Answer:
MnCO3 + 2H2O ⇄ MnO2 + HCO3- + 2e- +3H+
Explanation:
<u>The</u> unbalanced equation
MnCO3 ⇄ MnO2 + HCO3-
In MnCO3, the oxidation number of Mn is +2
In Mno2, the oxidation number of Mn is +4
The change from +2 to +4 requires an addition of 2 electrons (to the right side).
MnCO3 ⇄ MnO2 + HCO3- + 2e-
The total charge now is -3 on the right side. To balance this we add 3 hydrogen atoms on the right side.
MnCO3 ⇄ MnO2 + HCO3- + 2e- +3H+
On the right side we have 4 hydrogen atoms in total. On the left side we have 0 hydrogen atoms. So to balance, we have to add 2H2O on the left side
MnCO3 + 2H2O ⇄ MnO2 + HCO3- + 2e- +3H+
Now the reaction is balanced.
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
125 Kelvin turns into -146 C
Answer:
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Explanation: