Answer:
Mass of H₂O is 3.0g
Explanation:
The reaction equation is given as:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
Parameters that are known:
Mass of CO₂ used = 7.3g
Unknown: mass of water consumed = ?
Solution
To solve this kind of problem, we simply apply some mole concept relationships.
- First, we work from the known to the unknown. From the problem, we have 7.3g of CO₂ that was used. We can find the number of moles from this value using the expression below:
Number of moles of CO₂ = 
- From this number of moles of CO₂, we can use the balanced equation to relate the number of moles of CO₂ to that of H₂O:
6 moles of CO₂ reacted with 6 moles of H₂O(1:1)
- We can then use the mole relationship with mass to find the unknown.
Workings
>>>> Number of moles of CO₂ =?
Molar mass of CO₂ :
Atomic mass of C = 12g
Atomic mass of O = 16g
Molar mass of CO₂ = 12 + (2 x16) = 44gmol⁻¹
Number of moles of CO₂ =
= 0.166moles
>>>>>> if 6 moles of CO₂ reacted with 6 moles of H₂O, then 0.166moles of CO₂ would produce 0.166moles of H₂O
>>>>>> Mass of water consumed = number of mole of H₂O x molar mass
Mass of H₂0 = 0.166 x ?
Molar mass of H₂O:
Atomic mass of H = 1g
Atomic mass of O = 16
Molar mass of H₂O = (2x1) + 16 = 18gmol⁻¹
Mass of H₂O = 0.166 x 18 = 3.0g
Answer:
The answer is 2
The maximum number an subshell can have is 2
Answer:
Soils are a function of the five soil-forming factors: climate, organisms, relief, parent material, and time. Each of these factors range on a continuum, so the different soils of the world number in the thousands. Soil scientists recognize 12 major orders of soils.
Explanation:
A mixture
having a uniform composition where the components can't be seen separately and
all components are in the same state best describes a solution.
<span>In chemistry, a </span>solution<span> is a
homogeneous mixture composed of two or more substances. In such a mixture, a
solute is a substance dissolved in another substance, known as a solvent.</span>
The correct answer between all
the choices given is the third choice or letter C. I am hoping that this answer
has satisfied your query and it will be able to help you in your endeavor, and
if you would like, feel free to ask another question.