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Rudik [331]
3 years ago
8

Simon has collected three samples from the coral reef where he observes marine life. He must determine whether each one is a pur

e substance or a mixture.
A 4-column table with 3 rows. The first column has entries sample A, sample B, sample C. The second column labeled appearance has entries clear liquid, clear blue liquid, opaque whitish liquid. The third column labeled when heated has entries evaporates completely at 70 degrees C, boils at 90 degrees Cleaving blue crystals behind, boils at 100 degrees leaving white crystals behind. The fourth column labeled when left over time has entries appearance does not change, appearance does not change, dust appears to settle at the bottom.

Sample A is a
, Sample B is a
, and Sample C is a
.
Chemistry
1 answer:
-Dominant- [34]3 years ago
3 0

Answer:

SAMPLE A - pure substance.

SAMPLE B - homogeneous mixture.

SAMPLE C - heterogeneous mixture.

Explanation:

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The Molar mass(M.w) in g/mol of 6.3 grams of an ideal gas that placed in 5.0 L tank at ST
BigorU [14]

Answer:

28 g/mol, N2

Explanation:

Given data:

Volume of gas = 5.0 L

Mass of gas = 6.3 g

Pressure = 1 atm

Temperature = 273 K

Molar mass of gas = ?

Solution:

We will calculate the density first.

d = mass/ volume

d = 6.3 g/ 5.0 L

d = 1.26 g/L

Molar mass:

d = PM/RT

M = dRT/P

M =  1.26 g/L× 0.0821 atm.L/mol.K × 273 K/ 1 atm

M = 28 g/mol

Molar mass of N₂ is 28 g/mol thus given gas is N₂.

7 0
3 years ago
Several properties of water are shown. Classify each property as a physical property or a chemical property.
max2010maxim [7]

can be split into hydrogen and oxygen             chemical property

is liquid at room temperature                              physical property

has a density of 1.0 g/cm³                                    physical property

reacts with certain metals                                  chemical property

Explanation:

The physical properties of matter are the properties that occur when no change is occurring to a matter. They are usually observed with our senses or some simple laboratory experiments.

Some of the physical properties are boiling point, density, state of matter, e.t.c

The chemical properties are the properties that occurs when a matter undergoes some chemical changes.

For example rusting, decomposition, reactivity e.t.c

Learn more:

Chemical properties brainly.com/question/2376134

#learnwithBrainly

8 0
3 years ago
ch3)2c(oh)ch2ch3 skeletal structure, Functional groups, compound type & alcohol class if it’s an alcohol
xxTIMURxx [149]

The skeletal structure of an organic compound is an abbreviated representation of its molecular structure, they are quick and easy to draw.

For example, the following image shows the skeletal structure of a compound:

The peaks represent the carbons. We must remember that carbon can have a maximum of 4 bonds.

Now, I will show you how is the structure of this specific compound:

This is ternary alcohol, called 2-methyl-2-butanol. If you see carefully, you will notice that each carbon has 4 bonds. The functional groups present will be OH. The skeletal structure will be:

6 0
1 year ago
Which element forms more compounds than all the other elements combined?
katen-ka-za [31]
The Element that form more compounds than all the rest is Carbon.

Elements of Carbon
<span>A molecule of the compound carbon dioxide contains one atom of the element carbon and two atoms of the element oxygen. Each oxygen atom shares a double bond with the carbon atom. Carbon is the sixth element in the periodic table and occurs in pure form as coal and diamonds.

Hopefully I helped ^.^ Mark brainly if possible~
</span>
5 0
4 years ago
The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in
astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}

For this, it is necessary to know the values in meters for any of these diameters:

\\ 1m = 10^{3}mm = 1e+3mm

\\ 1m = 10^{12}pm = 1e+12pm

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m

<h3>Diameter of a biscuit in meters</h3>

\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0
3 years ago
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