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
Given:
257J of heat
5500g of mercury
increase by 5.5
degrees Celsius
Required:
Specific heat of
mercury
Solution:
H
= mCpT
257J = (5500g of
mercury) Cp (5.5 degrees Celsius)
Cp = 8.5 x 10^-3
Joules per gram per degree Celsius
<span>the table say that at 20 degree celcius 88.0g of NANO3 will remain dissolved in
100 gm of H2O
so at 20 degree celcius 80.0g of H20 will dissolve
(88.0g)x(80g/100g)=70.4g of NANO3
so at 20 degree celcius
86.3g-70.4g= 15.9 gram of NANO3 will come out of solution.</span>
1. Berkelium(Berkeley, CA) 2. Dubnium(Dubna, Russia) 3. Darmstaditum (Darmstadt, Germany) 4. Erbium(Ytterby, Sweden) 5. Strontium(Strontian, Scotland) 6. Terbium(Ytterby, Sweden) 7. Yttebium(Ytterby, Sweden) 8. Yttrium(Ytterby, Sweden)