Answer:
1/t means that the order of reaction is a first order. Meaning that the rate of reaction is directly proportional to reactant concentration. Scientists work with the standard units, therefore 1/t is 1 divide by 1 second.
Explanation:
Question: What is shown on the photo above?
Choices: A) An aquifer B) A sinkhole C) A watershed D) Karst topography
Answer: <u>It's C) A watershed because A. is where the water is underground, B. is a sinkhole and it has a big hole that is big or sometimes small, and D. Is where it has a little crack in it and it goes underground. </u>
<em>Hope this helps!.</em>
<em>~~~~~~~~~~~~~~~~~~~</em>
<em>~A.W~ZoomZoom44</em>
Let's rewrite the reaction for clarity:
2 SO₂(g) + O₂(g) ⇆ 2 SO₃(g) δhºrxn = –198 kj/mol
The equilibrium constant of a reaction is the ratio of the concentration its products to its reactants which are raised to their respective stoichiometric coefficients. For this reaction, the K would be
K = [SO₃]²/[SO₂]²[O₂]
To get a larger K, the products must be greater than the reactants. This means that the forward reaction must be favored to yield more of the product SO₃. There are different ways to do this: by manipulating the pressure, concentration or temperature.
For the concentration, you should add more amounts of the reactants. For the pressure, we should increase it. This is because the product side has only 2 moles of gas compared to 3 moles of gas in the reactants. So, it wall have more room for the product even at a higher pressure. Lastly, since the reaction is exothermic manifested by the negative sign of δhºrxn , the reaction would favor the forward reaction at high temperatures.
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Answer:
A saturated solution can become supersaturated when it is cooled. The solubility of solid solutes in liquid solvents increases as the solvent is warmed up. For example, you can dissolve more sugar in warm water as opposed to cold water.
