To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy.
The amount of energy required by an isolated, gaseous molecule in the electronic state of the ground to absorb in order to discharge an electron and produce a cation has been known as the ionization energy. The amount of energy required for every atom in a mole to drop one electron is most often given as kJ/mol.
Anything that causes electrically neutral atoms and molecules to gain or lose electrons in order to become electrically charged atoms as well as molecules .
Therefore, the "To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy."
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Answer:
130 Liters
Explanation:
if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)
Answer:
Change in chemical properties
Explanation:
There are 1.92 × 10^23 atoms Mo in the cylinder.
<em>Step 1</em>. Calculate the <em>mass of the cylinder
</em>
Mass = 22.0 mL × (8.20 g/1 mL) = 180.4 g
<em>Step 2</em>. Calculate the<em> mass of Mo
</em>
Mass of Mo = 180.4 g alloy × (17.0 g Mo/100 g alloy) = 30.67 g Mo
<em>Step 3</em>. Convert <em>grams of Mo</em> to <em>moles of Mo
</em>
Moles of Mo = 30.67 g Mo × (1 mol Mo/95.95 g Mo) = 0.3196 mol Mo
<em>Step 4</em>. Convert <em>moles of M</em>o to <em>atoms of Mo
</em>
Atoms of Mo = 0.3196 mol Mo × (6.022 × 10^2<em>3</em> atoms Mo)/(1 mol Mo)
= 1.92 × 10^23 atoms Mo
Data Given:
Time = t = ?
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 107.86/1 = 107.86 g
Amount Deposited = W = 17.3 g
Solution:
According to Faraday's Law,
W = I t e / F
Solving for t,
t = W F / I e
Putting values,
t = (17.3 g × 96500) ÷ (10 A × 107.86 g)
t = 1547.79 s
t = 1.54 × 10³ s
