To answer this question, it's important to remember the details of combustion reactions. In the presence of oxygen and a sufficient amount of energy to initiate the process, combustion reactions that go to completion will result in the complete oxidation and breakdown of the initial reactant.

Theproduced from the reaction contains all of the carbon in the original reactant. Likewise, theproduced contains all the hydrogen in the original compound. The oxygen content in the original compound is everything left over.

The first step then is to use the mass of each product to find the mass of each element in the original compound. To do this, we'll need to use the molar mass for each of the products as well as the individual elements.

The combined mass of carbon and hydrogen in the original compound is.

The remaining mass will come from oxygen. Thus, the original compound will containof oxygen.

Once we have the mass of each element, we can use that information to calculate the number of moles of each element in the original compound.

Now that we have the relative molar amounts of each element in the original compound, we can find the empirical formula by finding the smallest whole number ratios.

From this, we can conclude that the empirical mass of the compound is.

Finally, since we are given the compound's molecular weight, we can use that information to find the molecular formula for the compound.

Start by writing out the equation that illustrates the plating out of the metal:

Note the stoichiometric ratio for the moles of electrons to the moles of the metal.

Next, recall the following equation:

, whereis the charge,is the current, andis the time.

Plug in the given information to find.

Next, recall Faraday's law of electrolysis that equates Faraday's constant and the amount of electrons together to find the charge needed to deposit one mole of a particular substance.

Thus, we can set up the following to take the charge of the electrolysis through to figure out the number of moles unknown metal that was plated out.

Now, since we have the number of grams of the metal deposited, we can find the molar mass of the unknown metal.

The molar mass indicates that the metal must be titanium.

Recall the following formula:

, whereis the charge in Coulombs,is the current in Amps,is the time in seconds,is the number of electrons, andis Faraday's constant.

Start by finding how many moles of scandium was plated out.

Now, use the stoichiometric ratio present in the half reaction to find the number of moles of electrons.

Now, use the above formula to solve for the time.

Now, convert the seconds to minutes. Remember that your answer should only havesignificant figures.

In this question, we're given an overall redox reaction as well as the relevant reduction potentials, and we're asked to solve for.

To begin with, we can show each of the individual half reactions to make it more clear.

For gold:

For bromine:

Note that the reduction potential for each of these half reactions is positive. This means that for both elements, their reduction is spontaneous. The more positive the voltage (or the less negative), the more spontaneous it is.

From the overall question given to us in the question stem, we see that bromine is not being reduced but, rather, it is being oxidized. Since oxidation is the reverse of reduction, the reduction potential maintains the same magnitude but the sign in front of it changes. Thus, the oxidation of bromine has a value of. Furthermore, because the overall reaction shows gold being reduced, we don't need to change the sign of gold's reduction potential.

To find the overallfor the reaction, we simply just add these two values together.

One very important thing to note is that we did not need to multiply the reduction potential for either of the half reactions. Even though the reduction reaction for gold needs two stoichiometric equivalents, and bromine's oxidation needs three stoichiometric equivalents, the values ofanddo not change. This is because the anyvalue represents an intrinsic property. In other words, the value is not dependent on the amount of material present. As you add more material, it is true that there will be greater electron flow. But at the same time, there will also be more energy change as these electrons flow. The consequence is that both of these values (electron flow and change in energy) change proportionately, such that their ratio will always equal the value ofthat is characteristic of that redox reaction.

In this question, we're given the standard reduction potential of silver and told that it acts as the anode in a voltaic cell. We're then asked to identify a compound that could act as the cathode.

First, there are a few things we need to recognize. For one thing, we're told this is a voltaic cell, which hosts spontaneous redox reactions. This means that energy will be liberated from the reaction and, thus, the overall potential of the cell needs to be positive.

Moreover, the anode is the half-cell where oxidation occurs. In this case, since silver is serving as the anode, it is silver that will be oxidized. The electrons released from this oxidation will travel through the wire to the cathode, liberating energy available to do work in the process. Once at the cathode, the electron will be used to reduce whichever compound is at the cathode.

Since we're given the standard reduction potential of silver and we know the overall reaction must be spontaneous, we can determine what kind of reduction potential the cathode would need to have. Remember that in an electrochemical cell, the cell potential can be expressed as follows.

Moreover, as was mentioned previously, since this is a voltaic cell the reaction will be spontaneous and thus thevalue must be positive.

Plugging in the value forgives us the following.

This shows us that the cathode half-cell needs to have a compound with a standard reduction potential that is greater than silver's. The only answer choice that fits this criteria is cobalt, which has a standard reduction potential of.

For this question, we're given a redox reaction occurring within a voltaic cell. We're also supplied with the reduction potentials for the elements in the reaction, and are asked to find the standard free energy change as well as the equilibrium constant of the reaction at a given temperature.

Recall that in a voltaic cell, the reaction does not consume energy but rather produces it. The anode is where oxidation occurs, while the cathode is where reduction takes place. Based on the reaction given to us in the question stem, the anode will consist of solid, while the cathode will consist of solid. As oxidation occurs in the anode, the electrons liberated from the oxidation of solidwill result in the production of,这将作为一个阳离子存在水溶解tion. Moreover, these electrons will flow spontaneously from the anode to the cathode, releasing energy in the process. When the cathode receives these electrons, thein the aqueous solution will be reduced and thus deposited onto the cathode terminal as solid.

