The molecular problem for aluminum oxide is Al2O3 Atomic mass of Al = 26.982 Atomic mass of O = 16 From the problem, 1 mole of AL2O3 contains: 2 g-atom of Al or 2 x 26.982 = 53.964 g 3 g-atom of O or 3 x 16 = 48 g the mass of 1 mole of Al2O3 is equal to 53.964 + 48 = 101.964 g
The answer is the option d. metabolism.
Metabolism is the set of the chemical reactions that happens in the organism to transform nutrients (food) in energy and the products that conform the cells and all the constituents of the body.
Answer:
(d) 3,7-dimethyl-4-nonyne.
Explanation:
Hello,
In this case, considering the attached picture on which you can see that the main chain has nine carbon atoms, one tripe bond at the fourth carbon and two methyl radicals at the third and seventh carbons respectively, by following the IUPAC rules, the name would be: (d) 3,7-dimethyl-4-nonyne since the chain must start at the side closest to the first triple bond due to its priority and subsequently considering the present radicals.
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Answer:
All compounds are molecules
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Explanation:
Answer:
247 ml
Explanation:
How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF 0.150 moles/ liter = 0.150/1000 moles/ml =0.000150 moles/ml
0.000150 x ? = 0.0370 moles HF
? = 0.0370/0.000150 = 247 ml
check
247 ml = 247/1000 L = 0.247
(0.247) x (0.150) =0.370 check