Answer: D
Explanation:
Because velocity is speed
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!
Answer:
The final and initial concentration of the acid and it's conjugate base are approximately equal, that is we use the weak acid approximation.
Explanation:
The Henderson-Hasselbalch is used to calculate the pH of a buffer solution. It depends on the weak acid approximation.
Since the weak acid ionizes only to a small extent, then we can say that [HA] ≈ [HA]i
Where [HA] = final concentration of the acid and [HA]i = initial concentration of the acid.
It also follows that [A^-] ≈ [A^-]i where [A^-] and[A^-]i refer to final and initial concentrations of the conjugate base hence the answer above.
Answer:
C. A linear, nonpolar molecule
Explanation:
Molecules which are alike usually have the same degree of pull which results in them sharing electrons. This sharing of electrons is known as the molecules exhibiting Covalent bonding between them.
The equal pull also results in the cancelling out of electrons and favoring non polar bonds due to the absence of free electrons which would have been able to interact with H2O in a polar binding system.
Answer:
1.4 × 10^-4 M
Explanation:
The balanced redox reaction equation is shown below;
5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O
Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol
Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles
Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M
Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M
Volume of Fe^2+ (VA)= 20.0 ml
Let the concentration of MnO4^- be CB (the unknown)
Volume of the MnO4^- (VB) = 12.6 ml
Let the number of moles of Fe^2+ be NA= 5 moles
Let the number of moles of MnO4^- be NB = 1 mole
From;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CB= CAVANB/VBNA
CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5
CB = 1.4 × 10^-4 M
