Chemistry Problem Solving | Site Berezan school chemistry teacher Vinokourov TK Ok'XIMI4KA
Home Page for the guests. Write your impressions and suggestions. Materials for lessons in grade 9 classes Interesting quiz materials for lessons in 8th grade chemistry lessons heat capacity table in grade 10 chemistry lessons in 7th grade when classes heat capacity table ended, I congratulate you on your site "Ok'XIMI4KA" class teacher Super-ximikus
"The final lesson on the topic of solutions, and Lomonosov Books Actions Comments RSS trackback Information Date: September 28, 2011 Categories: Teaching Tips 243 responses June 11, 2011
Interaction with 100ml rastvora sulfate ferrous metal with yzbыtkom heat capacity table rastvora heat capacity table sulfyda sodium vыpal sediment, kotorыy not rastvoryaetsya in hlorydnoy kysloty.vydfyltrovanyy and vыsushennыy It would prokalennoho sediment oxygen in the atmosphere at etom obrazovalos 9.6 g of the Black powder and 2.24 liters of SO2. sulfate kotorogo heat capacity table metal bыl in rastvora heat capacity table initially, after calcination If no ego valence yzmenylas ???
Feature - a very long time to write. Take your notebook for 8th grade and find the plan that is characterized by an element. Regarding the second problem nonpolar covalent called * Connection with oxygen and nitrogen, polar covalent - water, hydrogen fluoride and ammonia. Those remaining are called ionic * Connection. Valence and degree of oxidation seek the textbook for Grade 8, last paragraph. And by the way, you seem to have forgotten to write "Please." Notice I did not folk symbol obliges you to answer.
Marina (21:30:59):
Help please Calculate the mass of nitric acid, which can be prepared from sodium nitrate masoyu17 d as a result of its interaction with kontsetrovanoyu heat capacity table sulfuric acid if the mass fraction of output nitric acid is 96%
2NaNO3 + H2SO4 = 2HNO3 + Na2SO4 M (NaNO3) = 85 g / mol, M (HNO3) = 63 g / mol. 17 g of sodium nitrate - a 0.2 mol. Now look at the reaction equation: heat capacity table 2 mol sodium nitrate obtained 2 M nitric acid (I think you know that all the coefficients in the equation). So 0.2 mol sodium nitrate form is also 0.2 mol of nitric acid is 0,2h63 = 12.6 g That is the weight of nitric acid could be obtained at the output 100%. But by the problem heat capacity table of the mass fraction of the output is 96%. So practically heat capacity table we get a 12,6h0,96 = 12.096 g of nitric acid.
March 10, 2014
In the periodic table is the general formula of higher oxides. For the sixth group is EO3. The relative atomic mass of oxygen 16. The formula of three, heat capacity table so that the Oxygen accounts for 3h16 = 48. By the condition it is known that the relative molecular mass of the compound heat capacity table is 6D. Then the relative atomic mass of an unknown element must be 60-48 = 12. But such an element in the sixth group, no! Rather, the error in the problem, so you never managed to loose this task.
6.3 g of nitric acid is 0.1 mol. Putting equation: Na2CO3 + 2HNO3 = 2NaNO3 + H2O + CO2 From the equation we see that the reaction came 2 M nitric acid and thus created on 1 mol of carbon dioxide (which show the coefficients in the equation of reaction). And if by nitratnoy heat capacity table acid was 0.1 mol? How many moles of CO2 formed? heat capacity table If you guessed - well done. 0.05 mol. What is this about * yemom can be determined by the formula V = nxVm where n - number of agents CO2, and Vm - molar volume v *, it is the same for all gases and is equal to 22.4 liters / mole.
Hello Ian. Let's look at the condition of your problem. At the end it says: ... from combustion of oxygen. That can not be. These other substances burn in oxygen. Oxygen also can burn only one single substance - in fluoride. So, first edit the problem.
Answer September 30, 2013
Help: Relative vapor density substance is hydrogen burning 805.Pry it a certain amount of substance formed 0.54 g. water and 0,224l. azotu.Vkazaty molecular formulu.Sklasty corresponding reaction equation.
Yuri, mark the relative atomic heat capacity table mass metal x metal molecular weight 32 + x and put the proportion usually do in a chemical equation problems (which, if known divalent metal Mg + S = MeS. Then do the same for the case of monovalent and trivalent metal with equations, respectively: heat capacity table 2Me + S = Me2S i 2Me + 3S = Me2S3. You will be only one suitable response in case of trivalent metal - 27. Metal with the relative heat capacity table atomic mass is, this AI.
First, multiply heat capacity table the amount of the substance to Avogadro's number and you will find the number of calcium ions, and results heat capacity table domnozhymo heat capacity table 2 and find out the number of chlorine ions, as in the formula for each ion of calcium ions account heat capacity table for 2 chlorine.
Vladimir, help reshyt one of two problems. FeS2 -> 2SO2 M (FeS2) = 120 g / mol; Vm (SO2) = 22,4 l / mol, respectively Proportsyya: x 120 = 26.88, 44.8 x 72 = December pyrite. Togda mass of impurities 79,2-72 = 7.2 oz. Avto to define Content impurities in%, 7,2: 79.2 = 0.09 ili 9%.
Good, pomohayu. BaO + CO2 = VaSO3 (obrazuetsja bars carbonate salt). Avto to define heat capacity table Massimo heat capacity table uh, need it to know, How reahyruyuschyh of substances before heat capacity table a yzbыtke.
