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:
Explanation:
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In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:
Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:
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Answer:
1.5g/cm³
Explanation:
density=mass÷volume
mass= 1.5kg (<em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>i</em><em>n</em><em>t</em><em>o</em><em> </em><em>g</em>) = 1500g
volume of the cube = 10×10×10 = 1000cm³
density= divide 1500g÷1000cm = 1.5g/cm³
<h2>
Density= 1.5g/cm³</h2>
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Answer:
Explanation:
Any densities less than 1g/cm3 will float, while objects with densities over 1g/cm3 will sink.
Answer:
1.03 atm
Explanation:
Primero <u>convertimos 21 °C y 37 °C a K</u>:
- 21 °C + 273.16 = 294.16 K
- 37 °C + 273.16 = 310.16 K
Una vez tenemos las temperaturas absolutas, podemos resolver este problema usando la<em> ley de Gay-Lussac</em>:
En este caso:
Colocando los datos:
- 294.16 K * P₂ = 310.16 K * 0.98 atm
Y <u>despejando P₂</u>:
