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Burka [1]
3 years ago
14

What volume (in liters) does 3.91 moles of nitrogen gas at 5.35 atm and 323 K occupy

Chemistry
1 answer:
ivanzaharov [21]3 years ago
4 0
Data Given:
Moles = n = 3.91 mol

Pressure = P = 5.35 atm

Temperature = T = 323 K

Volume = V = ?

Formula used: Ideal Gas Equation is used,

P V = n R T
Solving for V,
V = n R T / P
Putting Values,
V = (3.91 mol × 0.0825 atm.L.mol⁻¹.K⁻¹ × 323 K) ÷ 5.35 atm

V = 19.36 L
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Answer:

3.90*10^{-25}kg/atom

Explanation:

The molar mass of uranium-235 is 235 g/mol. So one mole of uranium-235 has a mass of 235 g. Put differently 6.022×10^23 atoms of uranium-235 have a mass of 235 g. Knowing that, how can we use that to find the mass of one atom?

mass of one atom = \frac{235 g}{1mol} *\frac{1 mol}{6.022*10^{23}atoms } *\frac{1kg}{1000g}= 3.90*10^{-25}kg/atom

7 0
2 years ago
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Answer:

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Vanadium (V)

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Which one of the following best represents the predicted approximate chemical shift and coupling:
mezya [45]

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C. 3.40 ppm, singlet

Explanation:

Given the information from the question .we have to select the best representation for represents the predicted approximate chemical shift and coupling for hydrogen(s). In this case, there is no neighboring hydrogen .Thus there won’t be any split .the best option answer is C. 3.40 ppm, singlet . Therefore the correct answer or option is C. 3.40 ppm, singlet.

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never [62]

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b. the maximum number of p-p bonds that the atom can form is 2

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Hybridization is the mixing of at least two nonequivalent orbitals, in this case, we have the mixing of one <em>s, 3 p </em> and <em> 2 d </em> orbitals. In hybridization the number of hybrid orbitals generated  is equal to the number of pure atomic orbital, so we have 6 hybrid orbital.

The shape of this hybrid orbital is octahedral (look the attached image) , it has 4 orbital located in the plane and 2 orbital perpendicular to it.

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4 0
3 years ago
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