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spin [16.1K]
3 years ago
6

Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The

energy needed to ionize a ground-state hydrogen atom is 2.18 x 10–18 J.
Chemistry
1 answer:
qaws [65]3 years ago
4 0

Answer:

E_{ionization}=5.45\times 10^{-19}\ J

Explanation:

E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

So, n_i=2 and n_f=\infty (As the hydrogen has to ionize)

Thus,

\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J

E_{ionization}=5.45\times 10^{-19}\ J

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4 years ago
How many grams are in 3.14 moles of PI₃?
OverLord2011 [107]

Answer:

\boxed {\boxed {\sf 1290 \ g \ PI_3}}

Explanation:

We want to convert from moles to grams, so we must use the molar mass.

<h3>1. Molar Mass</h3>

The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.

  • Phosphorus (P): 30.973762 g/mol
  • Iodine (I): 126.9045 g/mol

Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.

  • I₃: 126.9045 * 3=380.7135 g/mol
  • PI₃: 30.973762 + 380.7135 = 411.687262 g/mol

<h3>2. Convert Moles to Grams</h3>

Use the molar mass as a ratio.

\frac {411.687262 \ g \ PI_3}{ 1 \  mol \ PI_3}

We want to convert 3.14 moles to grams, so we multiply by that value.

3.14 \ mol \ PI_3 *\frac {411.687262 \ g \ PI_3}{ 1 \  mol \ PI_3}

The units of moles of PI₃ cancel.

3.14 *\frac {411.687262 \ g \ PI_3}{ 1 }

1292.698 \ g\ PI_3

<h3>3. Round</h3>

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.

  • 1292.698

The 2 in the ones place tells us to leave the 9.

1290 \ g \ PI_3

3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>

4 0
3 years ago
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