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Strike441
2 months ago
5

For each pair of gases, select the one that most likely has the highest rate of effusion. Use the periodic table if necessary. O

xygen (O2) or hydrogen (H2): Methane (CH4) or carbon tetrachloride (CCl4): Nitrogen (N2) or ammonia (NH3): Fluorine (F2) or chlorine (Cl2):
Chemistry
2 answers:
Alekssandra [3K]2 months ago
7 0

1) H2

2) CH4

3) NH3

4) F2

=)

lorasvet [2.7K]2 months ago
7 0
<paccording to="" graham="" law="" the="" effusion="" rate="" of="" a="" gas="" is="" inversely="" related="" square="" root="" its="" molecular="" weight.="">

The relationship can be expressed as:

Rate of effusion ∝  1/√M

To determine which gas has the highest effusion rate, we will compare the molar masses of the given gases. The lighter gas will have a higher effusion rate.

1) The molar mass of oxygen (O₂) is 32 g, while hydrogen (H₂) has a molar mass of 2.01 g. Thus, H₂ will exhibit the greatest effusion rate.

2) The molar mass of methane (CH₄) is 16.05 g [12.01 + 4(1.01)], and the molar mass of carbon tetrachloride (CCl₄) is 154 g [12 + 4(35.45)]. Therefore, methane will show the highest effusion rate.

3) The molar mass of nitrogen (N₂) is 28 g, while ammonia (NH₃) has a molar mass of 17.03 g [14 + 3(3.01)]. Thus, NH₃ will have the highest effusion rate.

4) The molar mass of fluorine (F₂) is 38 g, and chlorine (Cl₂) has a molar mass of 71 g. Hence, F₂ will have the greatest effusion rate.


</paccording>
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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in
eduard [2782]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K

Starting temperature of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

mC \Delta T = mC \times \Delta T

10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

T_{f} - 100^{o}C = 0^{o}C - T_{f}

2T_{f} = 100^{o}C

T_{f} = 50^{o}C

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

\Delta H = mC \Delta T

= 10000 g \times 0.385 J/K g \times 323 K

= 1243550 J

or, = 1243.5 kJ

Next, determine the entropy change for block 1 as follows.

\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

= 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

= 10000 g \times 0.385 J/K g \times -0.143

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

                   = -554.12 J/K + 647.49 J/K\Delta S_{total} = \Delta S_{1} + \Delta S_{2}

           = 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

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Greetings,


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I hope this was useful!

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