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Nikitich [7]
3 years ago
9

Compared to most substances, water is unusual because it ____ when it goes from the liquid to solid state.

Chemistry
2 answers:
yuradex [85]3 years ago
6 0

Answer:

expands

Explanation:

Most substances become increasingly compressed when frozen solid, but water is a famous exception.

Ira Lisetskai [31]3 years ago
5 0
Compared to most substances, water is unusual because it EXPANDS when it goes from the liquid to solid state
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The American Heart Association recommends to eat no more than 2,301mg of sodium per day. Convert the mass of
castortr0y [4]

The mass of 2,301 grams of sodium in ounces is 0.0811757609 ounces.

The amount of sodium recommended by American heart association is 2301 mg. This limit should not be crossed in a day.

We have to convert Mass of sodium from mg to ounces.

We know,

1 ounce = 28.3459 grams.

We also know,

1 gram = 1000 milligrams.

So,

28.3459 grams = 28.3459 x 1000 milligrams.

28.3459 grams = 28345.9 milligrams.

1 ounce = 28345.9 mg.

1 mg = 1/28345.9 ounces

Weight of sodium 2301mg in ounces,

2301mg = 2301/28345.9 ounces

Dividing till last significant figure,

2301 mg = 0.0811757609 ounces.

Mass of sodium in ounces is 0.0811757609.

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brainly.com/question/97386

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4 0
1 year ago
NEED HELP ASAP! Newton's Law of Gravity says that any the effect of gravity between two objects depends on what two factors?
kari74 [83]

Answer:

C. Distance and mass.

Explanation:

Objects that have greater mass have more force of gravity between them.

3 0
3 years ago
Any two substances which are odourless​
STatiana [176]

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!

8 0
3 years ago
Please help!!! I'll give brainliest to the most helpful answer
horsena [70]
87) A
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3 0
3 years ago
At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.600 M . N 2 ( g )
alexandr1967 [171]

Answer:

0.78 M

Explanation:

First, we need to know which is the value of Kc of this reaction. In order to know this, we should take the innitial values of N2, O2 and NO and write the equilibrium constant expression according to the reaction. Doing this we have the following:

N2(g) + O2(g) <------> 2NO(g)   Kc = ?

Writting Kc:

Kc = [NO]² / [N2] * [O2]

Replacing the given values we have then:

Kc = (0.6)² / (0.2)*(0.2)

Kc = 9

Now that we have the Kc, let's see what happens next.

We add more NO, until it's concentration is 0.9 M, this means that we are actually altering the reaction to get more reactants than product, which means that the equilibrium is being affected. If this is true, in the reaction when is re established the equilibrium, we'll see a loss in the concentration of NO and a gaining in concentrations of the reactants. This can be easily watched by doing an ICE chart:

      N2(g) + O2(g) <------> 2NO(g)

I:      0.2        0.2                 0.9

C:     +x         +x                   -2x

E:    0.2+x    0.2+x             0.9-2x

Replacing in the Kc expression we have:

Kc =  [NO]² / [N2] * [O2]

9 = (0.9-2x)² / (0.2+x)*(0.2+x)   ----> (this can be expressed as 0.2+x)²

Here, we solve for x:

9 = (0.9-2x)² / (0.2+x)²

√9 = (0.9-2x) / (0.2+x)

3(0.2+x) = 0.9-2x

0.6 + 3x = 0.9 - 2x

3x + 2x = 0.9 - 0.6

5x = 0.3

x = 0.06 M

This means that the final concentration of NO will be:

[NO] = 0.9 - (2*0.06)

[NO] = 0.78 M

8 0
3 years ago
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