Answer:
46.40 g.
Explanation:
- It is a stichiometric problem.
- The balanced equation of the reaction: 4K + O₂ → 2K₂O.
- It is clear that 4.0 moles of K reacts with 1.0 mole of oxygen produces 2.0 moles of K₂O.
- We should convert the mass of K (38.5 g) into moles using the relation:
<em>n = mass / molar mass,</em>
n = (38.5 g) / (39.098 g/mol) = 0.985 mole.
<em>Using cross multiplication:</em>
4.0 moles of K produces → 2.0 moles of K₂O, from the stichiometry.
0.985 mole of K produces → ??? moles of K₂O.
∴ The number of moles of K₂O produced = (0.985 mole) (2.0 mole) / (4.0 mole) = 0.4925 mole ≅ 0.5 mole.
- Now, we can get the mass of K₂O:
∴ mass = n x molar mass = (0.5 mole) (94.2 g/mol) = 46.40 g.
Answer:
0.3192 M
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 5.32 mL Molarity of stock solution (M1) = 6 M
Volume of diluted solution (V2) = 100 mL
Molarity of diluted solution (M2) =?
We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:
M1V1 = M2V2
6 × 5.32 = M2 ×100
31.92 = M2 × 100
Divide both side by 100
M2 = 31.92 / 100
M2 = 0.3192 M
Therefore, the molarity of the diluted solution is 0.3192 M.
Answer:
B- fission breaks apart nuclei; fusion puts them back together
3 Chlorine ions are required to bond with one aluminum ion.
In ionic bonds, metals atoms loses all its outermost shell electrons to form a cation. While, non metal atoms gains however many electrons in order to make its outermost electron shell be 8 (or 2 if there's only one shell).
Therefore, form the periodic table, we can see that aluminum has a atomic number of 13, which makes its electron arrangement be 2,8,3. So, in order to form a aluminum ion, an Al atom must lose 3 electrons. On the other hand, Chlorine has a atomic number of 17, which means it has the electron configuration of 2,8,7. It has to gain only 1 electron to have 8 outermost shell electron.
Thereofre, 3 Chlorine atom are required to gain all 3 electrons given out by just 1 aluminum ion.
