Answer:A mole is an arbitrary number of molecules in a single unit - refer to avogadro's number. Essentially, 1 mole is 6.022x10^23 molecules for ALL molecules or atoms, however one must remember that not all atoms/molecules are the same size, this is where mass comes into play. When you measure out 2 grams of carbon powder, there will be a lot more molecules present than if you weighed out 2 grams of thorium powder; this is because carbon is much smaller - kind of like a car filled with clowns, one given car can hold a lot of small clowns but only a few big ones; so the same volume is occupied but the amount of substance (clowns) varies on their own size. The arbitrary mass (relative to the hydrogen atom) for a molecule is the sum of its atomic components' atomic masses; e. g. C2H6's will have 2x12.00 (carbon) + 6x1.01 (hydrogen) = ~30 grams / mole.
Explanation:
Answer:
It will turn red
Explanation:
Bases will turn the litmus paper (impregnated with an acid-base indicator) blue.
Acids will turn turn the litmus paper (impregnated with an acid-base indicator) red.
Since the produced HCl is a strong acid, the litmus paper will turn red, when touching the HCl.
The red shows the presence of an acid, in this case HCl.
I would say the chain rings
Answer:
39.3%
Explanation:
CaF2 + H2SO4 --> CaSO4 + 2HF
We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:
For CaF2;
Number of moles reacted= mass/molar mass
Molar mass of CaF2= 78.07 g/mol
Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass
Molar mass of hydrogen fluoride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%
40 g Ar.......6.023*10^23 atoms
x g Ar.......3.8*10^24 atoms
x=40*3.8*10^24/(6.023*10^23)=252,365g Ar
