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:
(NH4)2Cr2O7
Explanation:
Hope this somehow helped.
Answer:The levels of classification are as follows (Broadest to
narrowest): Domain, Kingdom, Phylum, Class, Order, Family, Genus,
Species
Explanation:
Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
Answer: On heating, Magnesium forms its oxide; while potassium manganate(VII) decomposes
Explanation:
Magnesium Mg, on heating forms Magnesium oxide
2Mg(s) + O2(g) --> 2MgO
Potassium permanganate KMnO4, on heating decomposes to potassium manganate K2MnO4, manganese dioxide MnO2, and Oxygen gas O2.
2KMnO4 --> K2MnO4 + MnO2 + O2
The difference in observation is that, on heating, Magnesium forms its OXIDE as product; while potassium manganate(VII) decomposes, giving OFF most of its constituents and reducing its weight.
