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2 years ago

Under what conditions do real gases behave most ideally.

Chemistry

1 answer:

katrin [286]2 years ago

3 0

Answer:

Under high temperatures and low pressure, gases behave the most ideal.

Explanation:

Low pressure reduces the effect of the finite size of real particles by increasing the volume around each particle, and a high temperature gives enough kinetic energy to the particles to better overcome the attractions that exist between real particles. (Prevents sticking.)

In summary, real gases behave more like ideal gases when they are far away from a phase boundary, (condensation or freezing).

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How many oxygen atoms are present in 3 grams of glucose

Kaylis [27]

Answer:

Because CLEARLY, each mole of glucose, C6H12O6 contains 6⋅mol oxygen atoms.

5 0

3 years ago

Which of these have the same number of particles as 1 mole of water H2O

Fed [463]

Answer:

It is equal to Avogadro's number (NA), namely 6.022 x1023. If we have one mole of water, then we know that it will have a mass of 2 grams (for 2 moles of H atoms) + 16 grams (for one mole O atom) = 18 grams.

Explanation:

The question is not very much clear.

If you are asking for molecules then 1 mole water= 6.023 * 10^23

If you are asking for atoms then 1 mole water= 6.023 * 10^23 * 3

If you are asking for particles then,

So, in your example you would have one mole of water molecules. If you dissociated those water molecules, than you would end up with 2 moles of hydrogen atoms, and one mole of oxygen atoms.

I hope that was helpful!

H=1 proton,1 electron

O=8 protons,8 neutrons and 8 electrons

total particles in one H2O molecule-28

total no. of particles in 1 mole of water- 6.023 * 10^23 * 28

8 0

3 years ago

What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l

Vladimir [108]

Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law: $PV=nRT$ where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units: $R=0.0821\frac{L\cdot atm}{mol\cdot K}$

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means $T=273.15K$ and $P=1atm$

Lastly, we must calculate the number of moles of $O_2(g)$ there are. Given 0.80g of $O_2(g)$ , we will need to convert with the molar mass of $O_2(g)$ . Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: $32g\text{ }O_2=1mol\text{ }O_2$

Thus, $\frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n$

Isolating V in the Ideal Gas Law:

$PV=nRT$

$V=\frac{nRT}{P}$

...substituting the known values, and simplifying...

$V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}$

$V=0.56L \text{ } O_2$

So, 0.80g of $O_2(g)$ would occupy 0.56L at STP.

5 0

2 years ago

Read 2 more answers

Is poverty a cycle? And how?

Shalnov [3]

It is a cycle because if your mum or dad was poor then you would be poor unless you were able to get a job too make enough money to be in middle or 1st class

3 0

3 years ago

Read 2 more answers

A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu

bogdanovich [222]

Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

Explanation:

The temperature of an object can be written using different temperature scales.

The two most important scales are:

- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).

- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

$T(K)=T(^{\circ}C)+273.15$

Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

6 0

4 years ago