Answer:
aqueous acid is used as a reagent.
Explanation:
Addition of Grignard reagent in aldehyde and followed by the acidification give rise to the primary or secondary alcohol. when the formaldehyde is used than the primary alcohol is formed otherwise secondary alcohol is formed.
in this reaction we also use the aqueous acid for the acidification as a reagent. We add aqueous acid when ethanol is present. This is because ethanol is get converted in the presence of aqueous acid into the chloroethane.
Answer:
Intertidal zone
Neritic zone
Open-ocean zone
Note: the correct questions are found below;
In which zone do you find marshes and mangrove forests?
In which zone are plankton plentiful, providing plenty of food for the fish that live there?
In which zone would you find very little plant or animal life compared to other zones?
Explanation:
The intertidal zone, sometimes called the littoral zone, is the area of the marine shoreline that is exposed to air at low tide, and covered with seawater when the tide is high. Intertidal zonation refers to the tendency of plants and animals to form distinct communities between the high and low tide lines. Some microclimates in the littoral zone are moderated by local features and larger plants such as mangroves.
The neritic zone is the region of shallow water (200 meters depth) above the continental shelf where light penetrates to the sea floor.
Due to the abundant supply of sunlight and nutrients such as plankton in this zone, it is the most productive ocean zone supporting the vast majority of marine life.
The open oceans or pelagic ecosystems are the areas away from the coastal boundaries and above the seabed. It encompasses the entire water column and lies beyond the edge of the continental shelf. It extends from the tropics to the polar regions and from the sea surface to the abyssal depths.
Answer:
-0.050 kJ/mol.K
Explanation:
- A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
- The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K
All in all, ΔG° = 0 at 400. K.
We can find ΔS° using the following expression.
ΔG° = ΔH° - T.ΔS°
0 = -20 kJ/mol - 400. K .ΔS°
ΔS° = -0.050 kJ/mol.K
1. 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
2. CH₄ + 2O₂ → CO₂ + 2H₂O
3. Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag
4. MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
5. Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃
1)
2)
CuSO_4+Cu_2Cl_2\neq>
