The kinetic energy of emitted electrons when cesium is exposed to UV rays of frequency 1.80 * 10 ^15 is 3.054 x 10 ^- 19 J.
Explanation:
To calculate the kinetic energy of emitted electrons,
It is given that the frequency is 1.80* 10^15 Hz
We have,
KE = E – Eo = hv –hvo
Where, h = 6.626 x 10^ - 34 Js
Given frequency = 1.4 x 10 ^ 15 Hz
vo (Threshold frequency) for cesium = 9.39 x 10 ^ 14 Hz
Applying in equation,
we get
KE = 6.626 x 10^34 (1.4 x 10^15 - 9.39 x 10^14)
KE= 3.054 x 10^-19 J
[Note: Here, threshold frequency of Cesium is not provided. Apply the correct threshold frequency from the part A]
Formula: molality, m = n solute / kg solvent
n solute = # of moles of solute = mass(g) / molar mass
Molar mass of Mg Br2 = 184.11 g/mol
m = [46g / 184.11 g/mol] / 0.5 kg = 0.50 mol/kg
Answer:
C₁₂H₂₂O₁₁ and CH₃OH
Explanation:
Sucrose and methyl alcohol are nonelectrolytes. They do not ionize or conduct a current in aqueous solution.
HC₂H₃O₂ is a weak electrolyte. It produces only a few ions and is a poor conductor of electricity in aqueous solution.
HC₂H₃O₂ + H₂O ⇌ H₃O⁺ + C₂H₃O₂⁻
H₂SO₄ is a strong electrolyte. Its first ionization is complete, so it is a good conductor of electricity in aqueous solution.
H₂SO₄ + H₂O ⟶ H₃O⁺ + HSO₄⁻
