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The use of silica to form slag in the production of phosphorus from phosphate rock was introduced by Robert Boyle more than 300 years ago. When fluorapatite $\left[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\right]$ is used in phosphorus production, most of the fluorine atoms appear in the
(a) If 15$\%$ by mass of the fluorine in $100 . \mathrm{kg}$ of $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}$ , F forms
$\mathrm{SiF}_{4},$ what volume of this gas is collected at 1.00 $\mathrm{atm}$ and the industrial furnace temperature of $1450 .^{\circ} \mathrm{C} ?$
(b) In some facilities, the $\mathrm{SiF}_{4}$ is used to produce sodium hexafluorosilicate (Na_ilif_ ) which is sold for water fluoridation: $2 \mathrm{SiF}_{4}(g)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$
$$
\mathrm{Na}_{2} \mathrm{SiF}_{6}(a q)+\mathrm{SiO}_{2}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{HF}(a q)
$$
How many cubic meters of drinking water can be fluoridated to a level of 1.0 ppm of $\mathrm{F}^{-}$ using the $\mathrm{SiF}_{4}$ produced in part ( a )?

Were given this reaction for the conversion of the hydrated form of any to eso for into its anhydrous form. We're also told the value of the equilibrium constant K p at a temperature of 25 degrees Celsius at the same temperature in part a. We wanted a German with the vapor pressure of water is since we have a value for KP, we can write out the equilibrium expression and we see that H 20 vapor is the only gashes species on the product side of this reaction. So that means that KP is equal to the partial pressure of H 20 to the power of 10. And now, since we know the value of K P, we can solve for the partial pressure of H 20 in the system, which will be equal to the vapour pressure at the temperature at which we know the value of K P. So the partial pressure of H 20 is equal to the 10th root of KP, or K P, to the power of 1/10 and that is equal to the 10th root of the given value of K P 4.8 times 10 to the negative 25. And that comes out to a vapor pressure of water in this system, a 25 degree Celsius to be about three 0.64 times 10 do the negative third atmospheres. And now in part B, we want to use the chandeliers principle in order to determine whether each one of the given disturbances to the system will increase, decrease or have no effect on the ratio of the hydrated form to the anhydrous form of any two s 04 So this is the hydrated form and this is the anhydrous form because the hydrated form has those 10 water is bonded to it. So we start with number one and part B. We're increasing the concentration of any to S 04 solid. This is our equilibrium expression. And we know that KP is only dependent on the partial pressure of of each 20 and so that is. That means that the reaction quotient is also only dependent upon that. So the equilibrium will only shift if we change The partial pressure of H 20 since in a two s set of four is a solid. If we add it to the system that will know it, shift the equilibrium and therefore will have no change on that ratio of the hydrated form to the anhydrous form. Number two, we decrease the volume. On the reactant side of this equation, we have zero moles of gas and on the product side, we have one mole of gas. If we decrease the volume, we know that that means that the pressure was increased and so the system will respond by decreasing the pressure. The way that we do that is by shifting to the left to decrease the total number of moles of gas. And so that means that more reactant swill be formed them products and the reactive has the hydrated form and the product side has the anhydrous form. So that means that the ratio of the hydrated form to the anhydrous form will increase. So we increase at ratio when we decrease. Of all you number three, we increase the amount of H 20 gas. So that means that we increase the partial pressure of of H 20 And so when we do that, when we increase that value and research to the power of 10 then the result for the reaction quotient is greater than the equilibrium. Constant. So cute is greater than K, and so that means that the reaction will respond by shifting to the left to reestablish equilibrium. So when that's the case again, just like in number two for Part B, we have more reactant than products, so we have more of the hydrated form of any to S 04 than the anhydrous forms. So that ratio increases again. And number four, we add, and to gas and to does not appear anywhere in a reaction system. So it is an inert gas. Since it is a gas, it will increase the pressure of the system. But to the equilibrium is only affected by the partial pressure of of water. And we did not change that by by adding an inert gas. We only increased the total pressure, so that will have no change on that ratio of the hydrated form to the anhydrous form, because we did not change the equilibrium by adding an inert

Parque loro. While chlorate salt this salt age our band bye elect two lights is of by electrolyte. Take electoral light pink oxidation of oxidation. Uh, a solution official lucian of Hello green of a solution of clothing salt. Sorry, it is chlorate salt, not clothing salt. It is chlorate salt. Therefore, according to the option of Cindy. Each correct.

I'm writing the reaction which is really event to the micro's cosmic salt bead aged. So just look at it carefully and I think that taste the relieving reaction, then see, oh, we'll react with anne B or three, then the product form it. Any issue, be all four. And according to the problem, this is the answer. So option B is only got it answer for this problem.

This time again. I have been given two columns off before and after diastolic blood pressure readings. My null hypothesis is going to be that there is no difference. And the alternative is that there is a difference and Moody is greater than zero again, I simply put these values in many tap into different columns 14 before and one for after and after that, we can get the results directly. And over here I can see that the T value that I get is 1.18 and the P value is 0.138 And if I want to do this by hand, what is going to be the formula that s t is equal to X'd bar? What is X'd bar? It is the difference of the means off before and after. In this case, this is one minus moody. Mute is what the difference of hypothesized mean, which is zero. So this term is zero upon upon the standard at it. And what a standard letter it is s d by route N S. D by route And what is SD? It is a standard deviation of the differences, which is 2.390 So s D by route and and N is Aidan Ah, Keys. So from here also, after substituting the values, the value that I get 40 is 1.18 So my teeth or the stick from this formula also is going to come out close to 1.18 and degree of freedom is at minus one, so degree of freedom will be seven and again with the help of these two also, if I calculate the P value manually, it is going to come out close to zero point 138 Now what does this mean? At 99% confidence or at Alfa is equal to 0.1 al physical, 0.1 Sorry. I can see that my P value is much greater than Alfa. Hence, I can say that I do not have enough statistical evidence to reject the null hypothesis, which means that there is no difference in before and after according to this test. So we fail to reject the null hypothesis we fail to reject. It's not


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