Na because its a metal. Metals are the best conductors. S and Ne are nonmetals. and Ge is a metalloid. (Metalloids are semi conductors)
<u>Answer:</u> The empirical formula for the given compound is 
<u>Explanation:</u>
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Conversion factor: 1 g = 1000 mg
Mass of 
Mass of 
Mass of compound = 2.78 mg = 0.00278 g
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
- <u>For calculating the mass of carbon:</u>
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.00632 g of carbon dioxide,
of carbon will be contained.
- <u>For calculating the mass of hydrogen:</u>
In 18g of water, 2 g of hydrogen is contained.
So, in 0.00258 g of water,
of hydrogen will be contained.
- Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 
For Carbon = 
For Hydrogen = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 3 : 6 : 1
Hence, the empirical formula for the given compound is 
It creates oxygen which every living thing needs to live. i hope this helps! :)
Answer:
It becomes a positive ion and its radius decreases
Explanation:
As per the Octet rule, Barium has 2 electrons in its outermost shell. When it loses the two electron it gains two positive charge i.e Ba2+. As the barium loses the two electron from its outermost shell, the outermost shell becomes vacant and thus is no more considered as a part of atomic geometry of the barium atom and since the outermost shell is considered negligible the radius of barium atom reduces automatically.
Let's think, if you have a candle ( that is not blown out ) the physical properties are the candles mass and hence ( hence of the candle is the stiffness of the candle), weight, length, density, surface friction ( force resisting the relative motion of solid surface), and the energy content. You then, need to go to bed, so, therefore, you want to blow the candle out. Once you blow the candle out, the candle is evidently going to have at least a couple of different physical properties, than before it was blown out. The physical properties are a different color, the length of the candle, the texture, you could also apply the mass of the candleholder, and then, the mass of the candleholder and the candle, last but not least, the mass of just the candle. Once you observe the candle, you should be able to plug in those observations into the physical properties. As to, because you asked' what are the physical properties of a candle that has been blown out... We are going to assume that we did observe the candle, and the length of the candle in cm, after being blown out is 30cm. (12 inches; customary). Next, that the color of the candle is the same (let us say the original color is taffy pink). We can then say that the texture of the candle is waxy and the top and smooth as you get to the bottom ( the texture depends on how long the candle was burning, but we are saying that we lit the candle, and then immediately blew the flame out ) . We now have the mass of the candleholder, which will scientificity stay the same. Now, for the mass of the candleholder and the candle, that all depends of how long you let it burn ( remember, we are saying we lit the wick and then immediately blew the fame out ). So, the candle really didn't change is mass, so, therefore, wouldn't affect the mass of the candleholder including the candle. That also goes to the mass of the candle.