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Vaselesa [24]
3 years ago
8

An astronomer studying planets outside our solar system has analyzed the atmospheres of four planets. Which of these planets’ at

mospheres would be most able to support a colony of humans? Planet C: 78% Nitrogen, 1% Oxygen, 21% Other Planet B: 82% Nitrogen, 11% Oxygen, 7% Other Planet A: 76% Nitrogen, 23% Oxygen, 1% Other Planet D: 27% Nitrogen, 3% Oxygen, 70% Other
Chemistry
1 answer:
Nata [24]3 years ago
3 0

Answer: Planet A: 76% Nitrogen, 23% Oxygen, 1% Other

Explanation: Hope this helps!

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25 </span><span>Sodium has one valence electron and chlorine has seven valence electrons. The Ione valence electron from sodium is lost and is transferred to the chlorine atom. The result is a sodium ion with a charge of 1+ and a chloride ion with a charge of 1-. The oppositely charged ions attract each other and the charges balance to make a compound that is electrically neutral.

26. </span><span>Ionic compounds have high melting points and high boiling points compared to molecular compounds. Ionic compounds that are dissolved in water or melted will conduct electricity. Molecular compounds do not conduct electricity in either case. 

27. </span><span>A covalent bond is formed when two atoms share valence electrons. Neither atom loses electrons or takes electrons from the other. No charged particles form. In an ionic bond, one or more electrons are transferred from one atom to another. Atoms that lose electrons become positively charged ions, and atoms that gain electrons become negatively charged ions. These oppositely charged particles then attract each other.

28. </span><span>A metal crystal consists of positively charged metal ions embedded in a "sea" of loosely held valence electrons that can move around easily. Heat travels through materials as the increased motion of the particles in the hotter parts of the material is passed along to the particles in the cooler parts. In a metal, since particles are easily set in motion, heat is easily transferred or conducted. The same is true for the conduction of electricity. Electricity can flow when charged particles, such as electrons, are free to move. Since the electrons in a metal crystal can move freely among the atoms, electricity is easily conducted.</span>
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4 years ago
convert 0.876 atm to mmHg. solution: multiply the atm value by 760.0 mmHg/ 1.0 atm 0875 atm x _______ = _______ mm Hg (3 sig fig
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\tt \dfrac{760~mmHg}{1~atm}\times 0.876~atm=665~mmHg

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2 years ago
A 4-inch, f/5 telescope has a 1-inch eyepiece focal. Its magnifying power is:
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Its magnifying power is: 4X 5X 9X 20X. A 4-inch, f/5 telescope has a 1-inch eyepiece focal. Its magnifying power is 9x. This answer has been confirmed as correct and helpful.

7 0
3 years ago
Read 2 more answers
The Doppler effect related change in wave frequency to what?
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Answer:

Relative motion of source and observer.

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8 0
3 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

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3 years ago
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