Answer:
longitudinal
Explanation:
A prime meridian is the meridian (a line of longitude) in a geographic coordinate system at which longitude is defined to be 0°. Together, a prime meridian and its anti-meridian (the 180th meridian in a 360°-system) form a great circle. This great circle divides a spheroid into two hemispheres.
Answer:
[O2(g)][SO2(g)]^2/[SO3(g)]^2
Answer: 1.59atm
Explanation:
We have that for the Question "Calculate the final pressure of the gas mixture, assuming that the container volume does not change."
it can be said that
The final pressure of the gas mixture, assuming that the container volume does not change =
From the question we are told
A container of N2O3(g) has a pressure of 0.265 atm. When the absolute temperature of the N2O3(g) is tripled, the gas completely decomposes, producing NO2(g) and NO(g).
Answer is: concentration of hydrogen iodide is 6 M.
Balanced chemical reaction: H₂(g) + I₂(g) ⇄ 2HI(g).
[H₂] = 0.04 M; equilibrium concentration of hydrogen.
[I₂] = 0.009 M; equilibrium concentration of iodine.
Keq = 1·10⁵.
Keq = [HI]² / [H₂]·[I₂].
[HI]² = [H₂]·[I₂]·Keq.
[HI]² = 0.04 M · 0.009 M · 1·10⁵.
[HI]² = 36 M².
[HI] = √36 M².
[HI] = 6 M.
Answer:
9 : 8
Explanation:
Aluminum oxide has the following formula Al₂O₃.
Next, we shall determine the mass of Al and O₂ in Al₂O₃. This can be obtained as follow:
Mass of Al in Al₂O₃ = 2 × 27 = 54 g
Mass of O₂ in Al₂O₃ = 3 × 16 = 48 g
Finally, we shall determine the mass ratio of Al and O₂. This can be obtained as follow:
Mass of Al = 54 g
Mass of O₂ = 48 g
Mass of Al : Mass of O₂ = 54 : 48
Mass of Al : Mass of O₂ = 54 / 48
Mass of Al : Mass of O₂ = 9 / 8
Mass of Al : Mass of O₂ = 9 : 8
Therefore, the mass ratio of Al and O₂ in Al₂O₃ is 9 : 8
