Answer:
Explanation:
Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.
The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:
Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:
The net equation is then:
However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:
Actual anode half-equation:
Actual cathode half-equation:
Actual net reaction:
Answer:
Are you in flvs, if so im prettyb sure if yo look on page 3 of lesson 1.04 it tells you the answer.
Explanation:
Potassium Cyanide, is the compound of KCN.
d = diameter of the seat = 12 inch
we know that , 1 inch = 0.0254 m
hence
d = diameter of the seat = 12 inch = 12 (0.0254) m = 0.305 m
r = radius of seat = diameter/2 = d/2 = 0.305/2 = 0.1525 m
A = area of seat = πr²
m = mass of the person = 65 kg
g = acceleration due to gravity = 9.8 m/s²
F = force due to weight of person on the seat = mg
pressure on the seat is given as
P = F/A
P = mg/A
P = mg/(πr² )
inserting the values
P = (65 x 9.8)/((3.14) (0.1525)²)
P = 8723.1 N/m²
Answer:
You need 8,53 L of ammonia
Explanation:
Global reaction of remotion of nitrogen oxide with ammonia is:
4 NH₃ + 6 NO ⇒ 5 N₂ + 6 H₂O
This balanced equation shows that 4 NH₃ moles reacts with 6 NO moles.
With 100% yield and temperature and pressure constants it is possible to apply Avogadro's law. This law is an experimental gas law relating the volume of a gas to the amount of substance of gas present. The formula is:
Where:
V₁ is the NO volume = 12,8L
n₁ are NO moles = 6
n₂ are NH₃ moles = 4
V₂ is NH₃ volume, the unknown.
Thus, V₂ are 8,53 L of ammonia
I hope it helps!
