Answer:
B
Explanation:
First of all it is important to know that a half filled orbital is particularly stable. In phosphorus all the electrons occur singly in the 3p sublevel minimizing inter electronic repulsion hence it is more difficult to remove an electron from this energetically stable arrangement. In sulphur, electrons are paired in one of the 3p orbitals thereby lowering the energy of that level due to instability caused by interelectronic repulsion between two electrons in the same orbital.
Answer:
Increase
Explanation:
If temperature is held constant, the equation is reduced to Boyle's law. Therefore, if you decrease the pressure of a fixed amount of gas, its volume will increase.
Answer:
i have taken environmental science so i can help
Yes it does dissolve hope this helps
A) Limiting reactant
You need the molar ratios (from the balanced chemical equation) and the molar masses of each compound (from the atomic masses)
a) Molar ratios:
6 mol HF : 1 mol SiO2 : 1 mol H2SiF6
2) Molar masses:
Atomic masses:
H: 1 g/mol
F: 19 g/mol
Si: 28 g/mol
O: 16g/mol
=>
HF:1g/mol + 19 g/mol = 20 g/mol
SiO2: 28g/mol + 2*16g/mol = 60 g/mol
H2SiF6: 2*1g/mol + 28g/mol + 6*19g/mol = 144g/mol
3) convert data in grams to moles
21.0 g SiO2 / 60 g/mol = 0.35 mol SiO2
70.5 g HF / 20 g/mol = 3.525 mol HF
4) Use the theorical ratios to deduce which is in excess and which is the limiting reactant.
6 mol HF / 1mol SiO2 < 3.525 mol HF / 0.35 mol SiO2 ≈ 10
=> There is more HF than the needed to react with 0.35mol of SiO2 =>
SiO2 is the limiting reactant (HF is in excess)
b) Mass of excess reactant.
1) Calculate how many grams reacted, which requires to calculate first the number of moles that reacted
0.35 mol SiO2 * 6 mol HF / 1 mol SiO2 = 2.1 mol of HF
2.1 mol HF * 20 g/mol = 42 gram of HF
2) Subtract the quantity that reacted from the original quantity:
70.5 g - 42 g = 28.5 g of HF in excess
c) Theoretical yield of H2SiF6
1 mol of SiO2 ; 1 mol of H2SiF6 => 0.35 mol SiO2 : 0.35 mol H2SiF6
Convert those moles to grams: 0.35 mol * 144 g/mol = 50.4 grams
d) % yield
% yield = actual yield / theoretical yield * 100 = 45.8 / 50.4 * 100 = 90.87%
