By not shaking it up or leaving it in the heat.
What is the following ??? Do you have any more information
Answer:
19 g
Explanation:
Data Given:
Sodium Chloride (table salt) = 50 g
Amount of sodium (Na) = ?
Solution:
Molecular weight calculation:
NaCl = 23 + 35.5
NaCl = 58.5 g/mol
Mass contributed by Sodium = 23 g
calculate the mole percent composition of sodium (Na) in sodium Chloride.
Since the percentage of compound is 100
So,
Percent of sodium (Na) = 23 / 58.5 x 100
Percent of sodium (Na) = 39.3 %
It means that for ever gram of sodium chloride there is 0.393 g of Na is present.
So,
for the 50 grams of table salt (NaCl) the mass of Na will be
mass of sodium (Na) = 0.393 x 50 g
mass of sodium (Na) = 19 g
Answer:
Kc = 3.1x10²
Explanation:
At equilibrium, the velocity of product formation is equal to the velocity of reactants formation. For a generic reaction, the equilibrium constant (Kc) is:
aA + bB ⇄ cC + dD
![Kc = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%2A%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%2A%5BB%5D%5Eb%7D)
Where [X] is the molar concentration of X, and the solid substances are not considered (because it's activity is 1, for the other substances, the activity is substituted for the molar concentration, which forms the equation above).
For the reaction given, let's make an equilibrium chart:
Fe³⁺(aq) + SCN⁻(aq) ⇄ FeSCN²⁺(aq)
1.1*10⁻³ 8.2*10⁻⁴ 0 <em> Initial</em>
-x -x +x <em>Reacts</em> (stoichiometry is 1:1:1)
1.1*10⁻³ -x 8.2*10⁻⁴ -x x <em> Equilibrium</em>
x = 1.8*10⁻⁴ M, so the molar concentrations at equilibrium are:
[Fe⁺³] = 1.1*10⁻³ - 1.8*10⁻⁴ = 9.2*10⁻⁴ M
[SCN⁻] = 8.2*10⁻⁴ - 1.8*10⁻⁴ = 6.4*10⁻⁴ M
[FeSCN⁺²] = 1.8*10⁻⁴ M
Kc = [FeSCN⁺²]/([Fe⁺³]*[SCN⁻])
Kc = (1.8*10⁻⁴)/(9.2*10⁻⁴*6.4*10⁻⁴)
Kc = 306 = 3.1x10²