Answer is: silicon isotope with mass number 28 has highest relative abundance, this isotope is the most common of these three isotopes.
Ar₁(Si) = 28; the average atomic mass of isotope ²⁸Si.
Ar₂(Si) =29; the average atomic mass of isotope ²⁹Si.
Ar₃(Si) =30; the average atomic mass of isotope ³⁰Si.
Silicon (Si) is composed of three stable isotopes, ₂₈Si (92.23%), ₂₉Si (4.67%) and ₃₀Si (3.10%).
ω₁(Si) = 92.23%; mass percentage of isotope ²⁸Si.
ω₂(Si) = 4.67%; mass percentage of isotope ²⁹Si.
ω₃(Si) = 3.10%; mass percentage of isotope ³⁰Si.
Ar(Si) = 28.086 amu; average atomic mass of silicon.
Ar(Si) = Ar₁(Si) · ω₁(B) + Ar₂(Si) · ω₂(Si) + Ar₃(Si) · ω₃(Si).
28,086 = 28 · 0.9223 + 29 · 0.0467 + 30 · 0.031.
If you know the parent genotypes then you can make the Punnett square
Answer:
Explanation:
H = 1
C = 12
O = 16
Acetylene, HC≡CH = 2+24 = 26
H2O = 2 + 16 = 18
In XS oxygen, one HC≡CH yields one H2O
26 g HC≡CH ==> 18 g H2O
2000 g HC≡CH ==> 2000*18/26 g H2O = 1384.6154 g H2O
With the given formula, we can calculate the amount of CO₂ using the balance equation but we first need the moles of CH₄
1) to find the moles of CH₄, we need to use the ideal gas formula (PV= nRT). if we solve for n, we solve for the moles of CH₄, and then we can convert to CO₂. Remember that the units put in this formula depending on the R value units. I remember 0.0821 which means pressure (P) has to be in atm, volume (V) in liters, the amount (n) in moles, and temperature (T) in kelvin.
PV= nRT
P= 1.00 atm
V= 32.0 Liters
n= ?
R= 0.0821 atm L/mol K
T= 25 C= 298 K
let plug the values into the formula.
(1.00 x 32.0 L)= n x 0.0821 x 298K
n= (1.00 x 32.0 L )/ (0.0821 x 298)= 1.31 moles CH₄
2) now let's convert the mole of CH₄ to moles to CO₂ using the balance equation
1.31 mol CH₄ (1 mol CO₂/ 1 mol CH₄)= 1.31 mol CO₂
3) Now let's convert from moles to grams using the molar mass of CO₂ (find the mass of each atom in the periodic table and add them)
molar mass CO₂= 12.00 + (2 x 16.0)= 44.0 g/mol
1.31 mol CO₂ ( 44.0 g/ 1 mol)= 57.6 g CO₂
Note: let me know if you any question.
