Answer;
The partial negative charge on oxygen would stick out less and be less able to participate in hydrogen bonding.
Explanation;
Water is a polar molecule because the electrons are not shared equally, they're closer to the oxygen atom than the hydrogen.
-Normally, the water molecule is a bent shape because of the pair of lone electrons - they repulse each other and exert a compression to the hydrogen atoms at a slight 104º angle. It is a bent molecular geometry that results from tetrahedral electron pair geometry.
-The 2 lone electron pairs exerts a little extra repulsion on the two bonding hydrogen atoms to create a slight compression to a 104 degrees bond angle. Therefore, the water molecule is bent molecular geometry because the lone electron pairs.
Thus, If water were a linear molecule like co2, electrostatic interactions between water molecules would be much weaker, then the partial negative charge on oxygen would stick out less and be less able to participate in hydrogen bonding.
Answer:
41 mL
Explanation:
Given data:
Milliliter of HCl required = ?
Molarity of HCl solution = 4.25 M
Mass of CaCO₃ = 8.75 g
Solution:
Chemical equation:
2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O
Number of moles of CaCO₃:
Number of moles = mass/molar mass
Number of moles = 8.75 g / 100.1 g/mol
Number of moles = 0.087 g /mol
Now we will compare the moles of CaCO₃ with HCl.
CaCO₃ : HCl
1 : 2
0.087 : 2/1×0.087 = 0.174 mol
Volume of HCl:
Molarity = number of moles / volume in L
4.25 M = 0.174 mol / volume in L
Volume in L = 0.174 mol /4.25 M
Volume in L = 0.041 L
Volume in mL:
0.041 L×1000 mL/ 1L
41 mL
Answer:
1. The dye that absorbs at 530 nm
Explanation:
With a larger HOMO-LUMO gap, there's also a higher absorption energy, so this means that the dye with the higher absortion energy has the larger HOMO-LUMO gap.
The relationship between energy and wavelenght can be expressed by the formula E = hc/λ, this means that the <em>lower</em> the wavelenght, the <em>higher</em> the energy is. So the dye that absorbs at a lower wavelenght has a larger HOMO-LUMO gap.
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.
