Because of a territorial reorganization
Answer:
11.6mL of the 0.1400M NaOH solution
Explanation:
<em>0.154g of chloroacetic acid diluted to 250mL. Titrated wit 0.1400M NaOH solution.</em>
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The reaction of chloroacetic acid, ClCH₂COOH (Molar mass: 94.5g/mol) with NaOH is:
ClCH₂COOH + NaOH → ClCH₂COO⁻ + Na⁺ + H₂O
<em>Where 1 mole of the acid reacts per mole of the base.</em>
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That means the student will reach equivalence point when moles of chloroacetic acid = moles NaOH.
You will titrate the 0.154g of ClCH₂COOH. In moles (Using molar mass) are:
0.154g ₓ (1mol / 94.5g) = <em>1.63x10⁻³ moles of ClCH₂COOH</em>
To reach equivalence point, the student must add 1.63x10⁻³ moles of NaOH. These moles comes from:
1.63x10⁻³ moles of NaOH ₓ (1L / 0.1400moles NaOH) = 0.0116L of the 0.1400M NaOH =
<h3>11.6mL of the 0.1400M NaOH solution</h3>
Answer:
Yes You use the nuclear transfuser to do that
Explanation:
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is shown below:
Thus, we obtain 0.0471 g of lead (II) sulfate, the iodide ions in the original solution are computed below by stoichiometry, taking into account that into 1 mole of lead (II) iodide there are 2 moles of iodide ions:
Best regards.
A 1 F solution stands 1 formula unit per litre and 0.01 F describes the concentration of solution with no deliberation for the real form of existence of species.
Molarity is defined as the ratio of moles of solute to the volume of solution in litres and it is used to describe formality. For accuracy, it is essential to expressed molarity of each species. In case of acetic acid, the molarity of acetic acid molecules is less than 0.01 M due to dissociation.
Thus, it is more precise to say that the concentration of a solution of acetic acid is 0.01 F instead of 0.01 M.
