5.512 litres is the volume of 15.2 grams of sulphur dioxide gas at STP.
Explanation:
Data given:
mass of sulphur dioxide = 15.2 grams
conditions is at STP whech means volume = 22.4 litres
atomic mass of sulphur dioxide = 64.06 grams/mole
Number of moles is calculated as:
number of moles = 
Putting the values in the equation:
number of moles = 
= 0.23 moles
Assuming that sulphur dioxide behaves as an ideal gas, we can calculate the volume as:
When 1 mole of sulphur dioxide occupies 22.4 litres at STP
Then 0.23 moles of sulphur dioxide occupies 22.4 x 0.23
= 5.152 litres is the volume.
B is correct
salt lowers the freezing point of water (colligative property) by lowering the interaction and intermolecular forces between water molecules
pressure......................................
Answer:
The beaker holds 307.94 mL
Explanation:
As we know that the volume that beaker hold is the volume of water that occupied by it.
For this first we have to find mass of the water in the beaker
This can be calculated by the subtraction of beaker's weight from the weight of beaker and water.
weight of water (m) = total weight - weight of beaker
Empty weight of beaker = 25.91 g
Weight of beaker with water = 333.85 g
Weight of water = 333.85 - 25.91 = 307.94 g
Density of water = 1 g/mL
We have
Mass = Volume x density
307.94 = Volume x 1
Volume = 307.94 mL
The beaker holds 307.94 mL
They turn litmus paper blue
