Answer:
Average atomic mass of carbon = 12.01 amu.
Explanation:
Given data:
Abundance of C¹² = 98.89%
Abundance of C¹³ = 1.11%
Atomic mass of C¹² = 12.000 amu
Atomic mass of C¹³ = 13.003 amu
Average atomic mass = ?
Solution:
Average atomic mass of carbon = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass of carbon = (12.000×98.89)+(13.003×1.11) /100
Average atomic mass of carbon= 1186.68 + 14.43333 / 100
Average atomic mass of carbon = 1201.11333 / 100
Average atomic mass of carbon = 12.01 amu.
From Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.
<h3>What is Avogadro's number?</h3>
Avogadro's number, or Avogadro's constant, is the number of particles found in one mole of a substance.
The Avogadro's number is given as 6.02 x 10²³.
Summary of Josef Loschmidt and Amedeo Avogadro Contribution to chemistry.
- Equal volumes of gas contain equal numbers of molecules,
- Elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.
Thus, from Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.
Learn more about Avogadro's here: brainly.com/question/1581342
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The solid compound, K2SO4 contains a cation called K+ and an anion called SO42-. In this case, there are 2 atoms of potassium, 1 atom of sulfur and 4 moles of oxygen. The compound also contains ionic bonds because of the composing non-metals and metal.
(ANS1)— P4 + 5O2 ---> 2P2O5
(ANS2)— C3H8 + 5O2---> 3CO2 + 4H20
(ANS3)— Ca2Si + 4Cl2 ---> 2CaCl2 + SiCl4
We know that, M1V1 = M2V2
(Initial) (Final)
where, M1 and M2 are initial and final concentration of soution respectively.
V1 and V2 = initial and final volume of solution respectively
Given: M1 = 12 m, V1 = 35 ml and V2 = 1.2 l = 1200 ml
∴ M2 = M1V1/V2 = (12 × 35)/ 1200 = 0.35 m
Final concentration of solution is 0.35 m
