Answer:
115.2 °C since melting point is an intensive property
Step-by-step explanation:
The melting point of a substance does not depend on how much you have.
For example, the melting point of water is 0 °C, whether it is an ice cube from the refrigerator or in the frozen pond outside.
The freezing point of a substance is an <em>intensive property</em>.
Thus, the melting point of 100 g of sulfur is 115.2 °C because melting point in an intensive property.
Answer:
Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt
The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1
E decreseas 3/2 as fast as G increases = 0.30 M/s
Explanation:
Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt
When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:
Given data = d[D]/dt = 0.10 M/s
-d[D] / 2dt = d[H]/dt
d[H]/dt = 0.05 M/s
The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1
When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:
d[G] / 2dt = -d[H]/3dt
E decreseas 3/2 as fast as G increases = 0.30 M/s
Answer:
C. 1 cubic foot of loose sand
Explanation:
For many objects having equal volume , surface area will be maximum
of the object which has spherical shape .
But when a sphere is broken into tiny small spheres , total surface area of all the small spheres will be more than surface area of big sphere .
Hence among the given option , surface area of loose sand will have greatest surface area . Loose sand is equivalent to small spheres .
Overuse of the same chemicals can result in the pest becoming immune to the pesticides.
Strongest reducing agents are in Group 1 . For example lithium. The strongest oxidising agents are in Group 7 , For example Fluorine.
