9 Ways to Prevent Rust
Use an Alloy. Many outdoor structures, like this bridge, are made from COR-TEN steel to reduce the effects of rust. ...
Apply Oil. ...
Apply a Dry Coating. ...
Paint the Metal. ...
Store Properly. ...
Galvanize. ...
Blueing. ...
Powder Coating.
Answer:
6Fe^2+(aq) -------> 6Fe^3+(aq) + 6e
Explanation:
The balanced oxidation half equation is;
6Fe^2+(aq) -------> 6Fe^3+(aq) + 6e
A redox reaction is actually an acronym for oxidation-reducation reaction. Since the both reactions are complementary, there can't be oxidation without reduction and there can't be reduction without oxidation.
The main characteristic of redox reactions is that electrons are transferred in the process. The number of electrons transferred is usually deduced from the balanced reaction equation. For this reaction, the balanced overall reaction equation is;
Cr2O7^2–(aq) + 6Fe^2+(aq) +14H^+(aq)→ 2Cr^3+(aq) + 6Fe^3+ (aq) + 7H2O(l)
It is clear from the equation above that six electrons were transferred. Thus six Fe^2+ ions lost one electron each in the oxidation half equation as shown in the balanced oxidation half equation above.
Answer:
In conclussion, 0.60 moles of HCOOH contains the greatest mass of O
Explanation:
Let's make some rules of three, to solve this problem:
1 mol of ethanol has 2 moles of C, 6 moles of H, and 1 mol of oxygen
Therefore, 0.75 moles of ethanol must have 0.75 mol of oyxgen
Let's convert the moles to mass → 0.75 mol . 16 g/ 1 mol = 12 g
1 mol of formic acid has 2 moles of H, 1 mol of C and 2 mol of oxygen
0.60 moles of formic acid must have (0.6 .2) / 1 = 1.2 mol of O
If we convert the amount to mass → 1.2 mol . 16 g/ 1mol = 19.2 g
1 mol of water has 1 mol of oyxgen
Therefore, we have 1 mol of oxygen with a mass of 16 g.
In conclussion, 0.60 moles of HCOOH contains the greatest mass of O
Answer:
0.558mole of SO₃
Explanation:
Given parameters:
Molar mass of SO₃ = 80.0632g/mol
Mass of S = 17.9g
Molar mass of S = 32.065g/mol
Number of moles of O₂ = 0.157mole
Molar mass of O₂ = 31.9988g/mol
Unknown:
Maximum amount of SO₃
Solution
We need to write the proper reaction equation.
2S + 3O₂ → 2SO₃
We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.
So we simply compare the molar relationship between sulfur and product formed to solve the problem:
First, find the number of moles of Sulfur, S:
Number of moles of S = 
Number of moles of S = 
= 0.558mole
Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:
2 mole of Sulfur produced 2 mole of SO₃
Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃
The volume of the dry gas at stp is calculated as follows
calculate the number on moles by use of PV =nRT where n is the number of moles
n is therefore = Pv/RT
P = 0.930 atm
R(gas contant= 0.0821 L.atm/k.mol
V= 93ml to liters = 93/1000= 0.093L
T= 10 + 273.15 = 283.15k
n= (0.930 x0.093) /(0.0821 x283.15) = 3. 72 x10^-3 moles
At STp 1 mole = 22.4L
what about 3.72 x10^-3 moles
by cross multiplication
volume = (3.72 x10^-3)mole x 22.4L/ 1 moles = 0.083 L or 83.3 Ml
