Answer:
mass of C₂H₄N₂ = 3.472 g
Explanation:
We have the following chemical reaction:
3 CH₄ + 5 CO₂ + 8 NH₃ → 4 C₂H₄N₂ + 10 H₂O
Using the masses given by the problem we calculate the number of moles for each reactant:
number of moles = mass / molecular weight
number of moles of CO₂ = 13.2 / 44 = 0.3 moles
number of moles of NH₃ = 2.18 / 17 = 0.13 moles
number of moles of CH₄ = 17 / 16 = 1.06 moles
We can see that the limiting reactant is ammonia NH₃. Now we can devise the following reasoning:
if 8 moles of NH₃ produces 4 moles of C₂H₄N₂
then 0.13 moles of NH₃ produces X moles of C₂H₄N₂
X = (0.13 × 4) / 8 = 0.062 moles of C₂H₄N₂
mass of C₂H₄N₂ = number of moles × molecular weight
mass of C₂H₄N₂ = 0.062 × 56 = 3.472 g
Approximately, or at least the test must be repeated until 3 times, to find an accurate average and hypothesis.
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.
Answer:
11.842g
Explanation:
To find mass the formula is volume times density.
Lar mass of Ca<span> = 40.08 </span>grams/mole 77.4 g Ca<span> * ( 1 </span>mole Ca<span>/ 40.08 ... n = m / M 1mol </span>Ca<span>weights 40 gmol-1 n = 77,4 / 40 = 1.93 </span>mol<span>.</span>
