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.
35°c is equal to 95°f
To do this multiply 35 and 1.8
35 x 1.8=63
Now add 32
Resulting in the answer 95
(The equation for to solve for c and f is c1.8+32=f
The answer is "chemical properties". The original water is split apart by the current
<span>to form its constituent elements, hydrogen & oxygen. This is a chemical change, as </span>
<span>the original water is lost and new substances, H2 & O2, are produced. </span>
<span>Hope this answers your question.</span>
Answer:
0.1082M of Barium Hydroxide
Explanation:
KHP reacts with Ba(OH)2 as follows:
2KHP + Ba(OH)2 → 2H2O + Ba²⁺ + 2K⁺ + 2P²⁻
<em>Where 2 moles of KHP reacts per mole of barium hydroxide</em>
<em />
To solve this question we must find the moles of KHP in 1.37g. With these moles and the reaction we can find the moles of Ba(OH)2 and its molarity using the volume of the solution (31.0mL = 0.0310L) as follows:
<em>Moles KHP -Molar mass: 204.22g/mol-</em>
1.37g * (1mol / 204.22g) = 0.006708 moles KHP
<em>Moles Ba(OH)2:</em>
0.006708 moles KHP * (1mol Ba(OH)2 / 2mol KHP) =
0.003354 moles Ba(OH)2
<em>Molarity:</em>
0.003354 moles Ba(OH)2 / 0.0310L =
<h3>0.1082M of Barium Hydroxide</h3>
