Answer:
K= -0.286 Mol^-1.S^-1
Explanation:
FIRSTLY, WE WRITE A CHEMICAL REACTION EQUATION FOR THE REACTION BETWEEN OXALIC ACID AND POTASSIUM PERMANGANATE.
KMnO2(aq) + H2C2O4(aq) ----> Mn2+(aq) + CO2(g) + H2O(l)
(purple) (colorless)
THE RATE OF THE REACTION IS MEASURED AND MONITORED BY CHECKING FOR THE DISAPPEARANCE OF THE PURPLE COLOR OF THE POTASSIUM PERMANGANATE.
The rate of the reaction is given by : -Δ[KMnO4]/ΔT = -Δ[ H2C2O4]/ΔT = Δ[ Mn2+]/ΔT
When all of the purple color is gone, we can say that the reaction is finished and that the final concentration of the KMnO4 is now zero. Since the initial time for ΔT is zero, the equation for the rate of
disappearance of the potassium permanganate becomes:
Rate = Δ[KMnO4]/ΔT= (0-[initial KMnO4]) ÷ (elapsed time -0) = [Initial KMnO4 ] ÷ elapsed time
Hence, the rate of the chemical reaction is thus;
-1.3/5 = -0.26M/s
Now, to get the rate constant for the reaction, we need to know the rate law and this can be expressed as follows:
For a reaction A+B---> C The general rate law would be: Rate = k[A]^m[B]^n
In this particular case,
Rate =k [KMnO4][ H2C2O4]
Kindly note that m and n in this case is 1 each. We have calculated our rate from above. Hence:
-0.26 = k(1.3)(0.7)
k = -0.26÷(1.3)(0.7)
K= -0.286 Mol^-1.S^-1
Answer:
I has 2 double carbon carbon bonds
Answer:
first of all is that really chemistry
Explanation:
The correct answer is C.
Through runoff on Earth's surface. Hope this helped!
The precipitate formed when a solution of manganese(II) nitrate and
potassium sulfide react is Manganese sulfide (MnS)
To obtain the precipitate formed when a solution of manganese(II) nitrate and potassium sulfide react, we shall write the balanced ionic equation. This is illustrated below:
Manganese (II) nitrate => Mn(NO₃)₂
Potassium sulfide => K₂S
In solution, they will react as follow:
Mn(NO₃)₂ (aq) —> Mn²⁺(aq) + 2NO₃¯(aq)
K₂S(aq) —> 2K⁺(aq) + S²¯(aq)
Mn(NO₃)₂ (aq) + K₂S(aq) —>
Mn²⁺(aq) + 2NO₃¯(aq) + 2K⁺(aq) + S²¯(aq) —> MnS(s) + 2K⁺(aq) + 2NO₃¯(aq)
Cancel out the spectator ions (i.e K⁺ and NO₃¯)
Mn²⁺(aq) + S²¯(aq) —> MnS(s)
From the net ionic equation above, we can see that MnS is insoluble.
Therefore, the precipitate formed when a solution of manganese(II) nitrate and potassium sulfide react is Manganese sulfide (MnS)
Learn more: brainly.com/question/21280827