Answer:
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Answer:
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Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
The solution would be like this for this specific problem:
<span>Given:
H2 = </span><span>2.6 atm
CL2 = 3.14 atm</span>
<span>
pressure H2 = 2.6 - x
pressure Cl2 = 3.14 - x
<span>pressure HBr = 2x = 1.13
x = 1.13 / 2 = 0.565
<span>pressure H2 = 2.6 - 0.565 = 2.035
pressure Br2 = 3.14 - 0.565 = 2.575
Kp = (1.13)^2 / 2.035 x 2.575</span></span></span>
= 1.2769 / (5.240125)
= 0.24367739319195629875241525726963
= 0.244
<span>Therefore, the Kp for the reaction at the given temperature
is 0.244.
To add, </span>the hypothetical pressure of a gas if
it alone occupied the whole volume of the original mixture at the same
temperature is called the partial pressure or Kp.
