Answer:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5.
Explanation:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5 because of the diagram.
They turn litmus paper blue
Hey there!
C₉H₂O + O₂ → CO₂ + H₂O
First let's balance the C.
There's 9 on the left and 1 on the right. So, let's add a coefficient of 9 in front of CO₂.
C₉H₂O + O₂ → 9CO₂ + H₂O
Next let's balance the H.
There's 2 on the left and 2 on the right. This means it's already balanced.
C₉H₂O + O₂ → 9CO₂ + H₂O
Lastly, let's balance the O.
There's 3 on the left and 19 on the right. So, let's add a coefficient of 9 in front of O₂.
C₉H₂O + 9O₂ → 9CO₂ + H₂O
This is our final balanced equation.
Hope this helps!
Hi,
To solve the question, first of all we will find out the no. of moles of H2SO4 in 19 g of sulfuric acid.
As we know:
No . of moles = Mass/ Molar mass
No. of moles= 19 g/98.08
g
No. of moles= 0.1937
Now we know the no of moles of H2SO4 that will react with 2LiOH. We also know the molar equivalence of H2SO4 , and 2LiOH that will react.
So, the water that will be produced will be 2H2O and 1 Li2SO4 when H2SO4 that will react with 2LiOH.
0.1937 x 2x 18.01
=6.977
=6.98
Therefore, approximately 6.98 grams of water will be produced from 19 g of sulfuric acid.
Hope it helps!
