What are two characteristics of an ideal gas that are not true of a real gas

Chemistry

1 answer:

MrRissso [65]3 years ago

5 0

There are many differences between ideal gas and real gas; some of the main differences are as following:

An ideal gas follows the formula PV=nRT but a real gas does not always follow this formula.
There is no attraction between the molecules of an ideal gas. A real gas has significant particle attractions.
The particles of an ideal gas lose no energy to its container. A real gas conducts and radiates heat, thereby losing energy.
An ideal gas is infinitely compressible, a real gas will condense to a liquid at some pressure.
Real gas particles have a volume and ideal gas particles do not.
Real gas particles collide in-elastically (loses energy with collisions) and ideal gas particles collide elastically.

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3 years ago

The metal tantalum becomes superconducting at temperatures below 4.483 K. Calculate the temperature at which tantalum becomes su

masha68 [24]

Answer:

The correct answer is "-268.667°C".

Explanation:

Given:

Temperature,

= 4.483 K (below)

Now,

The formula of temperature conversion will be:

⇒ $T(^{\circ} C)=T(K)-273.15$

By putting the values, we get

⇒ $=4.483-273.15$

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Thus the above is the correct answer.

3 0

3 years ago

If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio

bulgar [2K]

Answer:

$m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

$n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2$

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

$m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2$