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serg [7]
4 years ago
15

Wind erosion occurs at a faster rate in the desert then in place of the pacific layer of vegetation covering the ground why do y

ou think this is the case
Chemistry
2 answers:
Fudgin [204]4 years ago
6 0
The wind isn't blocked at all in a desert. Also in deserts the sand is easily picked up by the wind hitting rocks further eroding them
cestrela7 [59]4 years ago
6 0

Answer:

Explanation:

when the soil is bare of vegetation the wind can pick up and carry sand, small rocks and soil. The sand being carried by the wind can act as an abrasive cutting into rocks, causing further erosion. This is nature's form of sand blasting.

Vegetation protects the soil, preventing the wind from picking up and carrying abrasives like sand and small pebbles. The wind by itself has little erosive power. Without the bare soil where wind can pick up sand there is little wind erosion.

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What is the molarity of a solution if you take 12.0 grams of ca(no3)2 and mix it with 0.105 l of water?
Lunna [17]
The  molarity  of  a solution  if it tale 12.0   grams  of  Ca(No3)2  is  calculated as below

molarity =  moles/volume in liters
 moles = mass/molar mass =  12.0 g/ 164 g/mol =  0.073 moles

molarity is therefore  = 0.073/0.105 = 0.7 M 
8 0
3 years ago
What is the Einstein theory of relevancy
Alinara [238K]
E=mc (square) E= mass times capacity squared
7 0
3 years ago
Please answer, please answer please answer please answer<br>my last 2 chemistry question. ​
scoundrel [369]

Answer:

1. filtration and evaporation

2. i) water is added to the sand and salt mixture

ii) then the mixture is filtrated and so the sand and the salt water was seperated

iii) the salt water is heated with the help of burner until the water gets evaporated

iv) after the water gets evaporated, the salt is remained in the container

3. observation:

  • on adding water to the mixture, the salt is completely dissolved in the water
  • when filtrated the sand is seperated from the salt water
  • now the salt water when heated with the burner until the evaporation, the water is evaporated
  • the salt is precipitated and remained in the container

4. cautions:

  • while using the burner, we should be cautious with fire
  • the container that is heated should be holded with the help of a cloth to avoid heat
7 0
3 years ago
Calculate the enthalpy of the reaction
harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)     \Delta H_A=+2035kJ

(2) 2B(s)+3H_2(g)\rightarrow B_2H_6(g)    \Delta H_B=+36kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_C=-285kJ

(4) H_2O(l)\rightarrow H_2O(g)    \Delta H_D=+44kJ

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D

\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)

\Delta H=-2552kJ

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0
3 years ago
How many molecules Of H20 are<br> equivalent to 97.2 g H20?<br> (H = 1.008 g/mol, O = 16.00 g/mol)
AleksandrR [38]

Answer:

m(H₂O) = 97,2 g.n(H₂O) = m(H₂O) ÷ M(H₂O).n(H₂O) = 97,2 g ÷ 18

Explanation:

8 0
2 years ago
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