Please help 3. The combustion of glucose is represented by the following balanced equation: C6H12O6 + 602 - 6H20 + 6CO2. Which reactant is the limiting reagent if there is 1 gram of both C6H12O6 and O2? (Hint: Must convert to moles to determine this.)
O a C. H 20
O bO2
O c H₂O
Od CO₂

Explanation:

C₆H₁₂O₆       +      60₂     =     6H₂0     +     6CO₂.

1 mole              6 mole

molecular weight of C₆H₁₂O₆ = 180 g

molecular weight of oxygen = 32 g

1 gram of glucose = 1 / 180 = 5.55 x 10⁻³ moles

1 gram of oxygen = 1 / 32 = 31.25 x 10⁻³ moles

1 mole of glucose reacts with 6 moles of oxygen

5.55 x 10⁻³ moles of glucose will react with 6 x 5.55 x 10⁻³ moles of oxygen

= 33.30 x 10⁻³ moles of oxygen .

But oxygen available = 31.25 x 10⁻³ moles

So available oxygen is less than required .

Hence oxygen is the limiting reagent .

b ) is the right option .

answer is: nuclear fusion produces elements that are heavier than helium.

nuclei combine to form a heavier nucleus, releasing energy.  

for example, fusion of hydrogen atoms at the sun.  

this fusion nuclear reaction is in the inner core of the sun and hydrogen (atomic number 1) is conveted into helium (atomic number 2).  

the inner core is the central region of the sun and all solar energy is produced in the core by nuclear fusion.