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3 years ago

4. How do minerals become part of the soil?

Chemistry

2 answers:

Gelneren [198K]3 years ago

5 0

I think it is C hope it helps

Goryan [66]3 years ago

4 0

Answer:

Explain how nearly anyone could become a miner under Spanish Mining Code.

Explanation:

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Nitric oxide is formed in automobile exhaust when nitrogen and oxygen in air react at high temperatures.N2(g) + O2(g) 2NO(g)The

yanalaym [24]

Answer : The correct option is, (E) 7.8 atm

Explanation :

The partial pressure of $N_2$ = 8.00 atm

The partial pressure of $O_2$ = 5.00 atm

$K_p$ = 0.0025

The balanced equilibrium reaction is,

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

Initial pressure 8.00 5.00 0

At eqm. (8.00-x) (5.00-x) 2x

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{NO})^2}{(p_{N_2})(p_{O_2})}$

Now put all the values in this expression, we get :

$0.0025=\frac{(2x)^2}{(8.00-x)\times (5.00-x)}$

By solving the terms, we get:

$x=0.15atm$

The equilibrium partial pressure of $N_2$ = (8.00 - x) = (8.00 - 0.15) = 7.8 atm

Therefore, the equilibrium partial pressure of $N_2$ is 7.8 atm.

3 0

3 years ago

Make an oblique drawing (page 9)

Olin [163]

Good luck folk :)!!!!!!!!

4 0

2 years ago

Consider the following reaction NHAHS(s)NH3(g) + H2S(g) If a flask maintained at 302 K contains 0.196 moles of NH4HS(s) in equil

quester [9]

Answer:

Kc = 3.72 × 10⁶

Explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

$Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2} \times 7.62 \times 10^{2}}{0.196} =3.72 \times 10^{6}$

7 0

3 years ago

Particle with more energy move _________ than particles with less energy.

s2008m [1.1K]

Faster and farther apart

5 0

3 years ago

Read 2 more answers

A mixture of three noble gases has a total pressure of 1.25 atm. The individual pressures exerted by neon and argon are 0.68

FinnZ [79.3K]

First of all, as you seen the gases are noble which means that will not react with each other and in this case each gas create individual pressure.

P $_{T}$ = total pressure

P $_{Ne}$ = pressure of neon

P $_{Ar}$ = pressure of argon

P $_{He}$ = pressure of helium {which is required}

P $_{T}$ = P $_{Ne}$ + P $_{Ar}$ + P $_{He}$

1.25 = 0.68 + 0.35 + P $_{He}$

P $_{He}$ = 1.25 - [0.68 + 0.35] = 0.22 atm

4 0

3 years ago