Answer:
We are in Tropical Rainforests.
Explanation:
Biome is basically an ecological region of earth that has specific type of flora and fauna (plants and animals). Every biome has specific abiotic factors like specific climate, temperature, geology, vegetation, soils and relief.
There are total 8 biomes in the world
- Desert
- Grassland
- Temperate Boreal Forest
- Chaparral
- Tropical Savanna
- Tropical Rainforest
- Tundra
- Temperate Deciduous Forest
Now in our question, we have been given some conditions and are asked to identify the biome. The annual precipitation is about 300 cm and the average temperature is around 25 degrees Celsius.
Now, if we study the characteristics of all these biomes, it is very simple to identify that the biome is Tropical Rainforest.
Important characteristics of Tropical Rainforest:
- They receive more than 200 cm of rain each year.
- The temperature of the biome is between 20 degree Celsius and 25 degree Celsius during the whole year.
- Around 50 percent of the animal specie of the world are found here.
<em>Note: </em><em>You can study the characteristics of world’s other biomes in the link :
http://www.physicalgeography.net/fundamentals/9k.html
</em>
Hope it helps! :)
Answer:
7.46 g.
Explanation:
- Firstly, we need to calculate the amount of heat needed to warm 5.64 kg of water from 21.0°C to 70.0°C using the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by water (Q = ??? J).
m is the mass of water <u><em>(m: we will determine).</em></u>
c is the specific heat capacity of water (c = 4.186 J/g.°C).
ΔT is the temperature difference (final T - initial T) (ΔT = 70.0 °C - 21.0 °C = 49.0 °C).
- To determine the mass of 1.76 L of water we can use the relation:
mass = density x volume.
density of water = 1000 g/L & V = 1.76 L.
∴ mass = density x volume = (1000 g/L)(1.76 L) = 1760.0 g.
∵ Q = m.c.ΔT
<em>∴ Q = m.c.ΔT </em>= (1760.0 g)(4.186 J/g.°C)(49.0 °C) = 360483.2 J ≅ 360.4832 kJ.
- As mentioned in the problem the molar heat of combustion of hexane is - 4163.0 kJ/mol.
<em>Using cross multiplication we can get the no. of moles of hexane that are needed to be burned to release 360.4832 kJ:</em>
Combustion of 1.0 mole of methane releases → - 4163.0 kJ.
Combustion of ??? mole of methane releases → - 360.4832 kJ.
∴ The no. of moles of hexane that are needed to be burned to release 360.4832 kJ = (- 360.4832 kJ)(1.0 mol)/(- 4163.0 kJ) = 0.0866 mol.
- Now, we can get the mass of hexane that must be burned to warm 1.76 L of water from 21.0°C to 70.0°C:
<em>∴ mass = (no. of moles needed)(molar mass of hexane)</em> = (0.0866 mol)(86.18 g/mol) = <em>7.46 g.</em>
The rate of Formation of Carbocation mainly depends on two factors'
1) Stability of Carbocation: The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.
2) Ease of detaching of Leaving Group: The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,
R-I > R-Cl > R-F
B > C > A