Knowing this, we can calculatefor the reaction.

Now that we havefor the reaction, we can plug this into the Nernst equation to solve for the equilibrium constant.

一个couple things to note. The value forused in the equation above isbecause this is the number of electrons transferred in the balanced reaction. Also, note that the value for this equilibrium expression is huge, meaning that the reaction is driven far to the right, making it spontaneous and capable of releasing a great deal of energy.

To find the value of, there are a few ways we can go about it. We can use the equilibrium value we just obtained, or we can use thevalue.

Using:

Using:

一个s shown above, either method leads us to the same value for. And once again, note how large of a negative value this term is. This is consistent with the large value for the equilibrium constant. The reaction equilibrium lies far to the right and produces an enormous amount of energy.

The oxidation number of each element in the compoundis:

ch的氧化值lorine is. There are 3 chlorine atoms present in the compound, so

Because this is a neutral atom, the overall charge is 0. Therefore we can set

Therefore,

We can then solve for the oxidation number for

Recall that a reducing agent is being oxidized and is losing electrons. Thus, when looking at the equation, look for which oxidation state is becoming more positive.

Start by assigning oxidation states to the elements.

For the reactants:

has an oxidation state of.

has an oxidation state of.

has an oxidation state of

has an oxidation state of.

Now, assign the oxidation state of the products.

has an oxidation state of.

has an oxidation state of.

has an oxidation state of.

has an oxidation state of.

Onlyundergoes a loss of electrons. It must be the reducing agent.

Let's start by looking at the equation and assigning oxidation numbers based off of the general rules we are given:

Let's start by looking at the reactants, namely. Knowing that an atom in its elemental form has an oxidation number of,has an oxidation number equal to.

Next let's look at. This is a neutral compound so know the oxidation number of the whole compound must equal. Going off of our general rules, the oxidation number of fluorine is equal. Since we have 3 fluorines we multiply the oxidation number byto get the cumulative oxidation number of all the fluorines together to get. So now we must solve for the oxidation of Cl, which is unknown. This is done through a simple algebraic equation:

so. Therefore the oxidation number of Chlorine is.

Now we have the oxidation numbers of all the elements present in this equation that are on the reactant side. I recommend writing down their oxidation numbers next to each other, so the elements that are oxidized and reduced can quickly be determined.

Oxidation numbers of elements in reactants:

Now let's look at the product side of the equation and determine the oxidation numbers of the elements there. First let's look at. Knowing the general rules of oxidation numbers we know that fluorine has an oxidation number of. Since there arefluorines we multiplywhich equals. We can now solve for the oxidation number of Uranium.

, thereforehas an oxidation number of.

Finally, let's look. We can once again use the general rule of oxidation numbers that fluorine has aoxidation number to simplify this. Therefore

, therefore Cl has an oxidation number ofin the product.

Let's now write down the oxidation numbers of all the elements in the products.

Oxidation numbers of elements in products:

Now let's compare the oxidation numbers of the elements in the reactants with those in the products.goes from aoxidation number to a, therefore it is the element that's oxidized (oxidation represents a loss in electrons).goes fromto, therefore it is reduced (reduction means gaining electrons). Finally's oxidation number isin both the products and reactants therefore it is neither oxidized nor reduced.This means thatis the only element that's oxidized.

Let's start by going through each answer case-by-case applying the elementary rules of oxidation states we are given. The most applicable of these rules in this problem are the oxidation state of hydrogen in a compound is generally equal to, and the oxidation state of oxygen in a compound is generally equal to.

Let's start by looking at. We know this compound as a whole is neutrally charged (equal tooverall charge), and applying our knowledge of the general oxidation state of hydrogen in a compound and knowing there arehydrogens we know that thehydrogens have a total charge oftogether, and nitrogen has an unknown oxidation state, so for our equation we have:

solving forgives, giving us the answer we need. Therefore the nitrogen atom inhas an oxidation state of,which is the correct answer.

Let's still look at the other cases, however:

一个s this compound is uncharged we know it's net charge is equal to

Using our knowledge of oxidation states of hydrogen and oxygen and counting the number of hydrogens and oxygen in this compound, we can determine that the total charge of all the hydrogens together is equal to, and that the total charge of the one oxygen is.

Just as we did above we can create an equation and solve for our unknown.

therefore, so the oxidation state of nitrogen inis. This means this isn't the right compound.

Next let's look at:

We can quickly apply our knowledge of oxidation states of oxygen knowing that the oxidation state of oxygen in compounds is generally. Since there areoxygens we must multiplyto find out the total charge contributed by all the oxygens, which is.

So our equation for this becomes:

thereforeand the oxidation state of nitrogen is, which isn't the answer we are looking for.

Finally the last compound is:

With this compound we can apply the rule we know about hydrogen's oxidation state in a compound being equal to. Since there arehydrogens, we must multiplywhich equals.

Since there are two nitrogens, our equation is:

and. Therefore the oxidation state of nitrogen is equal to.