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astra-53 [7]
3 years ago
15

Desertification

Chemistry
2 answers:
madam [21]3 years ago
8 0
A.) desertification is when you remove all the trees and most of the plant life in an ares
Reptile [31]3 years ago
6 0
A. <span>will typically increase the amount of erosion in an area
desertification will destroy the land</span>
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Can u pls help me with this question ​
lorasvet [3.4K]

Answer:

b i im pretty sure or a

Explanation:

f y bib0f84f69g85

8 0
3 years ago
A tau lepton decays into an electron, an electron antineutrino and a tau neutrino. Write out this reaction in symbolic (equation
Dmitry [639]

Answer:

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Explanation:

hhidhwvagaue hrurifv3v2vd

5 0
3 years ago
Explain why photosynthesis is much slower in underwater plants.
Viktor [21]

Answer:

Gases such as carbon dioxide diffuse much more slowly in water than in air. Plants that are fully submerged have greater difficulty obtaining the carbon dioxide they need. To help ameliorate this problem, underwater leaves lack a waxy coating because carbon dioxide is easier to absorb without this layer.

Explanation:

8 0
3 years ago
Read 2 more answers
Calculate delta Gº for the following system:
Llana [10]

Answer:

ΔG = - 442.5  KJ/mol    

Explanation:

Data Given

delta H = -472 kJ/mol

delta S = -108 J/mol K

So,

delta S = -0.108 J/mol K

delta Gº = ?

Solution:

The answer will be calculated by the following equation for the Gibbs free energy

                    G = H - TS

Where

G = Gibbs free energy

H = enthalpy of a system (heat

T = temperature

S = entropy

So the change in the Gibbs free energy at constant temperature can be written as

                ΔG = ΔH - TΔS . . . . . . (1)

Where

ΔG = Change in Gibb’s free energy

ΔH = Change in enthalpy of a system

ΔS = Change in entropy

if system have standard temperature then

T = 273.15 K

Now,

put values in equation 1

                  ΔG = (-472 kJ/mol) - 273.15 K (-0.108 KJ/mol K)

                 ΔG = (-472 kJ/mol) - (-29.5 KJ/mol)

                 ΔG = -472 kJ/mol + 29.5 KJ/mol

                 ΔG = - 442.5  KJ/mol          

7 0
3 years ago
What volume would 2.25 moles of Ne gas occupy at STP?
Dafna1 [17]
As we know that one mole of any Ideal gas at standard temperature and pressure occupies exactly 22.4 dm³ volume.

Solution for problem:

When 1 mole Neon (Ne) occupies 22.4 dm³ at STP then the volume occupied by 2.25 moles of Neon is calculated as,

                                             = ( 22.4 dm³ × 2.25 moles ) ÷ 1 mole
                
                                             = 50.4 dm³                   1dm³ = 1 L

Result:

So
, 50.4 dm³ (Liter) volume will be occupied by 2.25 moles of Neon gas if it acts ideally at STP.
8 0
3 years ago
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