A.)49.4974874 moles or 49.5 moles
B.)2.980808730172671e+25 or 3e+25
<span>London dispersion forces
is the weakest intermolecular force. It is a temporary force that happens when
electrons of two adjacent atoms occupy positions that make atoms form dipoles
which are temporary dipoles. This is also referred as dipole-dipole attraction.</span>
<u>Answer:</u> The equilibrium constant for the above reaction is 1.31
<u>Explanation:</u>
We are given:
Initial moles of nitrogen gas = 0.3411 moles
Initial moles of hydrogen gas = 1.661 moles
Equilibrium moles of nitrogen gas = 0.2001 moles
For the given chemical reaction:
<u>Initial:</u> 0.3411 1.661
<u>At eqllm:</u> 0.3411-x 1.661-3x 2x
Evaluating for 'x', we get:
Volume of the container = 2.50 L
The expression of 
for the above equation follows:
![K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ctimes%20%5BH_2%5D%5E3%7D)
We are given:
![[NH_3]=\frac{2\times 0.141}{2.50}=0.1128M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D%5Cfrac%7B2%5Ctimes%200.141%7D%7B2.50%7D%3D0.1128M)
![[N_2]=\frac{0.2001}{2.5}=0.08004M](https://tex.z-dn.net/?f=%5BN_2%5D%3D%5Cfrac%7B0.2001%7D%7B2.5%7D%3D0.08004M)
![[H_2]=\frac{1.661-(3\times 0.141)}{2.5}=0.4952M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%5Cfrac%7B1.661-%283%5Ctimes%200.141%29%7D%7B2.5%7D%3D0.4952M)
Putting values in above expression, we get:
Hence, the equilibrium constant for the above reaction is 1.31
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1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.
How to calculate number of atoms?
The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.
However, the number of moles of oxygen in glycine can be calculated using the following expression:
Molar mass of C₂H5O2N = 75.07g/mol
Mass of oxygen in glycine = 32g/mol
Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine
Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles
Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms
Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.
Learn more about number of atoms at: brainly.com/question/8834373
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