Particles in a gas are far apart compared to a solid or liquid, allowing it not to have a definitive shape or volume. This also means that gases can fill any container and be easily compressed.
Answer:
1)Reactants
2)Light
3)An item that can increase reaction rates
4)Reactants must collide with each other
Less molecules lower the chance for collisions
The more collisions there are the higher the reaction rate
Freezing rain is the term given to the precipitation of the rain sustained at temperatures below freezing by the ambient air mass, which results in freezing on encountering with the surfaces. Freezing rain is formed completely of liquid droplets, unlike sleet, ice pellets, or hail.
The freezing rain falls and solidifies on a wire fence as the rain is changing from a liquid to a solid state by withdrawing thermal energy, ultimately resulting in freezing.
Answer:
202 L
Explanation:
Step 1: Write the balanced equation
C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)
Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆
The molar mass of C₆H₁₂O₆ is 180.16 g/mol.
270 g × 1 mol/180.16 g = 1.50 mol
Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose
The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol
Step 4: Calculate the volume of 9.00 moles of CO₂ at STP
The volume of 1 mole of an ideal gas at STP is 22.4 L.
9.00 mol × 22.4 L/mol = 202 L
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
