The line segment is describe as a line which joints two points. The word segment is denotes the end points of the line. The symbol of the line segment is AB. This symbol is indicating a line from the point A to the point B.
- Generally, the line segment has two endpoints. The meeting point of two line segments is called as the vertex and also it is considered as a common endpoint on the other hand line is a straight path of points that goes on and on in two directions
- Vertex is a point where two rays meet OR where the sides of a polygon meet OR the point where three or more edges of a solid figure meet and the angle is Formed by two rays that have the same endpoint
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In three hours and twenty minutes from 1:40 will be 5:00
2Cu + S = Cu₂S
S⁰ + 2e⁻ = S⁻²
Cu⁰ - 1e⁻ = Cu⁺¹
A sulfur atom gains two electrons.
Answer:
5000 and 
indicate that there is more B than A at equilibrium
Explanation:
For the given reaction: ![K=\frac{[B]}{[A]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
where [B] and [A] represents equilibrium concentration B and A respectively. K represents equilibrium constant
More B than A at equilibrium means, [B] > [A]
So, ![K=\frac{[B]}{[A]}>1](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D%3E1)
As, both 5000 and are greater than 1 therefore these two K values indicate that there is more B than A at equilibrium
Answer:
True
Explanation:
In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].
The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.
This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.
Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.
