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3 years ago

What are two roles living things play in soil formation?

1 answer:

Studentka2010 [4]3 years ago

8 0

Soil Formation:- Living Organisms. Plants, animals, and micro-organisms (fungi and bacteria) all affect soil formation by producing or contributing to humus production. The amount of humus in a soil is a result of how much plant material has been incorporated into it.

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At 100 degrees Celsius, the ion product for pure water is Kw= 51.3*10^-14. What is the pOH of water at this temperature? A) 7.00

Levart [38]

7.86 is the pOH of water at this temperature of 100 degrees celsius.

Option E is the right answer.

Explanation:

Data given:

Kw = 51.3 x $10^{-14}$

pOH = ?

we know that pure water is neutral and will have pH pf 7.

The equation for relation between Kw and H+ and OH- ion is given by:

Kw = [H+] [OH-}

here the concentration of H+ ion and OH- ion is equal

so, [H+]= [OH-]

Putting the values in the equation of Kw

pKw = -log[Kw]

pKw = -log [51.3 x $10^{-14}$ ]

pKw = 12.28

since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14

Now, pOH is calculated by using the equation:

14 = pOH + pH

14- 6.14 = pOH

pOH = 7.86

8 0

3 years ago

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th

Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

$\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M$

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially conc. 0 0 0.1576

At eqm. (x) (x) (0.1576-2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}$

By solving the term, we get:

$x=0.0742,0.0839$

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

5 0

3 years ago

Copper is formed when aluminum reacts with copper (II) sulfate in a single-replacement reaction. How many moles of copper can be

serg [7]

Answer:

The answer to your question is the limiting reactant is CuSO₄ and 0.975 moles of Cu were obtained

Explanation:

moles of Copper = ?

mass of Aluminum = 29 g

mass of CuSO₄ = 156 g

Limiting reactant = ?

Balanced Chemical reaction

3 CuSO₄ + 2 Al ⇒ 3 Cu + Al₂(SO₄)₃

Calculate the moles of reactants

CuSO₄ = 64 + 32 + (16 x 4) = 160g

Al = 27 g

160 g of CuSO₄ ----------------- 1 mol

156 g ----------------- x

x = (156 x 1) / 160

x = 0.975 moles

27 g of Al -------------------------- 1 mol

29 g of Al -------------------------- x

x = (29 x 1)/27

x = 1.07 moles

Calculate proportions to find the limiting reactant

Theoretical 3 moles CuSO₄/2 moles Al = 1.5 moles

Experimental 0.975 moles CuSO₄/1.07 moles = 0.91

The experimental proportion was lower than the theoretical proportion that means that the limiting reactant is CuSO₄.

3 moles of CuSO₄ ------------------ 3 moles of Cu

0.975 moles of CuSO₄ --------------- x

x = (0.975 x 3)/3

x = 0.975 moles of Cu were obtained.

3 0

3 years ago

Read 2 more answers

What does the symbol (–Delta.Hfus) indicate in a phase change?

Rufina [12.5K]

Answer:

changes from a solid to a liquid

Explanation:

8 0

3 years ago

Can you please help. I don't know how to do it.

Rainbow [258]

The area used to be covered by an ocean

3 0

3 years ago