Answer:
NH3(aq)
Explanation:
Gold III hydroxide is an inorganic compound also known as auric acid. It can be dehydrated at about 140°C to yield gold III oxide. Gold III hydroxide is found to form precipitates in alkaline solutions hence it is not soluble in calcium hydroxide.
However, gold III hydroxide forms an inorganic complex with ammonia which makes the insoluble gold III hydroxide to dissolve in ammonia solution. The equation of this complex formation is shown below;
Au(OH)3(s) + 4 NH3(aq) -------> [Au(NH3)4]^3+(aq) + 3OH^-(aq)
Hence the formation of a tetra amine complex of gold III will lead to the dissolution of gold III hydroxide solid in aqueous ammonia.
The answer should be 479 i could possibly be wrong but that’s what i got because one mole is 47.90 grams
Answer:
Explanation:
1. Please provide the enthalpy info - I will work on it with the info
2.
i) Reaction a should be modified to match the number of S in equation:
2S + 2O2 -> 2SO2 deltaH = -370kJ
ii) Reaction b should be written reversely to match the reactants of SO2:
2SO2 + O2 -> 2SO3 deltaH = 256kJ
iii) Adding the equations together:
2S + 3O2 -> 2SO3
iv) Enthalpy of the combined reaction = -370+256 = -114kJ
It is negative so the reaction is exothermic.
<span>The problem has to do with oxidation states of the matter. The oxidation state of oxygen will always be -2 with the exception of peroxides which will have a state of -1. The overall balanced state of chemical compounds will be 0, so the oxidation state of Mn in MnO2 will be +4. The oxidation state of MnO4- will then be +7 to balance out to the negative one charge. The state change from +4 to +7 is 3, thus three electrons have to be lost in order for this to happen; a loss of a charge of -3 results in an increase of charge of 3. Oxidation is always the process of 'losing' electrons.
</span><span>E] MnO2(s) MnO4-(aq</span>
