A 5.0 gram sample of Cu(NO3) 2 . n H2O is heated, and 3.9 gram sample of the anhydrous salt remain.what is n?
445882
the formula mass of Cu(NO3)2 is 187.55g/mol\n\nthe mass of anhydrous salt is 3.9g therefore the number of moles of Cu(NO3)2 is 0.02079mols.\n\nThe mass of water lost was 1.1g, as the formula mass of water is 18.02g/mol then 0.0610mols of water were released.\n\n0.0610/0.02079=2.9\n\nSo the answer is 3.
706927
a. Silver nitrate\nb. isopropanol? methanol?\nc. Potassium chloride\nd. Benzene\ne. Carbon monoxide\nf. Sodium hydroxide
822606
CuSO4 + NH4S = CuS + NH4SO4\n\n1)Write a rough equation without balancing anything. The 2 after copper, which I presume is to be written as copper (II), indicates that copper has a charge of +2. It is a transition metal and can have a differing charge.\n\n2)Because ammonium has a charge of +1 and the sulfate has a charge of -2, there needs to be 2 ammoniums. SAme thing is for the left side. The ammonium is +1 and the sulfide is -2. There needs to be two ammoniums on the left side. \nCuSO4 + 2NH4S = CuS + 2NH4SO4\n\n3) Because there are two sulfides on the left side, there needs to be two sulfides on the right side. \n\nCuSO4 + 2NH4S = 2CuS + 2NH4SO4\n\n4) Now there are two sulfates on the right side, but not on the left. So we place a two in front of copper (II) sulfate.\n\n2CuSO4 + 2NH4S = 2CuS + 2NH4SO4\n\n5) I believe that we are done with balancing so let's check our work.\n\n6) 2 coppers are on the left and right side.\n2 sulfates are on the left and right side.\n2 ammoniums are on the left and right side. \n2 sulfides are on the left and right side.\n\n7) In most cases, not all compounds will have the same number on each side. It's just a coincidence that there are all 2 of each. If you need anymore help, email me ^_^
117425
Silver nitrate. \nThe Ag(+) is silver, and the NO3(-) is nitrate. \nDon't get this stuff on your hands; it stains!
1080354
NaCl IS A MOLECULE CALLED TABLE SALT!\nNH3 is not a molecule because the nitrogen atom is not bonded to another atom/atoms using all of its valence electrons and is therefore not a molecule, it is an ion.
698205
Zinc. \nBut your units are wrong. It should be Fahrenheit not Celsius.
38819
Transition metal.\n\nName: Copper \nSymbol: Cu \nAtomic Number: 29 \nAtomic Mass: 63.546 amu \nMelting Point: 1083.0 °C (1356.15 K, 1981.4 °F) \nBoiling Point: 2567.0 °C (2840.15 K, 4652.6 °F) \nNumber of Protons/Electrons: 29 \nNumber of Neutrons: 35 \nClassification: Transition Metal \nCrystal Structure: Cubic \nDensity @ 293 K: 8.96 g/cm3 \nColor: red/orange
513498
salt.\n\n(Sodium Chloride)
834399
Your answer depends on the nature of your problem, so you'll want to give more info if you want a complete answer.  The specific heat capacity you use depends on which substance is being heated up or cooled off.  The mass will be the mass of the substance that is being heated up or cooled off.\n\nOne exception that frequently occurs is when you are dissolving a substance in water.  Most problems will assume that the solution has the density and specific heat capacity of pure water.  So in other words, if you dissolve 3 grams of ammonium chloride in 100 mL of water and watch the temperature go down, most instructors would want you to use a mass of 100 grams (instead of 103) and the specific heat capacity of water 1.00 cal/g C or 4.184 J/g C if you're working in joules.\n\nHope this helps...this is the best I can do without more info about the problem.
