B). light energy is not required to proceed
Explanation:
In the Calvin cycle of photosynthesis, light energy is not required. The Calvin cycle is light independent and it is made up of a series of redox reactions.
- During photosynthesis reactions, green plants manufacture their food using carbon dioxide, sunlight and water.
- During the Calvin cycle aspect, light energy is not required for chemical reactions to take place. The light energy helps to move electrons.
- The cycle is also known as dark reactions.
- It is at this stage that carbon dioxide combines with water to form glucose.
- The reaction is initiated with light energy which produces NADPH and ATP.
- The Calvin cycle follows by using the NADPH and ATP to produce glucose in the dark phase.
Learn more:
ATP brainly.com/question/2953868
Light dependent reactions brainly.com/question/6866300
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Answer:
Volume of acid, Va=250mL; Volume of quinine,Vb=20mL; Molarity of acid, Ma=0.05M.
Molar mass of acid= H2+S+O4= 2+32+(16X4)= 2+32+64=98g
Concentration of acid, Ca= Molar mass of acid/ Ma =98/0.05=1960g/mol
Explanation: To calculate concentration of quinine, Cb is as follow
Va*Ca=Vb*Cb
∴ Cb=Va*Ca/Vb =250*1960/20 =24500g/mol
See image below for the lewis structure of acrolein
This is a straightforward question related to the surface energy of the droplet.
<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>
<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>
<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>
<span>The five smaller droplets need to have the same volume as the original. Therefore </span>
<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>
<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>
<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>
<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>
<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
