Find if the example is correct or not and explain using CER please. Will give 5/5
1 answer:
The Lewis structure the student drew is incorrect. The total number of electrons in bonds and lone pairs is equal to the total number of valence electrons available, then a valid Lewis structure is produced.
The correct structure is the following:
The structure that was drawn by the student was incorrect because the total number of electrons around Oxygen were incorrect, there were 6 instead of 8, which was against the octet rule. The correct structure should have 8 electrons around each atom.
You might be interested in
Answer: 1.06 kg
Explanation:
Density is defined as the mass contained per unit volume.
Given :
Density of mercury =
volume of cylinder = volume of mercury =
where r = radius of cylinder =
h= height = 25.0 cm
Putting in the values we get:
volume of mercury =
(1kg=1000g)
Thus mass of mercury required to fill a hollow cylinder having an inner diameter of 2.00 cm to a height of 25.0 cm is 1.06 kg.
Answer:
17.65 grams of O2 are needed for a complete reaction.
Explanation:
You know the reaction:
4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O
First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:
- N: 14 g/mol
- H: 1 g/mol
- O: 16 g/mol
So, the molar mass of the compounds in the reaction is:
- NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
- O₂: 2*16 g/mol= 32 g/mol
- NO: 14 g/mol + 16 g/mol= 30 g/mol
- H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol
By stoichiometry, they react and occur in moles:
- NH₃: 4 moles
- O₂: 5 moles
- NO: 4 moles
- H₂O: 6 moles
Then in mass, by stoichiomatry they react and occur:
- NH₃: 4 moles*17 g/mol= 68 g
- O₂: 5 moles*32 g/mol= 160 g
- NO: 4 moles*30 g/mol= 120 g
- H₂O: 6 moles*18 g/mol= 108 g
Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?
mass of O₂≅17.65 g
<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>