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3 years ago

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A 2.5-liter sample of a gas has 0.30 mole of the gas. If 0.15 mole of the gas is added, what is the final volume of the gas? Tem

perature and pressure remain constant. 3.4 liters 3.8 liters 4.2 liters 4.7 liters

1 answer:

alisha [4.7K]3 years ago

8 0

Answer:

It is 4.2 liters

Explanation:

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Read 2 more answers

Given the following information: Mass of proton = 1.00728 amu Mass of neutron = 1.00866 amu Mass of electron = 5.486 × 10^-4 amu

lesya692 [45]

<u>Answer:</u> The nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

<u>Explanation:</u>

For the given element $_3^6\textrm{Li}$

Number of protons = 3

Number of neutrons = (6 - 3) = 3

We are given:

$m_p=1.00728amu\\m_n=1.00866amu\\A=6.015126amu$

M = mass of nucleus = $(n_p\times m_p)+(n_n\times m_n)$

$M=[(3\times 1.00728)+(3\times 1.00866)]=6.04782amu$

Calculating mass defect of the nucleus:

$\Delta m=M-A\\\Delta m=[6.04782-6.015126)]=0.032694amu=0.032694g/mol$

Converting this quantity into kg/mol, we use the conversion factor:

1 kg = 1000 g

So, $0.032694g/mol=0.032694\times 10^{-3}kg/mol$

To calculate the nuclear binding energy, we use Einstein equation, which is:

$E=\Delta mc^2$

where,

E = Nuclear binding energy = ? J/mol

$\Delta m$ = Mass defect = $0.032694\times 10^{-3}kg/mol$

c = Speed of light = $2.9979\times 10^8m/s$

Putting values in above equation, we get:

$E=0.032694\times 10^{-3}kg/mol\times (2.9979\times 10^8m/s)^2\\\\E=2.938\times 10^{12}J/mol$

Hence, the nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

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