Answer:
C. U-235
Explanation:
A fission reaction is one in which an unstable radioisotope breaks down into smaller nuclei when bombarded with energetic particles.
The highly unstable isotope that is often used in fission reactions is U-235. When bombarded with neutrons that atom breaks down to form krypton and barium along with the release of a large amount of energy. The reaction is given as:
₉₂U²³⁵ + ₀n¹ → ₃₆Kr⁸⁹ + ₅₆Ba¹⁴⁴ + 3₀n¹ + 210 Mev energy
Under which color of light will plants be least likely to make food?
red
blue
orange
green
Answer:
The color of light in which plants are less likely to produce food is green.
Explanation:
Photosynthesis is the process by which plants take advantage of sunlight to convert it into chemical energy, necessary to manufacture their nutrients, their growth and development.
Plants do not absorb all the colors of the light spectrum, since due to their pigments they absorb colors with specific wavelengths. Green plants do not absorb the color green, but reflect it —due to the chlorophyll pigment— which gives the green color to those leaves.
For this reason, the light that plants take least advantage of to produce food is green light.
Other options:
The colors blue, orange and red can be absorbed by the plant, to a lesser or greater degree.
Answer:
it would be green
Explanation:
Answer:
-196 kJ
Explanation:
By the Hess' Law, the enthalpy of a global reaction is the sum of the enthalpies of the steps reactions. If the reaction is multiplied by a constant, the value of the enthalpy must be multiplied by the same constant, and if the reaction is inverted, the signal of the enthalpy must be inverted too.
2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ
S(s) + O₂(g) → SO₂(g) ΔH = -297 kJ (inverted and multiplied by 2)
2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ
2SO₂(g) → 2S(s) + 2O₂(g) ΔH = +594 kJ
-------------------------------------------------------------
2S(s) + 3O₂(g) + 2SO₂(g) → 2SO₃(g) + 2S(s) + 2O₂(g)
Simplifing the compounds that are in both sides (bolded):
2SO₂(g) + O₂(g) → 2SO₃(g) ΔH = -790 + 594 = -196 kJ
The enthalpy of the reaction where sulfur dioxide is oxidized to sulfur trioxide is -395 kJ.
The calculation of the enthalpy change of the reaction in which sulfur dioxide is oxidized to sulfur trioxide involves Hess's Law, which states that the enthalpy change of a chemical reaction is the same whether it takes place in one step or several steps. This can be solved by comparing the enthalpy changes given in the two reactions presented.
First, consider the reactions given:
2S(s) + 3O₂(g) → 2SO₃(g), ΔH = -790 kJ
S(s) + O₂(g) → SO₂(g), ΔH = -297 kJ
From these reactions, it is seen that the first reaction can be re-written as:
2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = -790 kJ
However, this reaction contains two moles of SO₂ whereas the reaction in question only requires one mole. Thus, the enthalpy change for the reaction becomes: ΔH = -790 KJ / 2 = -395 kJ.
#SPJ3