Iodine is decolorized.
The first reaction stated in the question occurs as follows;
2 KI (aq) + 2 H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (s) + 2 H2O (l)
The reaction here is the formation of iodine from MnO2 and KI in the presence of dropwise H2SO4.
Hypo is the common name of sodium thio-sulphate or sodium hypo-sulfite.
The equation of the titration reaction is;
2Na2S2O3 + I2→ Na2S4O6 + 2NaI
When this reaction takes place, iodine is decolorized due to its reduction to I^-.
Answer:
A. The reaction will proceed forward forming more CH4
B. The reaction will proceed forward forming more CH4
C. Since the reaction is exothermic, raising the temperature will cause the reaction to proceed backward, thus forming C and H2.
D. Lowering the volume makes the gas particles to be more close together thereby enhancing their collisions leading to reaction. Therefore the reaction will proceed forward forming more CH4
E. Catalyst only reduce the activation energy so the reaction can proceed faster. The reaction will proceed forward forming.
F. The following will favour CH4 at equilibrium
i. Catalyst to the reaction mixture,
ii. Both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture
iii. Adding more C to the reaction mixture.
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Answer:
~0.14
Explanation:
There are roughly ~0.14 moles in 4.5g of Sulfur. 1 mol = 32.06 g of Sulfur, so we just find the ratio and then multiply.
Ngl, there is no picture.
