Sulphur Dioxide. Toxic. Don't eat it.
0.300 M IKI represents the
concentration which is in molarity of a potassium iodide solution. This means
that for every liter of solution there are 0.300 moles of potassium iodide. Knowing
that molarity is a ratio of solute to solution.
By using a conversion factor:
100 ml x (1L / 1000 mL) x (0.300
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g
Therefore, in the first
conversion by simply converting the unit of volume to liter, Molarity is in L
where the volume is in liters. The next step is converted in moles from volume
by using molarity as a conversion factor which is similar to how density can be
used to convert between volume and mass. After converting to moles it is simply
used as molar mass of Kl which is obtained from periodic table to convert from
mole to grams.
In order to get the grams of IKI
to create a 100 mL solution of 0.600 M IKI, use the same formula as above:
100 ml x (1L / 1000 mL) x (0.600
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g
Answer:
hydrogen and cloride + hydrogen
Answer:
1.13 × 10⁶ g
Explanation:
Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.
Al³⁺ + 3 e⁻ → Al
We can establish the following relations:
- 1 Ampere = 1 Coulomb / second
- The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons circulate
- The molar mass of Al is 26.98 g/mol.
The mass of aluminum produced under these conditions is:
Answer is: 25.84 milliliters of sodium metal.
Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂.
d(Na) = 0.97 g/mL; density of sodim.
m(NaOH) = 43.6 g; mass of sodium hydroxide.
n(NaOH) = m(NaOH) ÷ M(NaOH).
n(NaOH) = 43.6 g ÷ 40 g/mol.
n(NaOH) =1.09 mol; amount of sodium hydroxide.
From chemical reaction: n(NaOH) : n(Na) = 2 : 2 (1: 1).
n(Na) = 1.09 mol.
m(Na) = 1.09 mol · 23 g/mol.
m(Na) = 25.07 g; mass of sodium.
V(Na) = m(Na) ÷ d(Na).
V(Na) = 25.07 g ÷ 0.97 g/mL.
V(Na) = 25.84 mL.
