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3 years ago

How many liters of radon gas would be in 3.43 moles at standard temperature and pressure (273 K and 100 kPa)?

Chemistry

1 answer:

MissTica3 years ago

5 0

Answer: Option B. 76.83L

Explanation:

1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.

If 1 mole of Radon = 22.4L

Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L

You might be interested in

How many moles of na contain 7.88x1021 atoms of na

labwork [276]

Answer:

the answer to the qustion is 0.013089701 na

Explanation:

n/a

7 0

2 years ago

A mixture initially contains AA, BB, and CC in the following concentrations: [A][A]A_1 = 0.550 MM , [B][B]B_1 = 1.40 MM , and [C

Alex787 [66]

Answer:

The value of the equilibrium constant KC is 1.244

Explanation:

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.550 M, [B] = 1.40 M, and [C] = 0.600 M. The following reaction occurs and equilibrium is established: A+2B<->C

At equilibrium, [A] = 0.430 M and [C] = 0.720 M. Calculate the value of the equilibrium constant, Kc

Step 1: The balanced equation

A+2B<->C

Step 2: The initial concentrations

[A] = 0.550 M

[B]= 1.40 M

[C] = 0.600 M

Step 3: The concentraions at equilibrium

[A] = 0.550 -X = 0.430 M

[B]= 1.40 -2X M

[C] = 0.600 + X = 0.720 M

X = 0.120 M

[A] = 0.550 - 0.120 = 0.430 M

[B]= 1.40 -2*0.120 = 1.16 M

[C] = 0.600 + 0.120 = 0.720 M

Step 4: Calculate Kc

Kc = [C] / [A][B]²

Kc = 0.720 / (0.430*1.16²)

Kc = 1.244

The value of the equilibrium constant KC is 1.244

5 0

3 years ago

Does gasoline (C8H12) have polar or non-polar molecules?

Archy [21]

Answer:

non polar molecules

Explanation:

Ethyl alcohol will dissolve in water in all proportions because the polar molecules interact with each other freely. In contrast, gasoline is a nonpolar molecule. hope that helped...

3 0

3 years ago

Entropy of a system decreases with decrease in Temperature T/F

Svet_ta [14]

Answer: The given statement is true.

Explanation:

Entropy means the measure of randomness present in a substance. That is, an increase in temperature will lead cause more motion in the particles of a substance more will be their kinetic energy.

As a result, there will occur more collisions due to which randomness of molecules will increase. Hence, there will be increase in entropy.

So, when we decrease the temperature then there will be decrease in motion of particles. As a result, lesser number of collisions will take place between them. Hence, degree of randomness will also decrease.

Thus, we can conclude the statement entropy of a system decreases with decrease in temperature, is true.

3 0

3 years ago

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867

Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

$1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O$

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of $CO_2$ gas and 4 moles of $H_2 O$ Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $$

6 0

3 years ago