At the complete combustion of butane the carbon dioxide and water are formed.
2C₄H₁₀ + 13O₂ = 8CO₂ + 10H₂O
C₄H₁₀ ⇒ 4CO₂, 5H₂O ⇒ 13O ⇒ 6.5O₂
we use integer coefficients: 8CO₂, 10H₂O, 13O₂
Answer:
Between 4s and 3s orbital , 3s has more energy .
Explanation:
According to the rule , the lower the value of (n+l) for an orbital , the lower is it's energy . And if two orbitals have the same value of (n+l), the orbital with lower value of n will have the lower energy .
Answer:
Average atomic mass = 17.5 amu.
Explanation:
Given data:
X-17 isotope = atomic mass17.2 amu, abundance:78.99%
X-18isotope = atomic mass 18.1 amu, abundance 10.00%
X-19isotope = atomic mass:19.1 amu, abundance: 11.01%
Average atomic mass of X = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100
Average atomic mass = 1358.628 + 181 +210.291 / 100
Average atomic mass = 1749.919 / 100
Average atomic mass = 17.5 amu.
P1v1/t1 = p2t2/t2
p is constant
v1=600, t1 =20c=293K
v2=?, t2=60c=333K
temperature must be in Kelvin
do the math
