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kolezko [41]
3 years ago
15

1pt Which statement best describes the relationship between the system and the surroundings?

Chemistry
1 answer:
Firlakuza [10]3 years ago
4 0

Answer:

Hi how are you doing today Jasmine

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IRINA_888 [86]

Answer:

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3 0
3 years ago
The decomposition of N2O4 into NO2 has Kp = 2. Some N2O4 is placed into an empty container, and the partial pressure of NO2 at e
Sunny_sXe [5.5K]

Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

           N_{2}O_{4} \rightarrow 2NO_{2}

Let us assume that x concentration of N_{2}O_{4} is present at the initial stage. Therefore, according to the ICE table,

                    N_{2}O_{4} \rightarrow 2NO_{2}

Initial :               x                   0

Change :       - 0.1        2 \times 0.1

Equilibrium : (x - 0.1)             0.2

Now, expression for K_{p} of this reaction is as follows.

     K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}

Putting the given values into the above formula as follows.

          K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}

                 2 = \frac{(0.2)^{2}}{(x - 0.1)}

                2 \times (x - 0.1) = (0.2)^{2}

                            x = 0.12

This means that P_{N_{2}O_{4}} = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.

6 0
3 years ago
HELP ASAP!<br><br> what is the molar mass of Na2CO3?
tekilochka [14]

The correct answer is 106.0 g/mol

5 0
3 years ago
Read 2 more answers
At 35°C, Kc = 1.6 multiplied by10-5 for the following reaction
Brums [2.3K]

Answer : The equilibrium concentrations of all species NO,Cl_2\text{ and }NOCl are, 0.05 M, 0.043 M and 0.975 M respectively.

Explanation : Given,

Moles of  NO = 2 mole

Moles of  Cl_2 = 1 mole

Volume of solution = 1 L

Initial concentration of NO = 2 M

Initial concentration of Cl_2 = 1 M

The given balanced equilibrium reaction is,

                            2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

Initial conc.          2 M            1 M            0

At eqm. conc.    (2-2x) M   (1-x) M         (2x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[NOCl]^2}{[NO]^2[Cl_2]}

The K_c for reverse reaction = \frac{1}{1.6\times 10^{-5}}

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

\frac{1}{1.6\times 10^{-5}}=\frac{(2x)^2}{(2-2x)^2\times (1-x)}

By solving the term 'x', we get :

x = 0.975

Thus, the concentrations of NO,Cl_2\text{ and }NOCl at equilibrium are :

Concentration of NO = (2-2x) M  = (2 - 2 × 0.975) M = 0.05 M

Concentration of Cl_2 = (1-x) M = 1 - 0.975 = 0.043 M

Concentration of NOCl = x M = 0.975 M

Therefore, the equilibrium concentrations of all species NO,Cl_2\text{ and }NOCl are, 0.05 M, 0.043 M and 0.975 M respectively.

6 0
3 years ago
According to early chemists, which substances were classified as elements? A. those that could not be broken down further by phy
Musya8 [376]
B. Those who could not be broken down chemically anymore than they already were
5 0
4 years ago
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