Molar mass:
O2 = 31.99 g/mol
C8H18 = 144.22 g/mol
<span>2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)
2 x 144.22 g --------------- 25 x 31.99 g
10.0 g ----------------------?? ( mass of O2)
10.0 x 25 x 31.99 / 2 x 144.22 =
7997.5 / 288.44 => 27.72 g of O2
hope this helps!
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Answer:
The mole ratio of the cation and the anion in a precipitate is a simple fraction. ( im sorry if this dosent help a lot.)
Explanation:
The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.
Isotope mass amu Relative abundance
1 77.9 14.4
2 81.9 14.3
3 85.9 71.3
Express your answer to three significant figures and include the appropriate units.
Answer: 84.2 amu
Explanation:
Mass of isotope 1 = 77.9
% abundance of isotope 1 = 14.4% = 
Mass of isotope 2 = 81.9
% abundance of isotope 2 = 14.3% = 
Mass of isotope 3 = 85.9
% abundance of isotope 2 = 71.3% = 
Formula used for average atomic mass of an element :

![A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2877.9%5Ctimes%200.144%29%2B%2881.9%5Ctimes%200.143%29%2B%2885.9%5Ctimes%200.713%29%5D)

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu
Answer:
Samarium
Explanation:
The element Sm describe is called Samarium. This element has unique sets of properties that makes it very unique and distinct.
The lanthanides are found in the f-block on the periodic table of elements.
This element is a moderately hard silvery metal that readily oxidizes in air. It assumes an oxidation state of +3. The element has an atomic number of 62