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german
3 years ago
8

Calculate the pressure in the atmosphere of 2.00 moles of helium gas in a 10.0 L container at 27 C

Chemistry
1 answer:
Jet001 [13]3 years ago
4 0

Answer:

The pressure is 4.926 atm

Explanation:

<u>Step 1:</u> Data given

Number of moles = 2.00 moles

Volume of the gas = 10.0 L

Temperature = 27 °C = 273 + 27 = 300 Kelvin

<u>Step 2:</u> Calculate the pressure

P*V = n*R*T

with P= TO BE DETERMINED

with V = 10.0 L

with n = 2.00 moles

with R = 0.00821 L*atm/ K*mol

T = 300 Kelvin

P = (n*R*T) / V = (2 * 0.00821 * 300) / 10 L = 4.926 atm

The pressure is 4.926 atm

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How many moles of bromine atoms are in 2.60×10^2 grams of bromine?
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Which of the following is a property of matter?
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Which element in period 3 has three times the electronegativity value compared to that of li in period 2
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Nitrogen

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7 0
3 years ago
Read 2 more answers
What is the longest wavelength in the Balmer series? (Hint: the Rydberg constant for Hydrogen is 1.096776×107 1/m, and the Balme
boyakko [2]

<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 1.096776\times 10^7m^{-1}

n_f = Higher energy level = n_i+1=(2+1)=3

n_i= Lower energy level = 2    (Balmer series)

Putting the values in above equation, we get:

\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m

Converting this into nanometers, we use the conversion factor:

1m=10^9nm

So, 6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm

Hence, the longest wavelength of light is 656.5 nm

4 0
3 years ago
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