V1/T1=V2/T2
V2=(V1)(T2)/T1
Plug in values given (for the temp you can either turn 300K to 27°C or turn 132°C into kelvin
V2= 4400 mL= 4.4L
First you need to know the molecular weight of sugar (C6H12O6) which is 180.156g/mol
You have half a mole so you have 90.078g
If you wanted to make 1L of a 1.2M solution of glucose you would need 180.156*1.2=216.1872g
But you only have 90.078g
So you need to figure out how much this 90.078g will make if the solution must be 1.2M:
90.078g/216.1872g=xL/1L
solve for the X and you get 0.416666666...
so 416.7ml or 0.417L
One mole of Fe(NO3)3, or iron(III) nitrate, has three moles of nitrate molecules, which have three moles of oxygen atoms each. We can show this mathematically:
1 mole Fe(NO3)3 * (3 moles NO3)/(1 mole Fe(NO3)3) = 3 moles NO3
3 moles NO3 * (3 moles Oxygen)/(1 mole NO3) = 9 moles Oxygen
9 moles of Oxygen in one mole Fe(NO3)3
