Answer:
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Question: A chemistry student weighs out 0.112g of acetic acid (HCH₃CO₂) into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1600 <em>M</em> NaOH solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits.
Answer: Volume of NaOH is 11.6 mL
Explanation:
The reaction of acetic acid with NaOH is as follows:
CH3COOH + NaOH -----> CH3COONa + H2O
M1V1 = M2V2
Here M1 V1 are molarity and volume of acetic acid.
M2, V2 are molarity and volume of NaOH.
Number of moles of acetic acid:
0.112 g CH3COOH × (1 mol / 60.05 g) = 0.001865 mol
Molarity = moles of solute / Liters of solution
Molarity = 0.001865 mol / 0.250 L = 0.00746 M
Hence,
M1 = 0.00746 M
V1 = 250 mL
M2 = 0.160 M
V2 = ?
V2 = M1V1 / M2
V2 = 0.00746 M × 250 mL / 0.160 M
V2 = 11.6 mL
Hence the volume of NaOH is 11.6 mL
Temperature creates more energy which increases movement thus causing it to speed up the reaction
It's known as chromium (III) sulphate which is green in colour.
Answer:
33
Explanation:
as the atomic number states the number of protons in each element hope this helps :)