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:
ice caps and glaciers
Explanation:
Over 68 percent of the fresh water on Earth is found in icecaps and glaciers, and just over 30 percent is found in ground water. Only about 0.3 percent of our fresh water is found in the surface water of lakes, rivers, and swamps.
The energy release when dissolving 1 mol of NaOH in water is 445.1 kJ
the mass of NaOH to be dissolved is 32.0 g
The number of NaOH moles in 32.0 g - 32.0 g / 40 g/mol = 0.8 mol
the energy released whilst dissolving 1 mol of NaOH - 445.1 kJ
when dissolving 0.8 mol - the energy released is 445.1 kJ/mol x 0.8 mol
therefore heat released is - 356.08 kJ
answer is -356.08 kJ
Answer:
Density, d = 1.779 g/cm³
Explanation:
The density of a material is given by its mass per unit volume.
Here, height of a piece of magnesium cylinder, h = 5.62 cm
Its diameter, d = 1.34 cm
Radius = 0.67 cm
Volume of he cylinder,


So, the density of the sample is 1.779 g/cm³.