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2 years ago

As a real gas deviates from ideal gas behavior, the particles _____

Chemistry

1 answer:

Alik [6]2 years ago

7 0

Answer:

Have some attraction towards each other

Explanation:

Gases deviate from the ideal gas behavior because their molecules have forces of attraction between them. At high pressure, the molecules of gases are very close to each other so the molecular interactions start operating and these molecules do not strike the walls of the container with full impact.

Hope this helps :-)

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<3 simplysun

ps. I do not own any of these answers so please don't give full credit to me

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Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol$

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

$P'V'=n'RT'$

$n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol$

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)$

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

$\frac{3}{1}\times 0.6402 mol=1.9208 mol$ of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

$\frac{1}{22.4 L}\times 26 L= 1.1607 mol$

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

$\frac{Experimental}{Theoretical}\times 100$

$\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%$

60.42% is the percent yield of the reaction.

8 0

3 years ago

X2o3 express you answer as a whole number

kobusy [5.1K]

<span>2.40 - 1.68 =0.72 g of oxigen
moles = 0.72/16 g/mol=0.045

moles x = 1.68/ 55.9=0.03

0.03/0.03 = 1 = x
0.045 / 0.03 = 1.5 = O

to get whole numbers multiply by 2

x2O3

X2O3 +3 CO = 2 X + 3 CO2</span>

4 0

3 years ago

A balloon contains 2.0 L of air at 101.5 kPa . You squeeze the balloon to a volume of 0.25 L.

8_murik_8 [283]

<h3>Answer:</h3>

812 kPa

<h3>Explanation:</h3>

According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.

Mathematically, $P\alpha \frac{1}{V}$

At varying pressure and volume;

P1V1=P2V2

In this case;

Initial volume, V1 = 2.0 L

Initial pressure, P1 = 101.5 kPa

Final volume, V1 = 0.25 L

We are required to determine the new pressure;

$P2=\frac{P1V1}{V2}$

Replacing the known variables with the values;

$P2=\frac{(101.5)(2.0L)}{0.25L}$

= 812 kPa

Thus, the pressure of air inside the balloon after squeezing is 812 kPa

8 0

3 years ago

Automotive airbags inflate when a sample of sodium azide, NaN3, is very rapidly decomposed.

Kobotan [32]

Answer:

148 g

Explanation:

Step 1: Write the balanced equation for the decomposition of sodium azide

2 NaN₃ ⇒ 2 Na + 3 N₂

Step 2: Calculate the moles corresponding to 95.8 g of N₂

The molar mass of N₂ is 28.01 g/mol.

95.8 g × 1 mol/28.01 g = 3.42 mol

Step 3: Calculate the moles of NaN₃ needed to form 3.42 moles of N₂

The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.42 mol = 2.28 mol.

Step 4: Calculate the mass corresponding to 2.28 moles of NaN₃

The molar mass of NaN₃ is 65.01 g/mol.

2.28 mol × 65.01 g/mol = 148 g

8 0

2 years ago

Phuong trinh phan ung ohc cho +02

scZoUnD [109]

Answer:

yes

hope this helps!!!!

5 0

3 years ago