When a substance undergoes a physical change does it create a new substance?

The dissociation of calcium carbonate has an equilibrium constant of Kp= 1.20 at 800°C. CaCO3(s) ⇋ CaO(s) + CO2(g)

Fed [463]

Explanation:

(a) Formula that shows relation between $K_{c}$ and $K_{p}$ is as follows.

$K_c = K_p \times (RT)^{-\Delta n}$

Here, $\Delta n$ = 1

Putting the given values into the above formula as follows.

$K_c = K_p \times (RT)^{-\Delta n}$

= $1.20 \times (RT)^{-1}$

= $\frac{1.20}{0.0820 \times 1073}$

= 0.01316

(b) As the given reaction equation is as follows.

$CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)$

As there is only one gas so ,

$p[CO_{2}] = K_{p}$ = 1.20

Therefore, pressure of $CO_{2}$ in the container is 1.20.

(c) Now, expression for $K_{c}$ for the given reaction equation is as follows.

$K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}$

= $\frac{x \times x}{(a - x)}$

= \frac{x^{2}}{(a - x)}[/tex]

where, a = initial conc. of $CaCO_{3}$

= $\frac{22.5}{100} \times 9.56$

= 0.023 M

0.0131 = $\frac{x^{2}}{0.023 - x}$

x = 0.017

Therefore, calculate the percentage of calcium carbonate remained as follows.

% of $CaCO_{3}$ remained = $(\frac{0.017}{0.023}) \times 100$

= 75.46%

Thus, the percentage of calcium carbonate remained is 75.46%.

3 0

4 years ago

Give the volume indicated. Be sure to include all significant figures.

miskamm [114]

Answer: -

71

Explanation: -

From the diagram, we see that volume goes from 70 to 75 in 5 markings.

Each marking is for 1.

It is the most accuracy possible as there is no smaller marking.

Significant figures expresses the required amount of accuracy.

Thus the volume indicated from the diagram is 71 .

8 0

3 years ago

Read 2 more answers

A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro

Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q

$Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J$

Heat released on combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows :

Heat released on combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q'

$Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J$

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

$q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J$

Heat released on combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0

3 years ago

I need an answer and explanation pls

OverLord2011 [107]

Answer:

Cl2 + 2NaBr --> 2NaCl + Br2

Explanation:

This is a single displacement reaction where one side of the ionic compound switches with the other.

So, Cl2 + NaBr ---> NaCl + Br2

This isolates the Bromine and puts the Chlorine in it's place.

Then, balance out the equation like so and you should get

Cl2 + 2NaBr --> 2NaCl + Br2

8 0

3 years ago

What inert (noble) gas structure do hydrogen and carbon atoms have in sodium chloride?

vampirchik [111]

The answer above is correct (I took a test on this)

7 0

3 years ago