Use the coefficients as a ratio. 1:6 = x:24. You can write these as fractions and cross multiply, 6x=24. Solve and x=4, so four molecules would be needed
Complete Question
A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.
Answer:

Explanation:
From the question we are told that:
Heat Capacity 
Mass of water 
Initial Temperature of Aluminium 
Initial Temperature of Water 
Final Temperature of Water 
Generally
Heat loss=Heat Gain
Therefore


Answer:
ok.. what is your question??? confused
Answer:
volume = 972.23ml
Explanation:
using general gas law
P1V1/T1 = P2V2/T2
765 x 585/293 = 443 x V2/282
1527.39 =443 x V2/282
1527.38 x 282 = 443 x V2
430695.78 = 443 x V2
V2 = 430695.68/443
V2 = 972.23mL