Answer:
11.9 g of nitrogen monoxide
Explanation:
We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:
Mass of NH₃ = 6.75 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 6.75 / 17
Mole of NH₃ = 0.397 mole
Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:
4NH₃ + 5O₂ —> 4NO + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted to produce 4 moles of NO.
Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.
Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:
Mole of NO = 0.397 mole
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO =?
Mass = mole × molar mass
Mass of NO = 0.397 × 30
Mass of NO = 11.9 g
Thus, the mass of NO produced is 11.9 g
I believe that crude oil is derived from plants because it is made up of a liquid that is found beneath the earth surface so I guess you could go with plants it is not an animal or rock or salt so.. could be brainliest not 100% sure so if you are not sure go with your gut
Answer: 13.9 g of 
will be produced from the given mass of oxygen
Explanation:
To calculate the moles :
The balanced chemical reaction is:
According to stoichiometry :
7 moles of 
produce = 6 moles of
Thus 0.900 moles of will produce =
of
Mass of 
Thus 13.9 g of will be produced from the given mass of oxygen
Answer:
The percent composition of fluorine is 65.67%
Explanation:
Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.
That is, the percentage composition is the percentage by mass of each of the elements present in a compound.
The calculation of the percentage composition of an element is made by:
In this case, the percent composition of fluorine is:
percent composition of fluorine= 65.67%
<u><em>The percent composition of fluorine is 65.67%</em></u>
