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:
-1.42, -0.375, 32.5% (.325), 3/8 (.375), √4 (2.0), 3 (3.0), 2³ (8.0)
Answer:
There are 8Si atoms and 16 O atoms per unit cell
Explanation:
From the question we are told that:
Edge length 
Density 
Generally the equation for Volume is mathematically given by



Where
Molar mass of (SiO2) for one formula unit


Therefore
Density of Si per unit length is


Molar mass of (SiO2) for one formula unit


Therefore
There are 8Si atoms and 16 O atoms per unit cell
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
Elements in the same group have D. Same number of valence electrons.