Answer:
Its when like two pure substances are like combined into one.
Explanation:
Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
Assuming that the solution is simply an aqueous solution
so that it is purely made of NaClO4 (the solute) and water (the solvent), then
I believe the dissolved species would only be the ions of NaClO4, these are:
Na+
ClO4 -
The ionization energy for a hydrogen atom in the n = 2 state is 328 kJ·mol⁻¹.
The <em>first ionization energy</em> of hydrogen is 1312.0 kJ·mol⁻¹.
Thus, H atoms in the <em>n</em> = 1 state have an energy of -1312.0 kJ·mol⁻¹ and an energy of 0 when <em>n</em> = ∞.
According to Bohr, Eₙ = k/<em>n</em>².
If <em>n</em> = 1, E₁= k/1² = k = -1312.0 kJ·mol⁻¹.
If <em>n</em> = 2, E₂ = k/2² = k/4 = (-1312.0 kJ·mol⁻¹)/4 = -328 kJ·mol⁻¹
∴ The ionization energy from <em>n</em> = 2 is 328 kJ·mol⁻¹
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390 because im cool and awsome
