The balanced equation given is:
4NH3 + 3O2 .....> 2N2 + 6H2O
From this equation, we can note that 4 moles of NH3 are required to produce 2 moles of N2.
Therefore, the mole ratio of NH3 to N2 is 4:2 which can be simplified into 2:1
Answer:
Yes, the chemist can determine which compound is in the sample.
Explanation:
In 1 mole of K₂O, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O is 94.2 g. The mass ratio of K to K₂O is 78.2 g / 94.2 g = 0.830.
In 1 mole of K₂O₂, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ is 78.2 g / 110.2 g = 0.710.
If the chemist knows the mass of K and the mass of the sample, he or she must calculate the mass ratio of K to the sample.
- If the ratio is 0.830, the compound is pure K₂O.
- If the ratio is 0.710, the compound is pure K₂O₂.
- If the ratio is not 0.830 or 0.710, the sample is a mixture.
Answer:
The actual number of moles is 9 moles.
It is less than 15
Number of moles needed is 9 moles
Explanation:
15H2 + 10N2 ——-> 10NH3
Now from the question, we can see that the percentage yield is 60%
The percentage yield can be calculated as actual moles of H2/Theoretical moles of H2 * 100%
From the equation, we can see that the theoretical number of moles of hydrogen is 15.
Now to get the actual : 60 = x/15 * 100
x = 9
The actual number of moles is 9 moles.
It is less than 15
Number of moles needed is 9 moles
Answer:
Zero
Explanation:
FrBr is an ionic compound
.
Fr is in Group 1. Br is in Group 17.
The charges on the ions are +1 and -1, respectively.
The compound consists of Fr⁺Br⁻ ions.
However, there are equal numbers of + and - charges, so
The overall charge of the compound is zero.
Ionic bond is formed due to the transfer of electrons from one atom to another so that all atoms involved in the bond would become stable (with 8 electrons in the outermost level)
Now, for bromine, it has 35 electrons. This means that bromine has 7 valence electrons in the outermost level. Therefore, bromine needs to gain one electron in order to become stable.
Bromine can react with elements from:
group 1: each element in group 1 needs to lose one electron to become stable. Therefore, one bromine atom can form an ionic bond when combined with an atom of an element from group 1 (element in group 1 loses its electron for bromine atom).
group 2: each element in group 2 needs to lose two electrons to become stable. Therefore, two bromine atoms can form ionic bonds when combined with an atom of an element from group 2 (element in group 2 loses two electrons, one for each bromine atom).
group 3: each element in group 3 needs to lose three electrons to become stable. Therefore, three bromine atoms can form ionic bonds when combined with an atom of an element from group 1 (element in group 3 loses three electrons, one for each bromine atom).
Since no choices are given , I cannot tell the exact choice. But the correct one would be the element from either group 1 , 2 or 3.
