Concentration of vinegar sample of 28mL which requires 21. 5 ml of 0.550 M NaOH to reach the endpoint in a titration is 0.42M.
<h3>How do we calculate the concentration of Vinegar?</h3>
Concentration of vinegar in the titration method will be used by using the below equation:
M₁V₁ = M₂V₂,
- M₁ & V₁ are the molarity and volume of acetic acid.
- M₂ & V₂ are the molarity and volume of vinegar sample.
By putting values on above equation from the question, we get
M₂ = (0.55M)(21.5mL) / (28mL) = 0.42 M
Hence, required concentration of vinegar is 0.42M.
To know more about concentration, visit the below link:
brainly.com/question/14469428
#SPJ4
Well, clearly the calculated value for the number of hydrating water molecules would increase above its true level, because the total weight loss would be greater than expected. This is of course undesirable, but may usually be avoided by careful application of the experimental procedures. The signs to look for include
<span>(a) loss of water of hydration usually occurs at a considerably lower temperature than decomposition of the salt, because the water molecules are not strongly bonded in the hydrated complex. Dehydration typically occurs in a broad range of temperatures, typically from 50°C to around 200°C, whereas decomposition of the dehydrated salt generally takes place at temperatures over 200°C and in some case over 1000°C. So dehydration should be performed with care - avoid over-heating the sample in order to ensure that all the water has been driven off. </span>
<span>(b) dehydration often results in a change of appearance of the sample, particularly the colour and particle size of crystalline hydrates. However, decomposition may be accompanied by an additional change at higher temperatures, which gives a warning of its occurrence. </span>
<span>(c) if it is suspected that decomposition is occurring, or that dehydration is not complete, exploratory runs of varying duration at a given temperature may be carried out. There are two criteria to judge the effectiveness of the procedure </span>
<span>(i) the weight of the sample decreases to a constant stable value: this is a sign that dehydration is complete and decomposition - which is usually a much slower process - is not occurring. </span>
<span>(ii) the calculated number of molecules of water lost should take an integer value. If it differs by more than, say, 0.1 from an integer than it is probable that one of these two undesirable effects is present. Some hydrates lose water in steps through intermediate compounds with a lower level of hydration. These may provide plateaus where the weight loss is stable but dehydration is not complete. These will, in general, not provide an integer value for the number of water molecules present (because the calculation is based on the assumption that the residual sample is completely dehydrated salt).</span>
Answer: 38.111111
Explanation: (100.6°F − 32) × 5/9 = 38.111°C
First let us calculate for the number of moles needed:
moles NaOH = 0.125 M * 0.500 L = 0.0625 mol
The molar mass of NaOH is 40 g/mol, hence the mass is:
mass NaOH = 0.0625 mol * 40 g/mol
<span>mass NaOH = 2.5 grams</span>
Question four is : Gaseous state
Question five is : A rise in temperature increases the kinetic energy and speed of particles; it does not weaken the forces between them. The particles in solids vibrate about fixed positions; even at very low temperatures. Individual particles in liquids and gases have no fixed positions and move chaotically.
Hope this helps!
Have a great day!
