The very first step would be to formulate a testable hypothesis.
Answer:
Valence of iron in FeS2
Let the valency of iron be x
x+2(oxidation number of Sulphur)=0 as their is no charge on FeS2
As sulphur belongs to Group VI (chalcogens) so the valency/oxidation number of Sulphur is -2
x+2(-2)=0
x-4=0
x=4
So the valency/oxidation number of iron in FeS2 is +4
Explanation:
Hope it will help you monkey ^_^ it's crab
Ca5(PO4)3(OH) + 9H2O →5Ca(OH)2 + 3H3PO4.
According to the information given, 135.62 g of hydroxyapatite were produced.
<h3>What is hydroxyapatite ?</h3>
Calcium apatite has the chemical formula Ca5(PO4)3(OH), however it is most commonly written Ca10(PO4)6(OH)2 to indicate that the crystal unit cell consists of two entities. Hydroxyapatite, also known as hydroxylapatite (HA), is a naturally occurring mineral form of calcium apatite. The complex apatite group's hydroxyl endmember is called hydroxyapatite. Fluorapatite or chlorapatite can be created by substituting fluoride, chloride, or carbonate for the OH ion. It forms a hexagonal crystal structure when it crystallizes.
Human bone contains up to 50% by volume and 70% by weight of bone mineral, a modified form of hydroxyapatite.
To learn more about hydroxyapatite from the given link:
brainly.com/question/27191380
#SPJ4
Answer:
82.08 %
Explanation:
- <em>The percent yield of the reaction = [(actual yield)/(calculated yield)] x 100.
</em>
Actual yield = 26.80 kg.
- <em><u>To get the calculated yield:
</u></em>
- The balanced equation of reacting N2 with H2 to produce NH3 is:
N₂ + 3H₂ → 2NH₃
- It is clear that 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- N₂ is present in excess and H₂ is the limiting reactant.
- We need to convert the mass of H₂ added (5.79 kg) to moles using the relation:
n = mass /molar mass = (5790 g) / (2.01 g/mol) = 2880.6 mol.
- We can get the no. of moles of NH₃ produced.
<u><em>Using cross multiplication:
</em></u>
3.0 mole of H₂ produce → 2.0 moles off NH₃, from the stichiometry.
2880.6 mol of H₂ produces → ??? moles of NH₃.
- The no. of moles of NH₃ produced = (2880.6 mol)(2.0 mol) / (3.0 mol) = 1920.4 mol.
- We can know get the calculated yield of NH₃ = no. of moles x molar mass = (1920.4 mol) (17.00 g/mol) = 32646.76 g ≅ 32.65 kg.
∴ <em>The percent yield of the reaction = [(actual yield)/(calculated yield)] x 100 = </em>[(26.8 kg) / (32.65 kg)] x 100 <em>= 82.08 %.</em>
Mixture is the right answer
