The expression of the Keq is [NH3]2/([N2][H2]3). The common formula of reaction: ax+by=cz is Keq - [z]c/([x]a*[y]b). Remember to use a balanced chemical reaction.
Answer:
Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)
Explanation:
Let's consider the molecular single displacement equation between Zn and Cu(NO₃)₂
Zn(s) + Cu(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Cu(s)
The complete ionic equation includes all the ions and insoluble species.
Zn(s) + Cu²⁺(aq) + 2 NO₃⁻(aq) ⇒ Zn²⁺(aq) + 2 NO₃⁻(aq) + Cu(s)
The net ionic equation includes only the ions that participate in the reaction and insoluble species.
Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)
Answer: The values for Ka and Kb for the species in solution must be known before a prediction can be made
Explanation:
1. If Ka is greater than Kb, the salt solution is acidic
2. If Ka is lesser than Kb, the salt solution is basic
3. If Ka is equal to Kb, the salt solution is acidic
Answer:
The answer to your question is [H₃O⁺] = 0.025 [OH⁻] = 3.98 x 10⁻¹³
Explanation:
Data
[H⁺] = ?
[OH⁻] = ?
pH = 1.6
Process
Use the pH formula to calculate the [H₃O⁺], then calculate the pOH and with this value, calculate the [OH⁻].
pH formula
pH = -log[H₃O⁺]
-Substitution
1.6 = -log[H₃O⁺]
-Simplification
[H₃O⁺] = antilog (-1,6)
-Result
[H₃O⁺] = 0.025
-Calculate the pOH
pOH = 14 - pH
-Substitution
pOH = 14 - 1.6
-Result
pOH = 12.4
-Calculate the [OH⁻]
12.4 = -log[OH⁻]
-Simplification
[OH⁻] = antilog(-12.4)
-Result
[OH⁻] = 3.98 x 10⁻¹³
Answer: 2 kg/L or 2 g/mL
Explanation:
litres are a cubic measure. “Cubic litres” is redundant at best, potentially confusing.
