Any two substances which are odourless

This question perplexes me somewhat. If we as humans are not able to detect a scent or odor from certain substances, that does not of necessity mean those substances are not giving off an odor, and are therefore not “truly” odorless, just odorless to the less sensitive noses of humans.

Water is supposedly odorless, but I know quite well of a test where subjects did a blind tasting, only instead of wine these people were tasting Australian waters, sourced from different locations around the country.

All of these tasters (professionals) were able to differentiate the waters and from what I recall, were all accurate. These wine buffs use their palates as well as their taste-buds, so, were there no odor, their task would have had a greater degree of difficulty.

My own baby sister has such a sensitive palate that she can tell you if you serve her a glass of water from the fridge whether that water was fresh, straight from the tap (unfiltered), filtered or had been boiled before and whether it had been boiled for more than five minutes or not. Truly.

So I would find it difficult to answer your question, “What substances are truly odorless?” , because I am not sure of the criteria you mean by “truly” odorless. In other words, If you are talking about zero degrees, are you talking of the freezing point of water, absolute zero or zero degrees Kelvin?

I’m sure there are substances that are odorless, to humans, but are those same substances “truly” odorless, or only to our perception of them?

Hope this helps, have a wonderful day/night, and stay safe!

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

12 is it do you think so