Data: molar mass 470 g/mol
Percent composition:
Hg = 85.0%
Cl = 15.0%
Solution:
1) Convert % to molar ratios
A. Base: 100 g
=> Hg = 85.0 g / 200.59 g/mol = 0.4235 mol
Cl = 15.0 g / 35.45 g/mol = 0.4231 mol
B. divide by the higher number and round to whole number
Hg = 0.4325 / 0.4231 = 1.00
Cl = 0.4231 / 0.4231 = 1.00
=> Empirical formula = Hg Cl
2) Find the mass of the empirical formula:
HgCl: 200.59 g/mol + 35.45 g/mol = 236.04
3) Determine how many times is the empirical mass contained in the molecular mass:
470 g/mol / 236.04 = 1.99 ≈ 2
=> Molecular formula = Hg2 Cl2.
Answers:
Empirical formula HgCl
Molecular Formula Hg2Cl2
B) Seawater. Because, it all has the same consistency. With the other choices, like vegetable soup, in one spoon full you may get a bit of potato but in another spoon full you may get a lima bean.
144 years, its simple math?
Answer:The correct answer is ;
The oxidation state of nitrogen in NO changes from +2 to 0, and the oxidation state of carbon in CO changes from +2 to +4 as the reaction proceeds.
Explanation:
In an oxidation recation addition of oxygen atom takes place or loss of electrons takes place.
In an reduction reaction removal of oxygen atom takes place or gain of electrons takes place.
In the given reaction , the nitrogen atom is present in +2 oxidation state in NO molecule and present in 0 oxidation state in 
molecule. Hence, nitrogen is getting reduced that is reduction reaction. NO is oxidizing agent
In the given reaction , the carbon atom is present in +2 oxidation state in CO molecule and present in +4 oxidation state in 
molecule. Hence ,carbon is getting oxidized that is oxidation reaction. CO is a reducing agent.
Answer:
43.75 g of Nitrogen
Explanation:
We'll begin by calculating the mass of 1 mole of NH₄NO₃. This can be obtained as follow:
Mole of NH₄NO₃ = 1 mole
Molar mass of NH₄NO₃ = 14 + (4×1) + 14 + (3×16)
= 14 + 4 + 14 + 48
= 80 g/mol
Mass of NH₄NO₃ =?
Mass = mole × molar mass
Mass of NH₄NO₃ = 1 × 80 = 80 g
Next, we shall determine the mass of N in 1 mole of NH₄NO₃.
Mass of N in NH₄NO₃ = 2N
= 2 × 14
= 28 g
Thus,
80 g of NH₄NO₃ contains 28 g of N.
Finally, we shall determine the mass of N in 125 g of NH₄NO₃. This can be obtained as follow:
80 g of NH₄NO₃ contains 28 g of N.
Therefore, 125 g of NH₄NO₃ will contain = (125 × 28) / 80 = 43.75 g of N.
Thus, 125 g of NH₄NO₃ contains 43.75 g of Nitrogen
