Answer:
the pressure will decrease
Explanation:
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.
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
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Answer:
plzz can you explain in detail this Q
