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Goryan [66]
3 years ago
7

Explain whether or not you expect the chaparral biome to be sensitive to the loss of a single species.

Chemistry
1 answer:
aleksley [76]3 years ago
3 0

Answer:

See the answer below

Explanation:

The chaparral biome is a temperate biome with a characteristic high temperature and dryness during summer and mild rainy winters and springs. The biome can be found in relatively small amounts in the major continents of the world with its rich plant and animal diversity who have successfully adapted to the conditions of the biome.

Due to the high biodiversity of the chaparral biome, <u>one would expect it to be resilient to the loss of a single species.</u> <em>The more the biodiversity of a biome or community, the more resilient such biome or community would be to the loss of species and lower the biodiversity, the more sensitive the community would be to the loss of species. </em>

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How do the isotopes of an element differ? How are they the same?
OverLord2011 [107]

Answer:

They differ by their atomic mass and have same proton number.

Explanation:

Isotopes are atoms of the same element with the same proton or atomic number but with different atomic masses.

7 0
3 years ago
A helium balloon containing 0.100 mol of gas occupies a volume of 2.4 L at 25 C and 1.0 atm. how many moles have we added if we
Vilka [71]

Answer: 0.13mol

Explanation:Please see attachment for explanation

6 0
4 years ago
If the pressure on the surface of water in the liquid stato
Novosadov [1.4K]

The water will boil at C) 80°C

<h3>Further explanation</h3>

Given

Vapour pressure of water = 47 kPa

Required

Boiling point of water

Solution

We can use the Clausius-Clapeyron equation :

\tt ln(\dfrac{P_1}{P_2})=\dfrac{-\Delta H_{vap}}{R}(\dfrac{1}{T_1}-\dfrac{1}{T_2})

Vapour pressure of water at boiling point 100°C=101.325 kPa

ΔH vap for water at 100°C=40657 J/mol

R = 8.314 J/mol K

T₁=boiling point of water at 101.325 kPa = 100+273=373 K

Input given values :

\tt ln\dfrac{101.325}{47}=\dfrac{-40657}{8.314}(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\0.7682=4890.1852(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\0.0001571=(\dfrac{1}{373}-\dfrac{1}{T_2})\\\\\dfrac{1}{T_2}=0.00283\rightarrow T_2=353~K=80^oC

6 0
3 years ago
How many moles are present in 63.80L of oxygen gas, O2?
matrenka [14]

Answer:

2.85moles of oxygen gas

Explanation:

Given parameters:

Volume of oxygen gas  = 63.8L

Unknown:

Number of moles  = ?

Solution:

We assume that the gas is under standard temperature and pressure. To find the number of moles, use the expression below:

     1 mole of a gas at STP occupies a volume of 22.4L

 So;

       63.8L of oxygen gas will take up a volume of

              \frac{63.8}{22.4}   = 2.85moles of oxygen gas

4 0
3 years ago
If a 0.2g of oil consumed 1ml of sodium thiosulphate, calculate its iodine value and classify the oil?
marta [7]

Iodine value is a measure of the degree of unsaturation in fats and oils. It is essentially the number of grams of iodine consumed by 100 g of fat. If the iodine number is in the range of 0-70 then it is a fat, any value above 70 is considered an oil.

Formula:

Iodine number = (ml of 0.1 N Thiosulphate blank- ml of 0.1N thiosulphate test) * 12.7 *100/1000* wt of sample

vol of thiosulphate required to titrate test sample (given oil) = 1 ml

wt of sample = 0.2 g

Information on the volume of thiosulphate required to titrate the blank solution is essential for calculation.

Iodine number = (X-1.0) * 12.7 * 100/1000* 0.2 = (X-1.0)*6.35

6 0
4 years ago
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