NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol
Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27%
3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3
0.8211 grams Na + 1.266 grams Cl = 2.087 grams
Answer: Fundamental, Key, vital, crucial
Explanation:
Answer is: the average atomic mass 217.606 amu.
Ar₁= 203.973 amu; the average atomic mass of isotope.
Ar₂ = 205.9745 amu.
Ar₃ = 206.9745 amu.
Ar₄ = 207.9766 amu.
ω₁ = 1.40% = 0.014; mass percentage of isotope.
ω₂ = 24.10% = 0.241.
ω₃ = 22.10% = 0.221.
ω₄ = 57.40% = 0.574.
Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.
Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.
Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.
Ar = 217.606 amu.
But abundance of isotopes is greater than 100%.
It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.
1- One mole is = 6.02 x 10^23 of anything, So one mole of atoms is 6.02x10^23.
2- when the balloon contains 0.15 moles of Co2 gas so:
the no.of molecules of Co2 = 0.15 x 6.02x 10^23
= 9.0 x 10^22
