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:
Translate from Ukrainian.
У цей приклад я візьму селекції пшениці. Коли ми будемо говорити про селекцію цього злака, то ми можемо отримати меньшу собівартість, та вартість вирощування. Збільшена урожайність, та менша потреба в гербіцидах, пестицидах, та інсектицидах це теж наслідок. Тому селекція це майбутнє!
I think the correct answer is the first option. It has nonpolar bonds and a symmetrical structure. The structure of a BF3 molecule shows a symmetrical trigonal geometry. The net dipole moment of the molecule is zero therefore it is polar.
1. At constant tempaerature and pressure, 3 tablets produce 600cm^3 of gas
Thus calculating for 1 tablet that produces 600 / 3 = 200 cm^3
So now two tablets produce 200 x 2 = 400 cm^3
2. We have the equation PV = nRT, n being the number of moles
Pressure P = 1,000 kPa
Volume V = 3 L
R = 8.31 L kPa/mol-K
Temperature T = 298 K
n = PV / RT = (1000 x 3) / (8.31 x 298) = 3000 / 2476.38 = 1.21 moles
Number of moles = 1.21 moles.
Answer : The new volume of the air is, 6.83 L
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:
Putting values in above equation, we get:
Therefore, the new volume of the air is, 6.83 L
