H2SO4 + 2 NaOH ----> Na2SO4 + 2 H2O.
0.085 L * 0.176 mol/L = 0.01496 mol H2SO4
is neutralised by 0.01496 mol * 2
= 0.02992 mol NaOH.
1000 mL of 0.492 M NaOH
contains 0.492 moles NaPH.
0.02992 / 0.452 * 1000 mL
= 66.19 = 66 mL
Answer:

Explanation:
Given that:
The work function of the rhodium = 480.5 kJ/mol
It means that
1 mole of electrons can be removed by applying of 480.5 kJ of energy.
Also,
1 mole = 
So,
electrons can be removed by applying of 480.5 kJ of energy.
1 electron can be removed by applying of
of energy.
Energy required = 
Also,
1 kJ = 1000 J
So,
Energy required = 
Also,
Where,
h is Plank's constant having value
c is the speed of light having value
So,



Also,
1 m = 10⁻⁹ nm
So,

Answer:
-431.5 J/g
Explanation:
Mass of solution = Mass of solute + mass of solvent
Solute is KOH while solvent is water.
Mass of KOH = 16.9 g
Mass of water = 90.8 g
Mass of solution = 16.9 + 90.8
= 107.7 g
Change in temperature (Δt) = 34.27 - 18.5
= 16.2 °C
Heat required to raise the temperature of water is released by dissolving KOH.
Therefore,
Heat released by KOH = m × s× Δt
= 107.7 × 4.18 × 16.2
= 7293 J
Heat released by per g KOH = 7293 J/16.9 g
= 431.5 J/g
As heat is released therefore, enthalpy change would be negative.
Enthalpy change of KOH = -431.5 J/g
The answer would be c I think k