Answer:The electromagnetic spectrum includes, from longest wavelength to shortest: radio waves, microwaves, infrared, optical, ultraviolet, X-rays, and gamma-rays.
Explanation:
Lewis Structure:
H -- O -- H (bent, Oxygen has 2 lone pairs)
Percent Composition
Hydrogen percent composition = [ 2 * (Hydrogen mass) ] / [Total mass of H2O]
Oxygen percent composition = [Oxygen mass] / [Total mass of H2O]
dose this make any sense...??
1) Balanced equation
C3H8 + 5O2 -> 3 CO2 + 4 H2O
2) 0.700 L C3H8
Given the pressure and temperature do not change, the molar ratio is equivalent to volume ratio
1molC3H8 / 5 mol O2 => 1 L C3H8 / 5 L O2
0.700 L C3H8 / x L O2 = 1 L C3H8 / 5 L O2 => x = 0.700 L C3H8 * 5 L O2 / 1 L C3H8
x = 3.500 L O2
3) CO2 produced
1 L C3H8 / 3 L CO2 = 0.700 L C3H8 / x L CO2 =>
x = 0.700 L C3H8 * 3 L CO2 / 1 L C3H8 = 2.100 L CO2
4) Water vapor produced
1) 1 L C3H8 / 4 L H2O = 0.700 LC3H8 / x L H2O =>
x = 0.700 L C3H8 * 4 L H20 / 1 L C3H8 = 2.800 L H2O
Answer:
Here's what I get
Explanation:
1. Water
The O-H bond in water is highly polar.
It has about one-third ionic character.
The partially-positive H atoms and partially negative O atoms strongly attract each other by hydrogen bonds.
It takes a large amount of energy to separate the water molecules from each other.
Thus, the boiling point of water is relatively high (100 °C).
2. Acetone
The C=O bond in acetone is much less polar.
It is less than 20 % ionic.
The dipole-dipole attractions in acetone are much weaker than the hydrogen bonds in water.
It takes much less energy to separate the acetone molecules from each other.
Thus, even though the molar mass of acetone is more than three times that of water, the boiling point of acetone is only 56 °C.
Hello!
The net chemical equation between Sodium Bicarbonate and H₂SO₄ is the following one:
2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂
To calculate the amount of Sodium Bicarbonate needed to neutralize 1000 L of 0,350 M H₂SO₄ we'll need to use the following conversion factor, to go from the volume of H₂SO₄ to grams of Sodium Bicarbonate:

So, to neutralize 1000 L of 0,350 moles of H₂SO₄ you'll need
58804,9 grams of NaHCO₃