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3 years ago

I'm afraid I won't pass. And very nervous. Can you help me please.

1 answer:

4 0

Just look at the number in front also called coefficient (you have to balance the equations first, but all the questions here are balanced, so no worries). for q1.
in the balanced equation, the number in front of aluminum oxide is 2 (2 - this number Al2O3) and for aluminium is 4 as in (4 Al). so the ratio is 2:4. simplified it is 1:2. or write it out fully
2 Al2O3: 4 Al
ignore everything after the number.
2:4
same as 1:2
Aluminium oxide to oxygen
2 Al2O3: 3 O2
2:3
aluminum to oxygen
4 Al: 3 O2
4:3

question 2
Mercury oxide to Mercury
2 HgO : 2 Hg
2:2
same as 1:1
Mercury oxide to oxygen
2 HgO : O2
since oxygen in this case does not have a number written in front of it, the default is 1.
2: 1.
you should be able to do the rest

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A researcher wants to determine if a unicellular organism he discovered is an autotroph or a heterotroph. He radioactively label

GaryK [48]

Answer:

Explanation:

Autotrophs utilize the energy from sunlight to reduce carbon dioxide to carbohydrates (glucose). The energy from the sunlight is used to split water into H+ and O2- and the H+ used in the reduction process. The labeled carbon in the carbon dioxide will, therefore, be incorporated by the autotrophs in the carbohydrates made in photosynthesis.

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3 years ago

Explain why scientists find the particulate matter of theory useful

BARSIC [14]

This idea has historical significance. The ancient Greek philosopher Democritus (born 460 BCE), who held that everything is composed of small particles moving in empty space, is credited with developing the first hypothesis we have about the microscopic universe. He had some concrete proof for this, such the fact that items like a new loaf of bread or a rose may give off a scent even when they are far from the source. Being a materialist, he thought that these odors originated from actual material particles released by the bread or the rose, rather than being purely a type of magic. He reasoned that these particles must float through the air, with some of them maybe landing in your nose where you can smell them immediately. This still makes sense in modern times. But many of us now have quite different perspectives on these "particles."

Thank you,

Eddie

6 0

2 years ago

Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)

saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if $\rm O_2$ is removed from the mixture. The reason is that collisions between $\rm SO_2$ molecules and $\rm O_2\!$ molecules would become less frequent.

The reaction would not be at equilibrium for a while after $\rm O_2$ was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of $\rm SO_3\, (g)$ to the system, the backward reaction would have broken down the same amount of $\rm SO_3\, (g)\!$ . So is the case for $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ .

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield $\rm SO_3\, (g)$ , only a fraction of the reactions will be fruitful.

Assume that $\rm O_2\, (g)$ molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer $\!\rm O_2\, (g)$ molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between $\rm SO_2\, (g)$ and $\rm O_2\, (g)\!$ molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after $\rm O_2\, (g)$ was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more $\rm SO_3\, (g)$ gets converted to $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ , the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0

2 years ago

How many molecules are there in 2.30g of NH3

Vinil7 [7]

Answer:

8.13x10^22 molecules

Explanation:

We can use the Avogadro's number(6.022 x 10^23 units / mole)

2.30 g NH3 (1 mol / 17.03 g ) (6.022 x 10^23 molecules / 1 mol ) = 8.13x10^22 molecules

Hope this helps! Feel free to ask any questions!

6 0

2 years ago

Aluminum is manufactured using electrolysis. Carbon electrodes are used. Describe the nature of the electrolyte.

Aleksandr [31]

Answer:

The traditional electrolyte for aluminium electrolysis is based on molten cryolite (Na3AlF6), acting as solvent for the raw material, alumina (Al2O3).Metals are found in ores combined with other elements. Electrolysis can be used to extract a more reactive metal from the ore.

Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions. As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere. In an aqueous solution, if the potential is high enough, OH− and O2− are generated at the anode, which can then react with the aluminum to produce aluminum oxide. Al^3+ can also be generated directly. The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the Al^3+ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode. In this way, coatings thicker than the normal passivation in air can be produced. However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made). This is the normal behaviour in aqueous solutions at near-neutral pH (5–7).

However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface. One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed OH−, leaving pores in the oxide layer through which the ions can travel and continue to react. These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.

3 0

2 years ago