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Lemur [1.5K]
3 years ago
7

What are the biotic and abiotic factors of a desert?

Chemistry
2 answers:
NeTakaya3 years ago
7 0
Biotic factors are the living components or parts of an ecosystem. So the biotic factors would be cacti, salamanders, lizards, and other animals living in a desert. Abiotic factors are the nonliving components in an ecosystem. Some abiotic factors are sand, soil, rocks, and minerals. Hope this helps. 
FrozenT [24]3 years ago
6 0
Biotic factors: a cactus, animals. They are living things. Abiotic factors: sand, rocks. They are non living things.
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Predict the sign of the entropy change of the system for each of the following reactions.
storchak [24]

Answer :

(a) The entropy change will also decreases.

(b) The entropy change will also increases.

(c) The entropy change will also decreases.

(d) The entropy change will also increases.

Explanation :

Entropy : It is defined as the measurement of randomness or disorderedness in a system.

The order of entropy will be,

As we are moving from solid state to liquid state to gaseous state, the entropy will be increases due to the increase in the disorderedness.

As we are moving from gaseous state to liquid state to solid state, the entropy will be decreases due to the decrease in the disorderedness.

(a) N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

In this reaction, the randomness of reactant molecules are more and as we move towards the formation of product the randomness become less that means the degree of disorderedness decreases. So, the entropy change will also decreases.

(b) CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

In this reaction, the randomness of reactant molecules are less and as we move towards the formation of product the randomness become more that means the degree of disorderedness increase. So, the entropy change will also increases.

(c) 3C_2H_2(g)\rightarrow C_6H_6(g)

In this reaction, 3 mole of gaseous C_2H_2 react to give 1 mole of gaseous C_6H_6 that means randomness become less that means the degree of disorderedness decreases. So, the entropy change will also decreases.

(d) Al_2O_3(s)+3H_2(g)\rightarrow 2Al(s)+3H_2O(g)

In this reaction, the number of moles of gases are same on both side of the reaction but 1 mole of solid Al_2O_3 react to give 2 moles of solid aluminium that means randomness become more that means the degree of disorderedness increases. So, the entropy change will also increases.

5 0
3 years ago
Who is good with science?
MrMuchimi
I can help with science.
7 0
3 years ago
How can a large body of water, such as an ocean, influence climate?
Nat2105 [25]
ANSWER- ocean winds can carry moisture with then mmm and can bring rain.

Hope this helps leave a thanks!
6 0
2 years ago
Read 2 more answers
If you repeat an experiment and the result are very different from the result you got the first time the next step would be to w
Bess [88]
You would want to make sure that you have controlled the variables properly, and if you determine that you did then you would repeat the experiment to be sure of the results.
4 0
3 years ago
The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat
guajiro [1.7K]
First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

                                             =  1 mol * 18 mol/g

                                            = 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

       = 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus 

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

     = 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g 

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

      = 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

                                                  = 1.881 KJ  +6.01 KJ + 4.514 KJ

                                                  = 12.405 KJ


5 0
3 years ago
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