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fomenos
2 years ago
11

If 1.76 g of an ideal gas occupy 1.0 L at standard temperature and pressure (STP), what is the molar mass of the gas?

Chemistry
1 answer:
ycow [4]2 years ago
8 0

Answer:

Explanation:

Whenever you see molar masses in gas law questions, more often than not density will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation PV=nRT to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:M=dRTPM = molar mass (g/mol)d = density (g/L)R = Ideal Gas Constant (≈0.0821atm⋅Lmol⋅K) T = Temperature (In Kelvin) P = Pressure (atm)As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the ideal gas law in which the identity of the gas plays a role in your calculations. Just something to take note of. Back to the problem: Now, looking back at what we're given, we will need to make some unit conversions to ensure everything matches the dimensions required by the equation:T=35oC+273.15= 308.15 KV=300mL⋅1000mL1L= 0.300 LP=789mmHg⋅1atm760mmHg= 1.038 atmSo, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:d=0.622g0.300L= 2.073 g/LNow, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.M=dRTP=(2.073)(0.0821)(308.15)1.038= 51 g/molRounded to 2 significant figuresNow if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol). Hope that helped :) 

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Hello!

A) When the temperature inside the container increases, the pressure inside the container increases as well. 

An example for that is when you have an aerosol can and start heating it. The pressure of the gas inside the aerosol can will start to increase, and that would lead to the exploding of the can if heating is kept for too long. Bombs work on this principle too: The heat from the violent chemical reaction inside the closed compartment increase the pressure of the gases until the fragments are ejected at high velocities. 

B) The effect of temperature on the pressure of a gas illustrates Gay-Lussac's Law.

This law was formulated by the famous French chemist that gives it its name. It relates the expansion of a gas with the increase in temperature when the volume is left constant. The Gay-Lussac's Law can be expressed as follows, for the case of this exercise:

\frac{P1}{T1}= \frac{P2}{T2} \\ \\ P2= \frac{P1}{T1}*T2=P1* \frac{80 degC}{25degC}= P1*3,2

You can see that the factor that is multiplying P1 is higher than 1 for the case of heating from 25 °C to 80 °C, so the pressure will increase.

C) At a molecular level, when the temperature is raised the kinetic energy of the molecules inside the container will increase. This increase in the kinetic energy will cause the molecules to move faster, and to hit the walls of the container more often. This causes an increase in the pressure inside the container because there more hits means more force on the walls of the container, and that is the definition of pressure. 

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A balanced reaction equation demonstrates atom conservation by displaying not only the several forms the reactants might take but also their amounts and quantities.

For instance, H + O H0 is the equation for the reaction between hydrogen and oxygen in its most basic form.

However, water is H₂O (the reaction proportion is 2:1, not 1:1), hence the equation needs an adjust: H₂ + O → H₂O.

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