The empirical formula for pyrite is FeS2.
HOW TO CALCULATE EMPIRICAL FORMULA:
- The empirical formula represents the simplest whole number ratio of constituents element of a compound. The empirical formula of pyrite can be calculated as follows:
46.5 mass % Fe = 46.5g of Fe
53.5 mass % S = 53.5g of S
- Next, we divide each element's mass value by its molar mass
Fe = 46.5g ÷ 56g/mol = 0.83mol
S = 53.5g ÷ 32g/mol = 1.67mol
- Next, we divide each mole value by the smallest (0.83mol)
Fe = 0.83mol ÷ 0.83 = 1
S = 1.67mol ÷ 0.83 = 2.014
Approximately, the ratio of Fe to S is 1:2. Therefore, the empirical formula of pyrite is FeS2.
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Answer:
See explanation below
Explanation:
First, we need to understand that the monochlorination of an alkane like this one, involves substitution of one of the atoms of hydrogen of the molecule for an atom of chlorine.
This reaction takes place when the alkane reacts with Cl₂ in presence of light or heat.
When this happens, the first step involves the breaking of the double bond of the chlorine to form the ion Cl⁻.
The next step involves the substraction of the hydrogen of the molecule by the Chlorine. This will leave the alkane with a lone pair available for reaction.
The third step, the alkane with the lone pair of electron substract a chlorine for the beggining and form the mono chlorinated product.
The final step involves forming the remaining products with the remaining reagents there.
In the picture attached you have the mechanism and product for this reaction:
The reaction that would beat represent a typical reaction mechanism is when A and B react with each other to produce certain products.
Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2
