Answer:
Explanation:
Using freezing point depression formula,
ΔTemp.f = Kf * b * i
Where,
ΔTemp.f = temp.f(pure solvent) - temp.f(solution)
b = molality
i = van't Hoff factor
Kf = cryoscopic constant
= 1.86°C/m for water
= (0 - (-5.58))/1.86
= 3.00 mol/kg
Assume 1 kg of water(solvent)
= (3.00 x 1)
= 3.00 mol.
Potential energy and height; best guess;)
They're only found in the nucleus and play an important role in keeping the atom stable because they carry a negative charge to counteract the proton's positive charge.
Answer:
Qm = -55.8Kj/mole
Explanation:
NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)
Qm = (mc∆T)water /moles acid
Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)
=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)
=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)
ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃
= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*
Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.
