<span>To calculate the average mass of the element, we take the summation of the product of the isotope mass and the percent abundance. In this case, it is 0.9963 * 14.003 amu + 0.0037* 15.00 amu. This is equal to an average mass of 14.00668889 amu.</span>
I would say the answer is A - a stars mass
The majority of stars in the galaxy, including our Sun, Sirius and Alpha Centauri A and B are all main sequence stars.
Mass is the key factor in determining the lifespan of a main sequence star, its size and its luminosity ( brightness)
Two characteristics define brightness: luminosity and magnitude. Luminosity is the amount of light that a star radiates. The size of the star and its surface temperature determine its luminosity. Apparent magnitude of a star is its perceived brightness, factoring in size and distance, while absolute magnitude is its true brightness irrespective of its distance from earth.
Answer:
it's urine option ( C ) .
Answer:
V₂ = 16.5 L
Explanation:
To solve this problem we use <em>Avogadro's law, </em>which applies when temperature and pressure remain constant:
V₁/n₁ = V₂/n₂
In this case, V₁ is 22.0 L, n₁ is [mol CO + mol NO], V₂ is our unknown, and n₂ is [mol CO₂ + mol N₂].
- n₁ = mol CO + mol NO = 0.1900 + 0.1900 = 0.3800 mol
<em>We use the reaction to calculate n₂</em>:
2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)
0.1900 mol CO * 
0.1900 mol CO₂
0.1900 mol NO * 
0.095 mol N₂
- n₂ = mol CO₂ + mol N₂ = 0.1900 + 0.095 = 0.2850 mol
Calculating V₂:
22.0 L / 0.3800 mol = V₂ / 0.2850 mol
V₂ = 16.5 L
