Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?
Solving:
Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT


If: ΔT (T final - T initial) = ?


Hello!
When NaOH is added to a buffer composed of CH₃COOH and CH₃COO⁻, the following reactions happen:
-First, the NaOH is neutralized by CH₃COOH:
NaOH + CH₃COOH → H₂O + CH₃COONa
-Second, the CH₃COONa dissociates in its ions:
CH₃COONa → CH₃COO⁻ + Na⁺
-Finally, CH₃COO⁻ (a weak base) reacts with water to form OH⁻ ions and regenerate CH₃COOH
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻
By this sequence of reactions, the buffer can mitigate the effect of the strong base added.
Have a nice day!
Answer:
Friction of wind against the ocean surface
Answer:
Removing the cap from a soft drink bottle releases pressure and causes the excess carbon dioxide molecules to come out as bubbles. however the drink is still supersaturated , and will release that carbon dioxide till it goes flat