
Strongest reducing agents are in Group 1 . For example lithium. The strongest oxidising agents are in Group 7 , For example Fluorine.
Answer:
Cost to supply enough vanillin is 
Explanation:
Threshold limit of vanillin in air is
per litre means there should be
of vanillin in 1L of air to detect aroma of vanillin.

So, 
So amount of vanillin should be present to detect = 
As cost of 50 g vanillin is
therefore cost of
vanillin = 