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.
Answer:
0.197 M
Explanation:
The reaction equation is:
H2SO4(aq) +2KOH(aq) ----> K2SO4(aq) + 2H2O(l)
number of moles of H2SO4 = 0.25 L * 0.45 M = 0.1125 moles
number of moles of KOH = 0.2 L * 0.24 M = 0.048 moles
since H2SO4 is the reactant in excess;
2 moles of KOH reacts with 1 mole of H2SO4
0.048 moles of KOH reacts with 0.048 * 1/2 = 0.024 moles of H2SO4
Amount of excess H2SO4 left unreacted = 0.1125 - 0.024 = 0.0885 moles
Total volume = 0.25 L+ 0.2 L = 0.45 L
concentration of H2SO4 = 0.0885/0.45 = 0.197 M
Answer:
Mass = 2.355 g
Explanation:
Given data:
Mass of K₂O needed = ?
Mass of KNO₃ produced = 5.00 g
Solution:
Chemical equation:
K₂O + Ca(NO₃)₂ → CaO + 2KNO₃
Number of moles of KNO₃:
Number of moles = mass/molar mass
Number of moles = 5.00 g/ 101.1 g/mol
Number of moles = 0.05 mol
now we will compare the moles of KNO₃ and K₂O.
KNO₃ : K₂O
2 : 1
0.05 : 1/2×0.05 = 0.025 mol
Mass of potassium oxide needed in gram:
Mass = number of moles × molar mass
Mass = 0.025 mol × 94.2 g/mol
Mass = 2.355 g
4 molecules bc the coefficient is 4
