Answer:
Here's what I get
Explanation:
1. Names
I. CH₃-CH₂-COOH = 49. propanoic acid
II. CH₃-CH₂-OH = 46. ethanol
III. CH₃-COO-CH₂-CH₂-CH₃ = 47. propyl ethanoate
IV. H-O-CH₂-CH₂-CH₃ = 48. propan-1-ol
V. H-COO-CH₃ = 51. methyl methanoate
VI. CH₃-COOH = 50. ethanoic acid
2. Precursors
52. methyl propionate ⇒ methanol + propanoic acid
53. ethyl methanoate ⇒ ethanol + methanoic acid
Answer:
Rb+
Explanation:
Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.
Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.
Thus our calculations are:
V = 17.0 mL x 1 L / 1000 mL = 0.017 L
2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol
0.0425 mol = 4.36 g/ MW XOH
MW of XOH = (atomic weight of X + 16 + 1)
so solving the above equation we get:
0.0425 = 4.36 / (X + 17 )
0.7225 +0.0425X = 4.36
0.0425X = 4.36 -0.7225 = 3.6375
X = 3.6375/0.0425 = 85.59
The unknown alkali is Rb which has an atomic weight of 85.47 g/mol
Answer:
A) It tends to increase from top to bottom of a group
