When 2 molecules of pyruvate from glycolysis enters the mitochondria,
2 molecules of CO2 are released
Answer:
-2
Explanation:
carbonate oxifation number is -2
Answer:
626.7nm
Explanation:
The energy of a photon is defined as:
E = hc / λ
<em>Where E is the energy of the photon, h is Planck constant (6.626x10⁻³⁴Js), c is speed of light (3x10⁸m/s) and </em>λ is the wavelength of light
The energy of 1 photon is:
(191000 J / mol) ₓ (1 mole / 6.022x10²³) = 3.1717x10⁻¹⁹ J
Replacing:
3.1717x10⁻¹⁹ J = <em>6.626x10⁻³⁴Jsₓ3x10⁸m/s / </em>λ
λ = 6.267x10⁻⁷m
as 1nm = 1x10⁻⁹m:
6.267x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =
<h3>626.7nm</h3>
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.
