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34kurt
3 years ago
8

Any two substances which are odourless​

Chemistry
1 answer:
STatiana [176]3 years ago
8 0

Answer:

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!

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A flask with a volume of 3.16 l contains 9.33 grams of an unknown gas at 32.0°c and 1.00 atm. What is the molar mass of the gas?
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Answer:

73.88 g/mol

Explanation:

For this question we have to keep in mind that the unknown substance is a <u>gas</u>, therefore we can use the <u>ideal gas law</u>:

PV=nRT

In this case we will have:

P= 1 atm

V= 3.16 L

T = 32 ªC = 305.15 ºK

R= 0.082 \frac{atm*L}{mol*K}

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So, we can <u>solve for "n"</u> (moles):

1~atm*3.16~L~=~n*0.082~\frac{atm*L}{mol*K}*305.15~K

n=\frac{1~atm*3.16~L~}{0.082~\frac{atm*L}{mol*K}*305.15~K}

n=0.126~mol

Now, we have to remember that the <u>molar mass value has "g/mol"</u> units. We already have the grams (9.33 g), so we have to <u>divide</u> by the moles:

molar~mass=\frac{9.33~grams}{0.126~mol}

molar~mass=73.88\frac{grams}{mol}

7 0
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Answer:

B. double-replacement RXN

Explanation:

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2 years ago
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murzikaleks [220]
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Answer:

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