Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
I believe the correct term that would fit the statement would be greater than. In a spontaneous fusion reaction, the total mass of the products is greater than the mass of the original elements. This nuclear reaction involves at least two nuclei that fuses to form one nuclei having larger mass than that of the reactant.
The chemical equilibrium will be unaffected
Answer:
Approximately 
.
Explanation:
Nitrogen 
reacts with hydrogen 
at a 
ratio to produce ammonia 
:
.
The ratio between the coefficient of and the coefficient of is:
.
Under the ideal gas assumptions, the same ratio would apply to the volume of and in this reaction:
.
.
Given that 
:
.
(Rounded to 
significant figures.)
Answer:
0.15 M KCl
Explanation:
(Step 1)
Convert the original volume from mL to L. Calculate the moles of the original solution.
75.0 mL / 1,000 = 0.0750 L
Molarity = moles / volume (L)
0.200 M = moles / 0.0750 L
0.0150 = moles
(Step 2)
Convert the final volume from mL to L. Calculate the final molarity.
100. mL / 1,000 = 0.100 L
Molarity = moles / volume (L)
Molarity = 0.0150 moles / 0.100 L
Molarity = 0.15 M
