Question
Fill in the following blanks:A) What Group 2 element reacts with water only in the form ofsteam and is the active site in chlorophyll: __________.B) What is name of the Group 2 element (M) whose sulfateMSO4 is highly insoluble in water and which forms themost soluble hydroxide, M(OH)2: __________.C) Which two ions are mainly responsible for hardwater:_________.D) What Group 2 element is fabricated as a window in X-raydevices: __________.
Fill in the following blanks: A) What Group 2 element reacts with water only in the form of steam and is the active site in chlorophyll: __________. B) What is name of the Group 2 element (M) whose sulfate MSO4 is highly insoluble in water and which forms the most soluble hydroxide, M(OH)2: __________. C) Which two ions are mainly responsible for hard water:_________. D) What Group 2 element is fabricated as a window in X-ray devices: __________.
Answers
Predict which of the following compounds are insoluble in water: (a) mercury(I) chloride, $\mathrm{Hg}_{2} \mathrm{Cl}_{2}$ (b) mercury(II) chloride, $\mathrm{HgCl}_{2}$ (c) silver bromide, $\operatorname{AgBr}$ (d) lead(II) iodide, $\mathrm{PbI}_{2}$
The question is which of the following molecules are insoluble in the worker? So we are going to discuss how can we identify insoluble molecules in the water? We have to learn table which is given in that vegetables. So in the question we have as our sio drink C. O. H. Car all boys nickel sulfide and N G B. Article. This is soluble all metals allied are soluble in water. But as A G food we are too will be insoluble in water. So keep learning this concept. Okay. Sulfide of nickel will be also insoluble. Hydroxide of calcium will be soluble and carbonate of this transition will be insoluble. So correct option will be A and steve what? Okay.
To answer this question, you must review the scalability rules. There's really no other way to answer this except knowing the scalability rules. The first compound is potassium nitrate. There's a scalability rule that states all potassium or alkali metal salts are soluble and one that states all nitrates are soluble so this compound would be soluble. Then we have led to chloride. There's a rule that states all chloride salts are soluble but lead to is an exception. So this would be insoluble. Then we have magnesium hydroxide. There's a rule that states all hydroxide are insoluble, so magnesium hydroxide is insoluble, Then we have sodium bromide. There's a rule that states all alkali metals and sodium is one of them. All alkali metal salts are soluble, so sodium bromide is soluble, then we have mercury one chloride. Again, there's a rule that states all chlorides are soluble, but mercury one is actually another exception. So this would be insoluble.
Mhm. Starting with option A. The insoluble products are B. A. Which to me plus late night. Right then we have B. B. And No. Three hold two plus H two. s. And lastly it was, it will be let and I tried plus MG. S. So forth for option B. The answer is barium hydroxide plus magnesia sulfide. And for options see it is very um hydroxide In combination with H two. s. so if you require.
For this problem. We have three equations, and for each equation we're going to name the water soluble product and we're going to write the Net Ionic equation. Let's start with our first equation by assigning states to each one of our products. If I check my scalability rules, I'll see that the copper to sulfide is insoluble water in water. Therefore, it is my precipitate, and hydrochloric acid is an acid, so it's a quiz. The water soluble product in problem one is hydrochloric asset. Now let's right a C I. D. Let's write that net Ionic equation. Look at the HCL, the quickest product, the acquis product yields to hydrogen ions and to chloride ions. That means that our two hydrogen ions and our two chloride ions are going to be our spectator ions. Writing the equation with the spectator ions left out will give us our Net Ionic equation. And here's our Net Ionic equation number one done for a second problem. We're going to sign states, and if we check our Celje bility, rules will see the potassium chloride is soluble in water, and calcium carbonate is not. It's art precipitate, So our water soluble product is potassium chloride. Okay, again, Our water soluble product contains, in this case, too potassium and to chloride ions. So our potassium and our claure are potassium and chlorides are Spectators. Let's rewrite this without the spectator ions and we get the calcium ion, the carbonate ion and our precipitate is calcium carbonate. And we have one more for 1/3 and final, um, reaction. We're going to check our soluble Celje bility rules and silver iodide is insoluble. It's or precipitate, and nitrates are always soluble. So our water soluble product is sodium nitrate. Let's look at our water soluble product, and it consists of one sodium ion and one nitrate ion, one sodium ion and one night trade. I on these are spectator ions. Okay, so we're going to rewrite our equation, excluding the Spectators. And my two ions are the silver ion of the iodide I on. And my product is the precipitate silver iodide. That's it for this problem