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3 years ago

How is everyone doing today?

Chemistry

2 answers:

anzhelika [568]3 years ago

4 0

Answer:Not so good, but I'll manage. What about you?

Explanation:

Luden [163]3 years ago

4 0

Doing ok. Getting tired of being at home though. Not been anywhere for 3 weeks

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Which of the following is true concerning gamma radiation?

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The answer should be D

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The most concentrated solution is b

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A certain atom has a nucleus containing six protons and eight neutrons and has six electrons orbiting the nucleus

zavuch27 [327]

Answer:

The answer is a carbon isotope.

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4 years ago

The density of copper decreases as temperature increases (as does the density of most substances). Which change occurs in a samp

azamat

Answer: Option (c) is the correct answer.

Explanation:

As we know that density the amount of mass present in per unit volume.

Mathematically, Density = $\frac{mass}{volume}$

So, it means that density is inversely proportional to volume. Hence, when there will be decrease in density of a substance then there will be increase in its volume. That is, expansion of substance will take place.

Also, boiling point of copper is 2,562 degree celsius but we are heating it up to a temperature of 95 degree celsius. This means that copper will remain in liquid state at this temperature.

Thus, we can conclude that a change which occurs in a sample of copper is that copper sample will expand.

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3 years ago

A 1.93-mol sample of xenon gas is maintained in a 0.805-L container at 306 K. Calculate the pressure of the gas using both the i

Alex

Answer : The pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

Explanation :

First we have to calculate the pressure of gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $Xe$ gas = ?

V = Volume of $Xe$ gas = 0.805 L

n = number of moles $Xe$ = 1.93 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $Xe$ gas = 306 K

Now put all the given values in above equation, we get:

$P\times 0.805L=1.93mole\times (0.0821L.atm/mol.K)\times 306K$

$P=60.2atm$

Now we have to calculate the pressure of gas by using van der Waals equation.

$(P+\frac{an^2}{V^2})(V-nb)=nRT$

P = Pressure of $Xe$ gas = ?

V = Volume of $Xe$ gas = 0.805 L

n = number of moles $Xe$ = 1.93 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $Xe$ gas = 306 K

a = pressure constant = $4.19L^2atm/mol^2$

b = volume constant = $5.11\times 10^{-2}L/mol$

Now put all the given values in above equation, we get:

$(P+\frac{(4.19L^2atm/mol^2)\times (1.93mole)^2}{(0.805L)^2})[0.805L-(1.93mole)\times (5.11\times 10^{-2}L/mol)]=1.93mole\times (0.0821L.atm/mol.K)\times 306K$

$P=44.6atm$

Therefore, the pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

5 0

3 years ago