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Question 2 (4 points) Write the equilibrium constant expression for the following reaction: Fe(OH):(s) + 3H+(aq) = Fe3t (aq) + HzO() K = 9.1 and compute the concent...

Question

Question 2 (4 points) Write the equilibrium constant expression for the following reaction: Fe(OH):(s) + 3H+(aq) = Fe3t (aq) + HzO() K = 9.1 and compute the concentration for Fe3+at pH-4.9.1X 10-49.1X 10-99.1x 10-109.1 X 10-1

Question 2 (4 points) Write the equilibrium constant expression for the following reaction: Fe(OH):(s) + 3H+(aq) = Fe3t (aq) + HzO() K = 9.1 and compute the concentration for Fe3+at pH-4. 9.1X 10-4 9.1X 10-9 9.1x 10-10 9.1 X 10-1



Answers

At $450^{\circ} \mathrm{C}$ , the value of the equilibrium constant for the following system is $6.59 \times 10^{-3} .$ If $\left[\mathrm{NH}_{3}\right]=1.23 \times 10^{-4} \mathrm{M}$ and $\left[\mathrm{H}_{2}\right]=2.75 \times 10^{-2} \mathrm{M}$ at equilibrium, determine the concentration of $\mathrm{N}_{2}$ at that point. $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)$

The first thing to do in this question is to write out your K equation. So here we have Kay is equal to concentration of N H three squared over the concentration of and two times the concentration of H two cubed. The question tells us that the cave value here is equal to 6.59 times 10 to the negative third, which we then no is equal to our concentration of any each three at equilibrium, which is 1.23 times 10 to the negative. Fourth squared over your concentration of end to which is the unknown times 2.75 times 10 to the negative. Third to the third. When you solve this math, you get concentration of end too is equal to 110 Moeller.

If we have this system in equilibrium and to four going to two moles and n 02 gas, the equilibrium expression would be the reactor. I'm sorry, the product No. Two raised to the second power because it has a coefficient of two divided by the concentration of n 204 This expression this ratio has a value equal to the equilibrium constant that is provided at 8.1 times 10 to the negative three. So if the concentration at equilibrium of n 204 is 5.4 times 10 to the negative four, we simply plug this value into the equilibrium, expression and solve for the concentration of co two. To do this will multiply both sides by 5.4 times 10 to the negative four and then take the square root of both sides to get the concentration of just know too. So n 02 concentration will be the square root of 4.37 times 10 to the negative six or 2.1 times 10 to the negative three molar

We will be doing a calculation to buy the concentration of the a. two plus and This one and see you any tree for that are pleasant that are present at equilibrium After dissolving five g of copper chloride. Got particular ride in one leader of 10.1 Mueller. Come on. Yeah. And let's get going our first order of business. Well to be to figure out our concentration of Yeah. Which one did I get here? five g. I was given 5.00 g of CUCL two. So let's get that 12 moles first off And I looked up the UCL to its 1:35 .45g per mole. But she gives us 3.69 times 10 to the -2 malls. And that's per liter because we have a leader of solution. Um What else? Let's draw. Let's write each of our equations here. Then we combine these two equations. Here's what we get. I just like to have my equations written out and we're also going to use our KF for For this one and that is five times 10 to the 12. And let's write our KF expression With the 4th. There. There we go. And let's figure out what we have here. I believe I was given that my concentration is .1 zero ammonia. So I'm given the following information on this. I've already calculated that my that this is the concentration of my um see you and all three for two plus will be since all of this is going to turn into our complex 3.69 times 10 to the -2. That's one of the things I know I was given that the concentration of my ammonia is 0.10 this was given. And what I'm trying to figure out is my concentration of my copper two plus. So I'm going to set this equation. I think I'll go to the next page when I saw this, I got 7.38 times 10 to the minus 11th And the one sick pig pig. This rounds to seven times 10 To the -11, vulgarity. And that is my concentration of Which one is this? Here. That is my concentration of See you two plus. Let me see if I've got this right now, That's my concentration of c. two plus. Let me see what else they want us to figure out here. I can't find my problem there. It is. Um concentration of that is the two plus and the concentration of the of the complex ion I have the the complex ion I had 3.69 Times 10 to the -2 malls. I believe. That's correct. I think that's it.

So for this question were given the following reversible reaction were also given that the concentration of each too is equal to 2.44 times 10 to the negative three at equilibrium. That I, too, is 7.18 times 10 to the negative five Mueller equilibrium and were given the cake you for the given temperature and were asked to find the concentration of theory reactant sat equilibrium. So the first thing I'm going to do is I'm going to actually write out the K e Q expression, which is that cake you is equal to the concentrations of the products raised to their coefficients over the concentration of the reactant raised to its coefficient. And so once I'm given this equation, I can just plug in the variables and use algebra to solve for my unknown. So I'm gonna say that the cake you, which is 1.67 times 10 to the negative too, is equal to the concentration of H two, which is 2.44 times 10 to the negative three times 7.18 times 10 to the negative five. And that is over my unknown. And if I rearrange these. I can say that my each eye is equal to over my cake you. And then because my age I was squared on one side. I have to take the square root of that. And if you plug that into a calculator, you're going to get that The concentration of H I is equal to 3.2 times 10 to the negative three Moeller at equilibrium.


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