The net ionic equation for the neutralization reaction involving equal molar amount amount of HNo3 and KoH is
H^+ + OH^- = H2O (l)
explanation
write the chemical equation
HNO3 (aq) + KOH(aq) = KNO3(aq) +H2O (l)
ionic eequation
H^+(aq) + NO3^- (aq) + K^+9aq) OH^-(aq) = K^+ (aq) + NO3^-(aq) + H2O(l)
cancel the spectator ions( ions which does not take place in equation ) for this case is NO3^- and No3^-
thus the net ionic is
H^+(aq) + OH^- (aq) = H2O(l)
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Answer:
0.99%
Explanation:
The Henderson-Hasselbalch equation relates the ratio of the ionized to the non-ionized form of an acid as follows:
pH = pKa + log([A⁻]/[HA])
Inserting the values for pH and pKa and solving gives:
7.4 = 5.2 + log([A⁻]/[HA])
2 = log([A⁻]/[HA])
[A⁻]/[HA] = 100/1
The percentage of the weak acid in non-ionized form (HA) is then calculated:
[HA]/([A⁻] + [HA]) = 1/(100+1) x 100% = 0.99&
Answer:
B. When it is obtained by rounding, as in 3.0
Explanation: