A homogeneous mixture<span> has the same uniform appearance and composition throughout. Many homogeneous </span>mixtures<span> are commonly referred to as </span>solutions<span>. A </span>heterogeneous mixture <span>consists of visibly </span>different<span> substances or phases. The three phases or states of matter are gas, liquid, and solid.</span>
1) Molar mass C5H12= 5*12 +1 *12=60+12=72 g/mol
2) 40g C5H12 * 1 mol C5H12/72 g C5H12 = 40/72 mol C5H12
3) C5H12 + 8O2 ------> 5CO2 + 6H2O
by reaction 1 mol 8 mol
from problem 40/72 mol x mol
x=(40/72) * 8/1=(40*8)/72=(40)/9 mol O2
4) M(O2)=2*16 g/mol =32 g/mol
5) (40)/9 mol O2 *(32 g O2/ 1 mol )=(40 * 32)/9 =142.2 g O2
Energy production
Thus most industries in working burn oxygen as fuel. Examples are steel production and production of chemicals
Answer:
Law of Definite Proportions states that in a given type of chemical substance, the elements are always combined in the same proportions by mass. ... Oxygen makes up 88.8% of the mass of any sample of pure water, while hydrogen makes up the remaining 11.2% of the mass
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>