Home Page for the guests. Write your impressions and suggestions. Materials for lessons in grade 9 classes Interesting quiz materials for lessons in 8th grade chemistry lessons heat capacity table in grade 10 chemistry lessons in 7th grade when classes heat capacity table ended, I congratulate you on your site "Ok'XIMI4KA" class teacher Super-ximikus
"The final lesson on the topic of solutions, and Lomonosov Books Actions Comments RSS trackback Information Date: September 28, 2011 Categories: Teaching Tips 243 responses June 11, 2011
Interaction with 100ml rastvora sulfate ferrous metal with yzbыtkom heat capacity table rastvora heat capacity table sulfyda sodium vыpal sediment, kotorыy not rastvoryaetsya in hlorydnoy kysloty.vydfyltrovanyy and vыsushennыy It would prokalennoho sediment oxygen in the atmosphere at etom obrazovalos 9.6 g of the Black powder and 2.24 liters of SO2. sulfate kotorogo heat capacity table metal bыl in rastvora heat capacity table initially, after calcination If no ego valence yzmenylas ???
Feature - a very long time to write. Take your notebook for 8th grade and find the plan that is characterized by an element. Regarding the second problem nonpolar covalent called * Connection with oxygen and nitrogen, polar covalent - water, hydrogen fluoride and ammonia. Those remaining are called ionic * Connection. Valence and degree of oxidation seek the textbook for Grade 8, last paragraph. And by the way, you seem to have forgotten to write "Please." Notice I did not folk symbol obliges you to answer.
Marina (21:30:59):
Help please Calculate the mass of nitric acid, which can be prepared from sodium nitrate masoyu17 d as a result of its interaction with kontsetrovanoyu heat capacity table sulfuric acid if the mass fraction of output nitric acid is 96%
2NaNO3 + H2SO4 = 2HNO3 + Na2SO4 M (NaNO3) = 85 g / mol, M (HNO3) = 63 g / mol. 17 g of sodium nitrate - a 0.2 mol. Now look at the reaction equation: heat capacity table 2 mol sodium nitrate obtained 2 M nitric acid (I think you know that all the coefficients in the equation). So 0.2 mol sodium nitrate form is also 0.2 mol of nitric acid is 0,2h63 = 12.6 g That is the weight of nitric acid could be obtained at the output 100%. But by the problem heat capacity table of the mass fraction of the output is 96%. So practically heat capacity table we get a 12,6h0,96 = 12.096 g of nitric acid.
March 10, 2014
In the periodic table is the general formula of higher oxides. For the sixth group is EO3. The relative atomic mass of oxygen 16. The formula of three, heat capacity table so that the Oxygen accounts for 3h16 = 48. By the condition it is known that the relative molecular mass of the compound heat capacity table is 6D. Then the relative atomic mass of an unknown element must be 60-48 = 12. But such an element in the sixth group, no! Rather, the error in the problem, so you never managed to loose this task.
6.3 g of nitric acid is 0.1 mol. Putting equation: Na2CO3 + 2HNO3 = 2NaNO3 + H2O + CO2 From the equation we see that the reaction came 2 M nitric acid and thus created on 1 mol of carbon dioxide (which show the coefficients in the equation of reaction). And if by nitratnoy heat capacity table acid was 0.1 mol? How many moles of CO2 formed? heat capacity table If you guessed - well done. 0.05 mol. What is this about * yemom can be determined by the formula V = nxVm where n - number of agents CO2, and Vm - molar volume v *, it is the same for all gases and is equal to 22.4 liters / mole.
Hello Ian. Let's look at the condition of your problem. At the end it says: ... from combustion of oxygen. That can not be. These other substances burn in oxygen. Oxygen also can burn only one single substance - in fluoride. So, first edit the problem.
Answer September 30, 2013
Help: Relative vapor density substance is hydrogen burning 805.Pry it a certain amount of substance formed 0.54 g. water and 0,224l. azotu.Vkazaty molecular formulu.Sklasty corresponding reaction equation.
Yuri, mark the relative atomic heat capacity table mass metal x metal molecular weight 32 + x and put the proportion usually do in a chemical equation problems (which, if known divalent metal Mg + S = MeS. Then do the same for the case of monovalent and trivalent metal with equations, respectively: heat capacity table 2Me + S = Me2S i 2Me + 3S = Me2S3. You will be only one suitable response in case of trivalent metal - 27. Metal with the relative heat capacity table atomic mass is, this AI.
First, multiply heat capacity table the amount of the substance to Avogadro's number and you will find the number of calcium ions, and results heat capacity table domnozhymo heat capacity table 2 and find out the number of chlorine ions, as in the formula for each ion of calcium ions account heat capacity table for 2 chlorine.
Vladimir, help reshyt one of two problems. FeS2 -> 2SO2 M (FeS2) = 120 g / mol; Vm (SO2) = 22,4 l / mol, respectively Proportsyya: x 120 = 26.88, 44.8 x 72 = December pyrite. Togda mass of impurities 79,2-72 = 7.2 oz. Avto to define Content impurities in%, 7,2: 79.2 = 0.09 ili 9%.
Good, pomohayu. BaO + CO2 = VaSO3 (obrazuetsja bars carbonate salt). Avto to define heat capacity table Massimo heat capacity table uh, need it to know, How reahyruyuschyh of substances before heat capacity table a yzbыtke.
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