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mezya [45]
3 years ago
6

Label the reactants, products, and intermediates.

Chemistry
1 answer:
Annette [7]3 years ago
5 0
Of what?i cant see any pic here?
You might be interested in
An acetylene tank has a volume of 390.0 L. It is stored at a temperature of 23.5 °C and has a
xeze [42]

Considering the ideal gas law,  there are 279.42 moles of acetylene in the tank.

<h3>Definition of ideal gas</h3>

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

<h3>Ideal gas law</h3>

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:

P×V = n×R×T

<h3>Moles of acetylene</h3>

In this case, you know:

  • P= 1765 kPa= 17.4192 atm (being 101.325 kPa= 1 atm)
  • V= 390 L
  • n= ?
  • R= 0.082 \frac{atmL}{molK}
  • T= 23.5 °C= 296.5 K (being 0 °C= 273 K)

Replacing in the ideal gas law:

17.4192 atm× 390 L = n×0.082 \frac{atmL}{molK}× 296.5 K

Solving:

n=\frac{17.4192 atmx 390 L}{0.082 \frac{atmL}{molK}x296.5 L}

<u><em>n= 279.42 moles</em></u>

Finally, there are 279.42 moles of acetylene in the tank.

Learn more about ideal gas law:

brainly.com/question/4147359

4 0
2 years ago
Ano ang pang uri pldssssssss pakasagot po plsss​
katrin [286]

Answer:

please if you don't have a question don't question it you wasting my time

3 0
2 years ago
A sample of argon gas, Ar(g). is placed in a 3.80 L container at 320 K. The gas pressure is 0.496 atm.
Lilit [14]

Answer:

n = 0.072 mol

Explanation:

Given data:

Volume of gas = 3.80 L

Temperature of gas = 320 K

Pressure of gas = 0.496 atm

Number of moles of Ar gas = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will put the values.

n = PV/RT

n = 0.496 atm ×3.80 L / 0.0821 atm.L/ mol.K ×  320 K

n = 1.88 /26.27 /mol

n = 0.072 mol

6 0
3 years ago
Read 2 more answers
PLEASE HELP ME FAST!! How many moles are contained in 500 mL of a 3.5M NaOH solution?
viva [34]

Answer:

1.75moles

Explanation:

Molarity = number of moles (n) ÷ volume (V)

According to the information given in this question,

V = 500mL = 500/1000 = 0.500L

n = ?

Molarity = 3.5M

Using Molarity = n/V

number of moles (n) = V × molarity

n = 0.5 × 3.5

n = 1.75moles.

5 0
2 years ago
A gas with a volume of 2 L at 25°C is placed into a container that is 4 L. What is the new temperature of the gas?
zalisa [80]

Answer:

Convert 539 torr to atm.

2.) A gas takes up 25.2 liters at 25oC. At 25oC, the gas can also take up 12.2 liters at 1500 torr. What was the pressure, in atm, of the original sample?

3.) A gas takes up 14.8 liters of 24oC. What temperature in kelvin is required to obtain a volume of 25.0 liters at constant pressure?

4.) How many moles of chlorine gas are present at 25oC, 762 torr, with a volume of 14.2 L?

5.) If "Bor" (thats the name of a person) has a sample of gas that has a volume of 8.2 liter at 25oC and 2 atm, how much volume will it take up if you decrease pressure to 1.5 atms and icrease temperature to 100oC?

6.) 25 liters of gas A is pumped into a container at 25oC and 760 torr with 20 liters of gas B at 25oC and 700 torr. Calculate the total pressure when both gases are pumped into a tank with 10 liters at 25oC.

7.) Calculate the mole fraction of oxygen when 200 torr of air (760 torr total) is oxygen.

8.) KClO3 is decomposed by the following reaction:

2KClO3(s) ---> 2KCl(s) + 3O2(g)

The O2 produced was collected by the displacement of water at 22oC at a total pressure of 760 torr. The volume of gas collected was 1.20 liters, and the vapor pressure of water at 22oC is 21 torr. Calculate the partial pressure of O2 in the gas collected and the mass of KClO3 in the sample that was decomposed.  

