I think this question is a true statement :)
Answer;
A) Stage 1: Chlorophyll captures light energy. Stage 2: Light energy is converted to chemical energy.
Explanation;
-Photosynthesis is the process by which green plants use energy from the sun, water and carbon dioxide to make organic compounds such as simple sugars together with release of oxygen.
-The process occurs in tow stages; light-dependent stage and light independent stage. During light dependent stage, chlorophyll absorbs sunlight and uses it to split water molecules into hydrogen ions and oxygen atoms. In the light independent stage carbon (iv) dioxide is fixed and the result is organic compound; the light energy is converted to chemical energy.
In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=

%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
Answer:
0.145 moles de AlBr3.
Explanation:
¡Hola!
En este caso, al considerar la reacción química dada:
Al(s)+Br2(l)⟶AlBr3(s)
Es claro que primero debemos balancearla como se muestra a continuación:
2Al(s)+3Br2(l)⟶2AlBr3(s)
Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.
¡Saludos!