Answer:
Based on compounds given, NO reaction occurs
Explanation
The compounds should exchange ions to generate a driving force that pulls the reaction to completion. => Example ...
The Molecular Equation is ...
NH₄Cl(aq) + AgNO₃(aq) => NH₄NO₃(aq) + AgCl(s)
Silver chloride forms in this reaction as a solid precipitate because of its low solubility and is the 'Driving Force' of the reaction. Driving Force is a more stable compound than any on the reactant side and when formed leaves the reaction system as a solid ppt, liquid weak electrolyte (i.e., weak acid or weak base) or a gas decomposition product of a weak electrolyte.
The Ionic Equation is ...
NH₄⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) => NH₄⁺(aq) + NO₃⁻(aq) + AgCl(s)
This shows all ions from reaction plus the Driving Force of the reaction.
The Net Ionic Equation is ...
Ag⁺(aq) + Cl⁻(aq) => AgCl(s)
The Net Ionic Equation shows only those ions undergoing reaction. The NH₄⁺ and NO₃⁻ ions are 'Spectator Ions' and do not react.
Attached is a reference sheet for determining the Driving Force of a Metathesis Double Replacement Reaction. Suggest reviewing acid-base theories and the products of decomposition type reactions.
Answer:
See explanation and image attached
Explanation:
This reaction is known as mercuric ion catalyzed hydration of alkynes.
The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.
Kiloliters and megaliters
I can’t see what image you have to refer too but India is now part of the rest of Asia due to continental plates running into eachother.
Answer:
31.9 °C
Explanation:
The formula for the heat q absorbed by an object is
q = mCΔT where ΔT = (T₂ - T₁)
Data:
q = 12.35 cal
m = 19.75 g
C = 0.125 cal°C⁻¹g⁻¹
T₂ = 37.0 °C
Calculations
(a) Calculate ΔT
q = mCΔT
12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT
12.35 = 2.406ΔT °C⁻¹
ΔT = 12.35/(2.406 °C⁻¹) = 5.13 °C
(b) Calculate T₂
ΔT = T₂ - T₁
T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C
The original temperature was 31.9 °C.