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Art [367]
3 years ago
5

Which substance has the highest melting point?

Chemistry
1 answer:
Andre45 [30]3 years ago
7 0
<span>A.H2O = water, so it is a liquid at room temperatures (melting point = 0C) and melting point must be below 20C

B.HCl = A gas at room temperatures, so melting point (-114C) is definitely lower than 20C.
Melting points of the aqueous solution (solution in water) at different concentrations are all below that of water, i.e. 0C.

C.CH4 = A gas at room temperatures, so melting point (-182C) must be below 20C.

D.MgCl2 = A white powdery solid at room temperatures, so its melting point (714C) must be above 20C (So this is the only one among the four choices that melts above 20 &deg; C).</span>
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Explanation:

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Have a nice day!

8 0
3 years ago
What inert (noble) gas structure do hydrogen and carbon atoms have in sodium chloride?
vampirchik [111]
The answer above is correct (I took a test on this)
7 0
3 years ago
How are covalent molecules named?
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APEX

5 0
3 years ago
Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o
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My Choice will be the sample C with a coefficient of absorption 62 % and the reason is that the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest. Because this sample has the highest percentage of the coefficient of absorption than the other samples, As the sound Absorption coefficient is a part of the sound energy that absorbed by a material.And Its value varies with the angles of incident and frequency. So sample C is the best to be used as a soundproof room.
4 0
4 years ago
What mass of water is produced from the complete combustion of 4.70×10−3 g of methane?
otez555 [7]
Since the equation is balanced we can use the molar ratios to jump between amounts of substances. 
CH4 + 2O2 ---> CO2 + 2 H2O 

1.You can use a molar ratio to convert between mols methane to mols CO2: 
4.70×10−3 mol CH4 X (1 CO2 / 1 CH4) = 4.7x10^-3 mol CO2 X (44g/1mol) = .2068g CO2

2. You can do the same technique with the water. 
4.70×10−3 mol CH4 X (2 H2O / 1 CH4) = .0094 mol H2O X (18g/1mol) = .1692g H2O 

3. You can do the same technique with the O2. 
4.70×10−3 mol CH4 X (2 O2 / 1 CH4) = .0094 mol O2 X (32g/1mol) = .3008g O2
8 0
4 years ago
Read 2 more answers
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