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GREYUIT [131]
1 year ago
5

How many moles of helium is required to blow up a balloon to 87.1 liters at 74 C and 3.5 atm?

Chemistry
1 answer:
marshall27 [118]1 year ago
8 0

Moles of helium is required to blow up a balloon to 87.1 liters at 74 C and 3.5 atm is  021.65 mole

Mole is the unit of amount of substances of specified elementary entities

According to the ideal gas law he number of moles of a gas n can be calculated knowing the partial pressure of a gas p in a container with a volume V at an absolute temperature T from the equation

n =pV/RT

Here given data is volume = 87.1 liters

Temperature = 74 °C means 347.15 k

Pressure = 3.5 atm

R = 0.0821

Putting this value in ideal gas equation then

n =pV/RT

n =  3.5 atm×87.1 liters / 0.0821 ×347.15 k

n = 021.65 mole

Moles of helium is required to blow up a balloon to 87.1 liters at 74 C and 3.5 atm is  021.65 mole

Know more about mole of helium

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11) If you have a density of 100 kg/L,, tell me the following:
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The unit of mass is 'Kilogram' which is written as 'kg' and volume, v = 10 L.

<h3>Equation :</h3>

To calculate the volume

Use formula,

density = mass / volume

density = 100 kg/L

mass  = 1000 kg

volume = mass / density

v = 1000/100

v = 10 L

<h3>What is density mass?</h3>

A substance, material, or object's mass density is a measure of how much mass (or how many particles) it has in relation to the volume it occupies.

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I understand the question you are looking for :

If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First what are the mass units? Secondly, what is the volume?

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1 year ago
Can light bend around corners to reach an object
Sholpan [36]

Answer: Yes, light can bend around corners. In fact, light always bends around corners to some extent.

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Does neon have a larger atomic radius than argon
Lera25 [3.4K]

Answer:

No

Explanation:

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For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80
Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

A\rightleftharpoons B

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%}  =0.25

Thus, by recalling the Van't Hoff's equation, we can write:

ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Hence, we solve for the enthalpy change as follows:

\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }

Finally, we plug in the numbers to obtain:

\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}

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