Answer:
THE MOLAR MASS OF THE UNKNOWN MOLECULAR SUBSTANCE IS 200 G/MOL.
Explanation:
Mass of the unknown substance = 0.50 g
Freezing point of the solution = 3.9 °C
Freezing point of pure benzene = 5.5 °C
Freezing point dissociation constant Kf = 5.12°C/m
First, calculate the temperature difference between the freezing point of pure benzene and the final solution freezing point.
Change in temperature = 5.5 -3.9 = 1.6 °C
Next is to calculate the number of moles or molarity of the compound that dissolved.
Using the formula:
Δt = i Kf m
Assume i = 1
So,
1.6 °C = 1 * 5.12 * x/ 0.005 kg of benzene
x = 1.6 * 0.008 / 5.12
x = 0.0128 / 5.12
x = 0.0025 moles.
Next is to calculate the molar mass using the formula, molarity = mass / molar mass
Molar mass = mass / molarity
Molar mass = 0.50 g /0.0025
Molar mass = 200 g/mol
Hence, the molar mass of the unknown compound is 200 g/mol
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The reason why you don’t see solid or liquid oxygen is because oxygen is a natural occurring gas. Gas can not be solid or liquid because there is too much energy to have volume/shape. That’s why you see oxygen tanks, as the tank keeps the gas contained and that is why you can’t see it in the air outside or in something like H20 (water).
Hope this helps :)
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Answer:
a)calculated molarity of NaOH would be lower
b) calculated molarity of NaOH would be lower
c) calculated molarity of NaOH would be lower
d) calculated molarity of NaOH would be unaffected
Explanation:
Let us recall that the reaction of NaOH and HCl is as follows;
NaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)
Since the reaction is 1:1, when the number of moles of HCl reacting with NaOH is low due to dilution, the calculated molarity of NaOH also becomes less than it's accurate value.
When 40mL of water is added to the titration flask rather than 25ml of water, the acid is more dilute hence less number of moles of acid than necessary reacts with the base thereby yielding a less than accurate value of the molarity of NaOH.
If the burette wet with water is not rinsed with NaOH solution, the concentration of the NaOH in the burette decreases due to dilution with water and a less than accuracy value is calculated for the molarity of NaOH.
If five drops of phenolphthalein is used instead of one or two drops, there is no qualms since enough phenolphthalein may be added to ensure that a sharp end point is obtained.
The compound that will have a sweet smell would be the one, whereby the molecular formula closely resembles that of an ether
R-O-R.
I believe the third one
