The types of intermolecular forces that occur in a substance will affect its physical properties, such as its phase, melting point and boiling point.
Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
Explanation : Given,
Mass of oxygen in sulfur dioxide = 3.49 g
Mass of sulfur in sulfur dioxide = 3.50 g
Mass of oxygen in sulfur trioxide = 9.00 g
Mass of sulfur in sulfur trioxide = 6.00 g
Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.
Mass of oxygen per gram of sulfur for sulfur dioxide = 
Mass of oxygen per gram of sulfur for sulfur dioxide = 
and,
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
<u><em>on the basis of the reaction of heat in plastic , their are two types of plastics : </em></u>
<h2><u>
<em>Thermoplastic & Thermosetting plastics </em></u></h2>
<u><em>Thermoplastics : Plastics which easily get deformed (become soft )on heating and also get bend easily are known as thermoplastics . </em></u>
<u><em>examples : polythene , polyvinyl chloride and polystyrene .</em></u>
<u><em>thermosetting plastics : plastics which once moulded into a shape do not become soft on heating and cannot be moulded again are called thermosetting plastics .</em></u>
<u><em>examples : bakelite , melamine and formica</em></u>
The enthalpy change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol. This means for every mole of ice we melt we must apply 6.02 kj of heat. We can calculate the heat needed with the following equation:
Q = N x ΔH
where:
Q = heat
N = moles
ΔH = enthalpy
In this problem we would like to calculate the heat needed to melt 35 grams of ice at 0 °C. This problem can be broken into three steps:
1. Calculate moles of water
2. multiply by the enthalpy of fusion
3. Convert kJ to J.
Step 1 : Calculate moles of water
![[ 75g ] x (\frac{1 mol}{18.02g} ) =](https://tex.z-dn.net/?f=%5B%2075g%20%5D%20x%20%28%5Cfrac%7B1%20mol%7D%7B18.02g%7D%20%29%20%3D)
Step 2 : Multiply by enthalpy of fusion
Q = N × ΔH = <em> [ Step 1 Answer ]</em> × 6.02 =
Step 3 : Convert kJ to J
![[ Step 2 Answer ] x (\frac{1000j}{1kJ} ) =](https://tex.z-dn.net/?f=%5B%20Step%202%20Answer%20%5D%20x%20%28%5Cfrac%7B1000j%7D%7B1kJ%7D%20%29%20%3D)
Finally rounding to 2 sig figs (since 34°C has two sig figs) we get
Q Would Equal ____