To compute for the empirical formula, assume there is 100 grams of the compound. That means there is 40.28 g B, 52.2 g N and 7.53 g H. Convert the mass into moles using their molar masses:
40.28 g B * 1 mol/10.811 g = 3.725835 mol B
52.2 g N * 1 mol/14 g = 3.72857 mol N
7.53 g H * 1 mol/1 g = 7.53 mol H
Divide all the moles by the smallest amount which is 3.725835 mol.
B: 3.725835/3.725835 = 1
N: 3.72857/3.725835 = 1
H: 7.53/3.725835 = 2
Therefore, the empirical formula is BNH₂.
Answer:
3-
Explanation:
Sodium aurothiosulfate is a salt with the formula Na₃Au(S₂O₃)₂. The cation of the salt is sodium ion, and the anion is aurothiosulfate ion. We can determine the charge of the aurothiosulfate ion, considering that the sum of the positive and negative charges must be equal to the charge of the compound, which is zero.
3 × Na⁺ + 1 × Au(S₂O₃)₂ⁿ⁻ = 0
3 × +1 + 1 × Au(S₂O₃)₂ⁿ⁻ = 0
Au(S₂O₃)₂ⁿ⁻ = 3-
Answer:
The pressure of N₂ gas in cylinder B when compressed at constant temperature increases due to the increase in the frequency of collision between the gas molecules with themselves and with the wall of their container caused by a decrease in volume of the container.
Explanation:
Gas helps to explain the behavior of gases when one or more of either temperature, volume or pressure is varying while the other variables are kept constant.
In the gas cylinder B, the temperature of the given mass of gas is kept constant, however, the volume is decreased by pushing the movable piston farther into the cylinder. According to the gas law by Robert Boyle, the volume of a given mass of gas is inversely proportional to its pressure at constant temperature. This increase in pressure is due to the increase in the frequency of collision between the gas molecules with themselves and with the wall of their container caused by a decrease in volume of the container. As the cylinder becomes smaller, the gas molecules which were spread out further become more packed closely together, therefore, their frequency of collision increases building up pressure in the process.
Answer:
1 B
2A
#D
3B
1D
2B
3C
4A
A,C,A,B
C,A,C,A
C,A,A,B
Explanation:
Easy ive done this brings back memories
Answer:
Boron and Aluminium
Explanation:
Boron and Aluminium are present in Group 13 of the modern periodic table. Group 13 (IUPAC System) can also be referred to as Group III-A. Logically, Boron and Aluminum can't be placed alongwith elements such as Yttrium as they don't exhibit properties of a transition metal.