9.) Calculate the root mean square velocity of hydrogen gas at 25oC.

10.) What is the ratio of effusion between fluorine and chlorine?

1.) Just set up a simple ratio.

539 torr * (1 atm/760 torr) = 0.709 atm

2.) Use this formula:

P1V1 = P2V2

460 atm  * 1.20 liters = 839 torr * V2

V2= 0.658 L

3.) You will use the ratio P1V1 = P2V2 in order to solve this equation.

P1V1 = P2V2

25.2 liters * P1 = 12.2 liters * (1500 torr / 760 torr)

Note that you have to convert the torr to atm. The problem asks for the final pressure in atms. You could of just solved the ratio without the conversion, and converted the answer later. Either way is acceptable.

25.2 liters * P1 = 24

P1 = 0.956 atm

4.) Use the formula:

V1/T1 = V2/T2

14.8 liters / 297 torr = 25.0 liters / x

14.8 x = 7425

x = 502 K

5.) PV = nRT (the ideal gas law! Note: you will see this a LOT in gases, so memorize it)

n = PV/RT

P = (762 torr/760 torr) = 1.00atm

n = (1.00 atm * 14.2 L) / (298oK * 0.08206)

n = 0.582 moles

6.) PV = nRT

n = PV/RT

= (2 atm * 8.2 L) / (0.08206 * 298oK) = 0.671 moles

Now, you must remember that you are taking this amount and changing it some more. So, you do this again, this time, solving for VOLUME. (No, the initial 8.2 doesn't count, because you are manipulating the environment)

PV = nRT

V = nRT/P

=(0.671 moles * 0.08206 * 373oK) / 1.5 atm = 13.76 liters

7.) You must solve this twice for both gas A and B, then combine the individual pressures to obtain the final pressure. Use the Ideal Gas Law.

ngas a = PV/RT = (1 atm * 25 liters) / (0.08206 * 298 K) = 1.02 moles

ngas b = PV/RT = (1 atm * 20 liters) / (0.08206 * 298) = 0.818 moles

You know how many moles you have. The second part of the problem calls for a manipulation of the environment, which means that you will be solving for PRESSURE.

Pgas a = nRT/V = (1.02 moles * 0.08206 * 298K / 10.0 liters) = 2.49 atm

Pgas b = nRT/V = (0.818 moles * 0.08206 * 298 K / 10.0 liters) = 2.00 atm

2.49 atm + 2.00 atm = 4.49 atm

8.) This is a simple mole fraction. You know that air has a total of 760 torr. Since we are given that O2 directly contributes 200 of that, we can set up this ratio:

O2 pressure/ total pressure = 200 torr / 760 torr = 0.263 = 26.3%

9.) Remember that:

Ptotal = PO2 + PH2O = PO2 + 21 torr = 760 torr

PO2 = 739 torr

PV = nRT or n=PV/RT

P = 739 torr/760 torr = 0.972 atm

n = (0.972 atm * 1.20 liters) / (0.08206 * 295 K) = 0.0482 moles O2

The final answer calls for mass of the KClO3, not the O2. So we must find the number of moles of KClO3.

0.0482 moles O2 * (2 mol KClO3/3 mol O2) = 0.0321 moles KClO3

0.0321 mol KClO3 * (122.6 g KClO3 / 1 mol KClO3) = 3.94 grams KClO3

10.) Remember that urms formula is:

So, using that, let's fill in our variables and constants. R, for this equation is 8.3145 J/K*mol and T = 298 K.  Calculating for M =

MH2 = 1 g/mol * 2 mol (since H2 is diatomic) * 1 kg/1000g = 0.002 kg

(3 * 8.3145 * 298/ 0.002) = 3716581.5 (you have to take the square root of this) -----> 1928 m/s

11.) Both fluorine and chlorine are diatomic. The equation relating effusion and diffusion is:

Taking this, you take the square root of 70.90 g (which is the weight of Cl2), and divide it by the square root of 37.996 g (which is the weight of F2). The answer is 1.366. This means that fluorine effuses 1.366 times as fast as chlorine, which makes sense because F2 is smaller.

Explanation:

3 0
2 years ago
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