Given:
Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL
The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:
45ml * 0.10 M = C analyte * 20 ml
C analyte = 0.225 M
Answer:
2. The metal would lose one electrons and the non metal would gain one electrons
Explanation:
An atom of a certain element reacts with the atoms of other elements in order to fullfill its outermost shell (called valence shell).
We notice the following:
- The elements in Group 1 (which are metals) have only 1 electron in their valence shell
- The elements in Group 17 (which are non-metals) have 1 vacancy (lack of electron) in their valence shell
This means that in order for both an atom of group 1 and an atom of group 17 to fullfill the valence shell, they have to:
- The atom in group 1 has to give away its only electron of the valence shell
- The atom in group 17 has to gain one electron in order to fullfill the shell
Therefore, the correct option is
2. The metal would lose one electrons and the non metal would gain one electrons
Answer:
Explanation:
The volume and amount are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
Data:
p₁ = 1520 Torr; T₁ = 27 °C
p₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 27 + 273.15) K = 300.15 K
T₂ = (150 + 273.15) K = 423.15 K
(b) Calculate the new pressure
(c) Convert the pressure to atmospheres
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
6 g S * 1 mol/32.06 g S = 0.187 mol S
Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂
