2Al + 3Cl₂ → 2AlCl₃
mol Al = 2/3 x 1.25 = 0.83
mass Al = 0.83 x 27 g/mol = 22.41 g
Answer is 14.5 g L⁻¹.
<em>Explanation;</em>
Here, the question says reduce the units as one.
The presented units are g/L. To reduce the units to one, what we can do is take L to the upper side.
This can be done according to the rules of indices;
1 / aˣ = a⁻ˣ
Like that, we can write 1 / L as L⁻¹.
Hence, the reduced unit is g L⁻¹.
But remember to keep a space between when writing two different units.
Actually, this is an unit for density.
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
Answer:
Halogens always form anions, alkali metals and alkaline earth metals always form cations. Most other metals form cations (e.g. iron, silver, nickel), whilst most other nonmetals typically form anions (e.g. oxygen, carbon, sulfur).
Explanation:
Examples: Sodium (Na+), Iron (Fe2+), Ammonium (NH4
