A solution with a higher concentration of hydroxide ions than hydrogen ions is basic solution.
This solution formed by Base dissolved in water and release hydroxide ions.
The PH of the solution is greater than 7
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:- partial pressure of Kr = 0.306 atm, partial pressure of oxygen = 0.264 atm and partial pressure of carbon dioxide = 0.396 atm
Total pressure is 0.966 atm
Solution:- moles of Kr = 21.7 g x (1mol/83.8g) = 0.259 mol
moles of oxygen = 7.18 g x (1mol/32g) = 0.224 mol
moles of carbon dioxide = 14.8 g x (1mol/44g) = 0.336 mol
Volume of container = 23.1 L and the temperature is 59 + 273 = 332 K
From ideal gas law equation, P = nRT/V
partial pressure of Kr = (0.259 x 0.0821 x 332).23.1 = 0.306 atm
partial pressure of oxygen = (0.224 x 0.0821 x 332)/23.1 = 0.264 atm
partial pressure of carbon dioxide = (0.336 x 0.0821 x 332)/23.1 = 0.396 atm
Total pressure of the gas mixture = 0.306 atm + 0.264 atm + 0.396 atm = 0.966 atm