617322
Althouigh I am sure you mean Sodium Chloride when you say "Salt", just to be perfectly clear....there are multiple types of "salt".\nA salt is an ionic compound made up of a matal cation and a non-metal anion. Sodium Chloride ("table salt") is an example of a salt, so is Potassium Chloride and Ammonium Nitrate.\n\nWhen you dissolve a salt in water, the freezing point of the solution is lowered.  Likewise, when dissolving a salt, the boiling point of the solution is raised.\n\nBut you asked about a temperature change.\nWhen dissolving some salts, the dissolving process is endothermic...meaning that it requires enegry input to proceed.  When a endothermic process occurs, the temperature of the surroundings (this case, the water) drops.\nThere are also salts which give off energy when they are dissolved in water.  Processes which give off excess energy are called exothermic and can raise the temperature of the surroundings.\nThere are also a few (not too many) "ideal" salts which neither take in or give off energy when they dissolve.\n\nWhen the salt dissolves, if the "product" (the aqueous salt) has a lower enthalpy than the undissolved salt, the solution is said to be exothermic.  If the enthalpy increases, the solution is said to be endothermic.\n\nSalts like Potassium Chloride, Ammonium Nitrate, and several others (probably NaCl too, which you refered to) are endothermic when dissolved in water, and thus, when dissolved cause the temperature of the solution to decrease slighty.\n\nSalts like Sodium or Potassium Hydroxide are great examples of salts which are exothermic (give off energy) when they are dissolved in water.  When dissolving these salts in water, a notible increase in temperature can be observed.
596804
3Cu+8HNO3=3Cu(NO3)2+2NO+4H2O
857898
2Al + 3*(CuO) -> Al2O3 + 3*Cu\n\nI2 + 2*(Na2S2O3) -> 2*(NaI) + Na2S4O6\n\nI2 + 5*(Cl2) -> 2*(ICl5)
707894
Cu 2+ ClO3 1-\n\nCu(ClO3)2
1029892
Na2S (aq) + 2H2O (l) = 2NaOH + H2S (g)
903130
Na2SO4(aq)+ 2 AgNO3(aq)-->Ag2SO4(s)+ 2 NaNO3(aq).
471526
I suspect that the product is AlSO4.\n\nI'm thinking that the Al swaps with the Cu.  That would leave a free Cu+ which might displace the H from H20 and form either CuOH or CuO, either one of which would probably produce H2.  \n\nSo, I guess you would first get CuOH and AlOH, as intermediate reactions.\n\nI think the equation is:\n\n                          NaCl (goes over the arrow)\n6CuSO4 + 4Al + 12H2O ---------------> 2Al2(SO4)3 + 6Cu(OH)2 + 6H2
1087032
try Vogel's "Qualitative Inorganic Analysis", I'm pretty sure it has all the answers 4 u.\n\ndont care editions, but for 7th ed, ISBN is 0582218667\n\n****\nokey then, from Vogels\n\n1) when heated, lead nitrate decomposes to white solid lead oxide (PbO), brown gas (NO2) and oxygen gas. Brown gas decolorizes when contact with water. \n\n2) ammonia odur. ammonium compounds are all volatile towards heat and decomposes to ammonia and an acid but quickly recombine if the acid, in this case, hydrochloric acid, is a strong acid.\n\n3) anhydrate copper sulphate is white, hydrate copper sulphate is blue. I have no idea which one you actually heated. If it's anhydrate, no change; if hydrate, blue ---> white.\n\n4) blue precipitate forms initally, but resolve upon excess reagent, clear blue soln results due to the formation of copper complex [blue precipitate: Cu(OH)2*CuSO4, complex: [Cu(NH3)4]^2+]\n\n5) brown precipitates froms [Cu(OH)2], no change with excess reagents.\n\n6) metal-gray stipe (magnesium stipe) gains a redish brown color due to formation of copper metal. Possible decolorization of the soln (depends on how much magnesium u actually put in)\n\n7) volatile reaction, solid metal solves,ignitable gas is formed. (if phenolphthalein is added to water initally, soln becomes pink)