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Elina [12.6K]
2 years ago
7

Passenger airplanes take off in the troposphere but can eventually move into

Chemistry
1 answer:
zhuklara [117]2 years ago
5 0
Well that is because the stratosphere has lower temperatures and low air density. This gives the airplane less air pressure and with low air density they can fly faster :) Hope this helps.
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What is the amount of aluminum chloride produced from 40 moles of chlorine and excess aluminum?
Vera_Pavlovna [14]

Answer: Molar mass of Al = 26.98 g/mol

mass of Al = 12 g

mol of Al = (mass)/(molar mass)

= 12/26.98

= 0.4448 mol

According to balanced equation

mol of AlCl3 formed = moles of Al

= 0.4448 mol

Answer: 0.445 mol

hope this help boo❤️❤️❤️

Explanation:

3 0
2 years ago
Identify one benefit of using a scientific name to classify an organism.
dusya [7]

Answer:

The organism can be easily categorised, which really helps making it easier to understand the characteristics of a specific organism

3 0
3 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
Yasir wants to determine if there is a relationship between room color and sleep. Based on his research, Yasir makes an educated
Helen [10]

Answer: B. An experiment that directly tests the hypothesis

Explanation:

Yasir didn’t really test his experiment he only ask people for the opinion on which color they like so he forget to actually do the experiment therefore the answer is b.

8 0
3 years ago
Describe how oxidation and reduction involve electrons, change oxidation numbers, and combine in
Sholpan [36]

Answer:

Redox

Explanation:

Reduction is gain of electrons

oxidation is loss of electrons

3 0
3 years ago
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