Copper nitrate and nitric oxide are produced in this reaction.
Answer:
36
Explanation:
Since the sample was undiluted the number of colonies is the number that grew on the nutrient agar which is 36 colonies. If it was diluted for example let say 0.1 ml from a dilution in which 1 ml of the sample was added to 9 ml of water, and it grew colonies then 0.1 ml yielded 6 colonies, 1 ml of the diluted sample will yield 60 colonies and 10 ml will have 600 colonies and therefore the 1 ml undiluted sample will have 600 colonies.
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
