For this problem, we use the Hess' Law.
ΔHrxn = ∑(ν*Hf of products) - ∑(ν*Hf of reactants)
The ν represents the corresponding stoichiometric coefficients of the substances, while Hf is the heat of formation. For pure elements, Hf = 0.
Hf of Al₂O₃ = <span>−1676.4 kJ/mol
</span>Hf of Fe₂O₃ = <span>-826.0 kJ/mol
Thus,
</span>ΔHrxn = 1*−1676.4 kJ/mol + 1*-826.0 kJ/mol
<em>ΔHrxn = -2502.4 kJ/mol</em>
The balanced chemical equation is written as :
2H3PO4 + 3Ca(OH)2 → 6H2O + Ca3(PO4)2.
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From the above equation we can see that 3 moles of Ca(OH)2 produces 1 mole Ca3(PO4)2 .
( Don't take tension about H3PO4 as it is present in excess )
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Further ,
we are given 78.5 g Ca(OH)2 .
then no of moles Ca(OH)2 given = 78.5/74 = 1.06 moles.
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3 moles Ca(OH)2 produces → 1 mole Ca3(PO4)2.
=> 1 mole Ca(OH)2 produces → 1/3 mole Ca3(PO4)2.
=> 1.06 mole Ca(OH)2 produces → (1/3)×1.06 mole Ca3(PO4)2.
Hence number of moles Ca3(PO4)2 produced = (1/3)×1.06 = 0.353 moles.
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Now gram molecular mass of Ca3(PO4)2 is = 310 g /mole .
hence grams of Ca3(PO4)2 produced
= 0.353× 310 = 109.43 g
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Hope it helps...
1 is 7 and a half!
I hope this helps you!
Answer:
Monosaccharides are the simplest form of carbohydrates that cannot be hydrolyzed to smaller compounds. Monosaccharides are the basic units of carbohydrates and are also known as simple sugars.
The monosaccharides are classified on the basis of number of carbon atoms present.
Triose is a type of monosaccharide molecule, which is composed of 3 carbon atoms.
Tetrose is a type of monosaccharide molecule, which is composed of 4 carbon atoms.
Pentose is a type of monosaccharide molecule, which is composed of 5 carbon atoms.
Hexose is a type of monosaccharide molecule, which is composed of 6 carbon atoms.
D-glucose is a hexose sugar and it is the <u>most abundant monosaccharide</u> in the nature.
<span>the answer is cyanobacteria</span>
