Answer:
The mass of third isotope is 25.98 amu.
Explanation:
Mass of Mg²⁴ = 23.985 amu
Mass of Mg²⁵ = 24.986 amu
Abundance of Mg²⁴ = 78.99 %
Abundance of Mg²⁵ = 10%
Abundance of third isotope = 11.01%
Atomic weight of magnesium = 24.305
Solution:
Atomic mass = ( % age of first isotope × its atomic mass) + (% age abundance of second isotope × its atomic mass) + ( % age of third isotope × its atomic mass ) / 100
Now we will put the values in formula.
24.305 = (78.99 × 23.985 ) + (10 ×24.986 ) + ( 11.01 × X) / 100
24.305 = 1894.56 + 249.9 +( 11.01 × X ) / 100
24.305 × 100 = 2144.46 + ( 11.01 × X)
2430.5 - 2144.46 = (11.01 × X)
286.04 / 11.01 = X
25.98 = X
The mass of third isotope is 25.98 amu.
True. Our earth Is getting hotter due to global warming.
Answer:
= 3.78 g H₂O
Explanation:
2C₂H₆ + 3O₂ => 4CO₂ + 6H₂O
2.1g C₂H₆ = 2.1g/30.0 g/mol = 0.07 mole ethane
3.68g O₂ = 3.68g/32 g/mol = 0.115 mole oxygen
Limiting Reactant:
A quick way to determine limiting reactant is to divide moles of reactant by its respective coefficient in the balanced molecular equation. The smaller value is the limiting reactant.
moles ethane = 0.07 mole / 2 (the coefficient in balanced equation) = 0.035
moles oxygen = 0.115 mole / 3 (the coefficient in balanced equation) = 0.038
Since the smaller value is associated with ethane, then ethane is the limiting reactant and the problem is worked from the 0.07 moles of ethane in an excess of O₂.
From the equation stoichiometry ...
2 moles C₂H₆ in an excess of O₂ => 6 moles H₂O
then 0.07 mole C₂H₆ in an excess of O₂ => 6/2(0.07 moles H₂O = 0.21 mole
Converting to grams of water produced
= 0.21 mole H₂O X 18 g/mol = 3.78 g H₂O
Potassium (K), Sodium(Na), Iron(Fe), Copper(Cu), Silver(Ag), Tin(Sn), Antimony(Sb), Tungsten(W), Gold(Au), Mercury(Hg), Lead(Pb)