1. 0.33 M
2. 0.278 M
<h3>Further explanation</h3>
Molarity is a way to express the concentration of the solution
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution
Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
1. 0.350 mol of NaOH in 1.05 L of solution.
n=0.35
V=1.05 L
Molarity :
2. 14.3 g of NaCl in 879 mL of solution.
mol NaCl(MW=58.5 g/mol) :
Molarity :
Democritus was the one who had theorized that atoms make up everything and they are indivisible.
Dalton was the creator of the first actual atomic theory, most of his research was on gasses and meteorology.
Thompson was the original man who put together the plum pudding model in which Rutherford later proved wrong during his career.
Rutherford had discovered the nucleus within an atom. He had put together gold foil experiment.
Bohr had developed the idea of neutrons and electrons surrounding the nucleus. He was also the creator of the planetary model we now use to calculate electrons with.
Answer:
none of the above
Explanation:
A system is said to have attained dynamic equilibrium when the forward and reverse reactions proceed at the same rate. That is;
Rate of forward reaction = Rate of reverse reaction
The implication of this is that the concentrations of reactants and products remain constant when dynamic equilibrium is attained in a system. This does not mean that the reactant and product concentrations become equal; it rather means that their concentrations do not significantly change once dynamic equilibrium has been attained.
Answer:
1. V₁ = 2.0 mL
2. V₁ = 2.5 mL
Explanation:
<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>
To calculate the volume required (V₁) in each dilution we will use the dilution rule.
C₁ . V₁ = C₂ . V₂
where,
C are the concentrations
V are the volumes
1 refers to the initial state
2 refers to the final state
<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>
C₁ . V₁ = C₂ . V₂
(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL
V₁ = 2.0 mL
<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>
C₁ . V₁ = C₂ . V₂
(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL
V₁ = 2.5 mL
