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r-ruslan [8.4K]

3 years ago

15

Ammonia and oxygen react to form nitrogen monoxide and water, like this:

1 answer:

olchik [2.2K]3 years ago

6 0

4NH3+5O2 <=>4NO + 6H2O

Using the definition of Kp, we have
Kp=(Pno^4*Ph2o^6)/(Pnh3^4*Po2^5)
where Pno=partial pressure of NO, etc.

The numerical value for a given temperature can be evaluated when the actual partial pressures are known.

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Mario places 10 mL of water in a test tube and heats the liquid over a Bunsen burner for 2 minutes. After removing the test tube

monitta

Mario places 10 mL of water in a test tube and heats the liquid over a Bunsen burner for 2 minutes. After removing the test tube from the Bunsen burner, there are 6 mL of water left in the test tube. This experiment is a good example of a <span>physical change involving phase changes. </span>

6 0

3 years ago

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Fynjy0 [20]

Answer:

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Explanation:

8 0

3 years ago

Condensation is a process for which a. ΔG is negative at high temperature but positive at low temperature. b. ΔH and ΔS are nega

IgorC [24]

Answer:

b. ΔH and ΔS are negative at all temperatures .

Explanation:

During the process of condensation ,

The gaseous state convert to liquid state ,

Hence , the entropy of the system reduces , i.e. , the randomness decreases .

And the value for entropy is negative ,

hence ,

Δ S = negative ,

Δ H = negative ,

Since ,

The heat is releasing from system .

hence , the most appropriate option will be ΔH and ΔS are negative at all temperatures .

7 0

3 years ago

Convert 6.93 x 1024 atoms of carbon to moles of carbon.

Aleksandr [31]

Answer: 11.5 moles of carbon

Explanation:

Based on Avogadro's law:

1 mole of any substance has 6.02 x 10^23 atoms

So, 1 mole of carbon = 6.02 x 10^23 atoms

Z moles = 6.93 x 10^24 atoms

To get the value of Z, cross multiply:

(6.93 x 10^24 atoms x 1mole) = (6.02 x 10^23 atoms x Z moles)

6.93 x 10^24 = (6.02 x 10^23 x Z)

Z = (6.93 x 10^24) ➗ (6.02 x 10^23)

Z = 1.15 x 10

Z = 11.5 moles

Thus, there are 11.5 moles of carbon.

7 0

3 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago