B.
I hope this will help a bit (Forgive me of its wrong pls)
Answer:
second law of thermodynamics.
Explanation:
The second law of thermodynamics deals with interconversion of energy from one form to another. Although energy can be converted from one form to another, this conversion is never 100% efficient because energy is lost in certain ways such as through heat. In a combustion engine, it is not possible to recover the energy from the gasoline 100% since energy must be lost along the way via such means as heat losses. Hence I will be skeptical about such an advert.
Steel Wool + Oxygen (Fire) The steel wool is a grayish color and has a rough surface. Oxygen is transparent It looks like a powder, and like if it was rusted steel wool but then broken apart and turned into a powder Well Iron oxide is 7 grams and steel wool combined is 7 grams so you can say that the mass was conserved Yes, this is a chemical change because the steel wool rusted and rust is a chemical change, so iron oxide is cause because if a physical change.
Egg + Heat The egg has a yellow yolk in the middle while there is some type of liquid at the border of the yolk, but it is clear, the heat is hot but transparent The egg turned into a delicious food called an omelet what was yolk staid as a liquid but got a tad harder, but the transparent border around it turned white It was conserved because the eggshell was 4 grams and the fried egg is 41 grams It is a chemical change because it results in the formation of new particles, and the chemical bonds break up and new ones are formed.
Water + Heat The water is a clear liquid, while the heat is very hot but transparent The water turned into a type of oxygen -Water Vapor- If 5 g of water becomes a gas it becomes 5 g of water vapor. The mass of the liquid water is simply transferred into the mass of the newly formed water vapor. There was no chemical reaction because the water vapor can be turned back to water also it just changed from a liquid to a gas and did not change its composition
Zinc + Hydrogen Chloride Zinc: a white/silver metallic solid. Hydrogen chloride (dilute hydroelectric acid) a transparent, colorless liquid with a very low pH (acidic). Zinc "dissolved" in hydrogen chloride, while emitting a colorless gas. The liquid remains a colorless liquid, possibly still having a low pH from the unused acid. The colorless gas collected in a test tube gives a popping sound when ignited with a burning wooden splint, so it is not air embedded in the zinc, nor dissolved in hydroelectric acid. Well although the zinc chloride is 12 grams and not 15 the gas that was released was 3 grams and as we all know 3 + 12 is 15 so you could say that the mass was conserved The production of a new substance (most probably hydrogen) from the reaction of the two reactants. When a few drops of the liquid product are evaporated on a watch glass, a white residue is left. When a few drops of the liquid hydroelectric acid are evaporated on a watch glass, there is no residue. This proves that a new product (hydrogen gas), (white powder, zinc chloride) is produced instead of zinc being physically dissolved in hydroelectric acid.
Sodium Hydroxide + Copper Sulfate Sodium hydroxide is a turbid solution and copper sulfate is in form of bright blue crystals. When their solutions are mixed with each other, a pale blue precipitate of basic copper hydroxide & a solution of neutral salt sodium sulfate will be formed. The sodium hydroxide and the copper sulfate combined make a total of 67 grams and the product is split because the sodium sulfate is 47 grams and the copper hydroxide is 20 grams but all together it is still 67 grams so you could say that the mass was conserved The proof of the reaction is the appearance of pale blue precipitates of basic copper hydroxide & a solution of neutral salt sodium sulfate.
Answer:
The standard cell potential is 2.00 V
Explanation:
<u>Step 1:</u> Data given
Cu is cathode because of
higher EP
Al3++3e−→Al E∘=−1.66 V anode
Cu2++2e−→Cu E∘=0.340 V cathode
<u>Step 2:</u> Balance both equations
2*(Al → Al3+-3e−) E∘=1.66 V
3*(Cu2++2e−→Cu) E∘=0.340 V
<u>Step 3:</u> The netto equation
2 Al + 3Cu2+ +6e- → 2Al3+ + 3Cu -6e-
2 Al + 3Cu2+ → 2Al3+ + 3Cu
<u>Step 4:</u> Calculate the standard cell potential
E∘cell = E∘cathode - E∘anode
E∘cell = E∘ Cu2+/Cu - E∘ Al3+/Al
E∘cell =0.340 V - (-1.66) = 2.00 V
The standard cell potential is 2.00 V
