Propose structures that meet the following descriptions. Use bond-line notation to draw your structural formulas. a. ketone with
five carbons and one alcohol c. alkane (no double bonds) with the formula, C6H12 e. nitrile with the formula, C5H9N g. phenol with a carboxylic acid group adjacent to hydroxyl i. amide with the formula, C3H7NO k. a secondary thiol with 3 carbons m. a primary alcohol with five carbons b. a-ketoester (i.e., a molecule with a ketone one carbon away from an ester) d. dialdehyde with the formula, C4H6O2 f. b-ketocarboxylic acid (i.e. a molecule with a ketone two carbons away from a carboxylic acid) h. amino acid (i.e. a carbon atom attached to both a carboxylic acid and an amine j. ether with the formula, C4H10O l. alkane with five carbons, one of which is a quaternary carbon n. hydroxy aldehyde (a compound containing an aldehyde and hydroxy group)
1 answer:
Answer:
See explanation
Explanation:
In putting up prospective structures for the compounds, we must consider the tetra valency of carbon. Again you must read the description carefully before drawing each structure.
In each of the proposed structures, the number of atoms present must reflect the condensed structural formula given in the question.
Proposed structures for each chemical specie in the question are shown in the images attached.
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Answer:
There are present 5,5668 moles of water per mole of CuSO₄.
Explanation:
The mass of CuSO₄ anhydrous is:
23,403g - 22,652g = 0,751g.
mass of crucible+lid+CuSO₄ - mass of crucible+lid
As molar mass of CuSO₄ is 159,609g/mol. The moles are:
0,751g × = 4,7052x10⁻³ moles CuSO₄
Now, the mass of water present in the initial sample is:
23,875g - 0,751g - 22,652g = 0,472g.
mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid
As molar mass of H₂O is 18,02g/mol. The moles are:
0,472g × = 2,6193x10⁻² moles H₂O
The ratio of moles H₂O:CuSO₄ is:
2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668
That means that you have <em>5,5668 moles of water per mole of CuSO₄.</em>
I hope it helps!
Answer:
1935100 Bq
Explanation:
Let us recall that:
If 1 μCi can be equivalent to 37000 Bq
Then; the activity of 52.3 μCi will be: