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Sedaia [141]
3 years ago
8

Each of the four liquids had the same volume. Why didn’t

Chemistry
1 answer:
Alexeev081 [22]3 years ago
5 0
Bdjdisisididiieddiidid
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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
A decomposition reaction always breaks a compound down into two pure elements. True or False.
skelet666 [1.2K]

False-- most of the time not enough energy is put into the system, so the compounds or whatever cant break down fully into their elements

6 0
3 years ago
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Cci(4) will be soluble in
Romashka-Z-Leto [24]
LIKE DISSOLVES LIKE. Since Ccl4 is non-polar, it'll be soluble in any non-polar solvent. Hope this helps you!
8 0
3 years ago
If I initially have a gas at a pressure of 12 atm, volume of 23 liters, and temperature of 200 K, and then I raise the pressure
CaHeK987 [17]

Answer : The volume of gas will be 29.6 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 12 atm

P_2 = final pressure of gas = 14 atm

V_1 = initial volume of gas = 23 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 200K

T_2 = final temperature of gas = 300K

Now put all the given values in the above equation, we get the final pressure of gas.

\frac{12atm\times 23L}{200K}=\frac{14\times V_2}{300K}

V_2=29.6L

Therefore, the new volume of gas will be 29.6 L

5 0
3 years ago
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Why does a can of diet cola float in water but a can of regular cola does not
maria [59]

Answer:

because there more acid in diet cola

7 0
3 years ago
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