The reactant being used up is called limiting reagent as it limits the total amount of product produced.
if 4 units of HCL gives 2 units of Cl therefore
4:2
0.98:x
x=(0.98*2) /4
x=0.49L
It has: 2 atoms of sodium (Na) 1 atom of Carbon 3 atoms of Oxygen bound together with ionic and polar-covalent bonds
Answer:
The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are
Ku = 38.252 W/mK
K lower = 0.199 W/mK
Explanation:
As we know
Ku = Vp * Kair + Vmagnesium * K metal
Ku = 0.10 *0.02 + (1-0.25) * 51
Ku = 38.252 W/mK
The lower limit
K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)
K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)
K lower = 0.199 W/mK
You have to use everything that is given since you have to know which is the limiting reactant. We find the limiting reactant by calculating the number of moles of each reactant and compare the number of moles. The limiting reactant would be the one that is consumed fully by the reaction.
Answer:
The number of protons you welcome
Explanation:
