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Maru [420]
3 years ago
11

ASAP MULTIPLE CHOICE WILL MARK BRAINLIEST

Chemistry
1 answer:
yaroslaw [1]3 years ago
6 0

Answer:

There are four electrons

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Why isn't every point exactly on the 'best fit line'? Give a reason and explain
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How does a concave lens differ from a convex lens? A) A convex lens is another way to describe a plane mirror; a concave lens is
elixir [45]
<h3><u>Answer;</u></h3>

C) A convex lens has a thick center and thin edges; a concave lens has a thin center and thicker edges.

<h3><u>Explanation;</u></h3>
  • Convex lens refers to the lens which merges the light rays at a particular point, that travels through it, while a concave lens can be identified as the lens which disperses the light rays around, that hits the lenses.
  • A convex lens is thicker at the center, as compared to its edges, while a concave lens is thinner at the center as compared to its edges.
  • A concave lens is thicker at the edges than in the middle and spreads light rays apart producing an image smaller than the actual object. A convex lens on the other hand, is thinner at the edges and thicker towards the center, that is they are bent towards a central point.
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In the figure shown, AD, CE, and FB intersect at point F.
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7 0
3 years ago
What kind of change occurs with no new substance being formed apex?
Andrew [12]
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3 years ago
What is the value of the equilibrium constant at 25 oC for the reaction between the pair: I2(s) and Br-(aq) Give your answer usi
insens350 [35]

Explanation:

Formula to calculate standard electrode potential is as follows.

          E^{o}_{cell} = E^{0}_{cathode} - E^{0}_{anode}

                             = 0.535 - 1.065

                             = - 0.53 V

Also, it is known that relation between E^{o}_{cell} and K is as follows.

            E^{o}_{cell} = \frac{RT}{nF} \times ln K

                 ln K = \frac{nFE^{0}_{cell}}{RT}      

Substituting the given values into the above formula as follows.

                 ln K = \frac{nFE^{0}_{cell}}{RT}    

                        =  \frac{2 \times 96485 C mol^{-1} \times -0.53 V}{8.314 l atm/mol K \times 298 K} \times \frac{1 J}{1 V C}  

                ln K = -41.28

                    K = e^{-41.28}    

                        = 1 \times 10^{-18}

Thus, we can conclude that the value of the equilibrium constant for the given reaction is 1 \times 10^{-18}.      

5 0
3 years ago
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