A tin ore contains 3.5% SnO2.  How much tin is produced by reducing 2.0 kg of the ore with carbon?
917635
The ore contains 3.5% SnO2, which is 70 gram. The atomic weight of tin is 118.7, while the  molecular weigth of SnO2 is 118.7 + 2*16 = 150.7. Therefore 70 gram of SnO2 contains 70* 118.7/150.7 = 55.1 gram tin.
171957
CO with a -2 charge\nsee below:
866210
2NO+O2--->2NO2\n\nCaO+H2O--->Ca(OH)2\n\nCaCO3--->CaO+CO2
823470
First write the two half reactions:\n\n3 e(-) + Bi(OH)3 ---> Bi\nSnO2(=) ---> snO3(=) + 2 e(-)\n\nMultiply the first equation by 2 and the second by 3 and sum to get:\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi\n\n12 O in the left, 9 on the right, so add 3H2O\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi + 3H2O\n\nCheck that H is OK.  Yes. Done\n\nSo the sum is 13. (d) is the answer.
938225
a compound contaning carbon
391116
2Ca(s) + O2(g) -> 2CaO(s)
825957
That depends on what you mean; Carbon is found in its pure form as graphite (pensil lead) and diamond.  It is a major component of coal, and a fairly large amount of every living organism on the planet.  For experiments using carbon (like production on nanotubes), pure carbon monoxide gas is typically used in a high-energy environment (like high power lasers, plasma gas, etc).  The carbon dioxide (CO) is produced from iron pentacarbonyl (Fe(CO)5).
377149
Nitrogen (N2) 78.084 % by volume\nOxygen(O2)20.9476 % by volume\nArgon(Ar)0.934 % by volume\nCarbon Dioxide(CO2)0.0314 % by volume
520980
nitrogen-80%, oxygen-18%, 2%trace elements
1103330
My thought is air as well. Air is composed of around 2/3s part of Nitrogen. Oxygen is a much smaller amount
502630
The Extraction of Iron \n \n\n\n The solid mixture of haematite ore, coke and limestone is continuously fed into the top of the blast furnace. \nThe coke is ignited at the base and hot air blown in to burn the coke (carbon) to form carbon dioxide in an oxidation reaction (C gains O). \nThe heat energy is needed from this very exothermic reaction to raise the temperature of the blast furnace to over 1000oC to effect the ore reduction. The furnace contents must be he \ncarbon + oxygen ==> carbon dioxide \nC(s) + O2(g) ==> CO2(g) \nat high temperature the carbon dioxide formed, reacts with more coke (carbon) to form carbon monoxide \ncarbon dioxide + carbon ==> carbon monoxide \nCO2(g) + C(s) ==> 2CO(g) \n(note: CO2 reduced by O loss, C is oxidised by O gain) \nThe carbon monoxide is the molecule that actually removes the oxygen from the iron oxide ore. This a reduction reaction (Fe2O3 loses its O, or Fe3+ gains three electrons to form Fe) and the CO is known as the reducing agent (the O remover and gets oxidised in the process). \nThis frees the iron, which is molten at the high blast furnace temperature, and trickles down to the base of the blast furnace. The main reduction reaction is ... \niron(III) oxide + carbon monoxide ==> iron + carbon dioxide \nFe2O3(s) + 3CO(g) ==> 2Fe(l) + 3CO2(g) \nnote, as in the two reactions above, oxidation and reduction always go together! \nOther possible ore reduction reactions are ... \nFe2O3(s) + 3C(g) ==> 2Fe(l) + 3CO(g) \n2Fe2O3(s) + 3C(g) ==> 4Fe(l) + 3CO2(g) \nThe original ore contains acidic mineral impurities such as silica (SiO2, silicon dioxide). These react with the calcium carbonate (limestone) to form a molten slag of e.g. calcium silicate. \ncalcium carbonate + silica ==> calcium silicate + carbon dioxide \nCaCO3 + SiO2 ==> CaSiO3 + CO2 \nthis is sometimes shown in two stages: \nCaCO3 ==> CaO + CO2 \nCaO + SiO2 ==> CaSiO3 \nThe molten slag forms a layer above the more dense molten iron and they can be both separately, and regularly, drained away. The iron is cooled and cast into pig iron ingots OR transferred directly to a steel producing furnace. \nIron from a blast furnace is ok for very hard cast iron objects BUT is too brittle for many applications due to too high a carbon content from the coke. So it is converted into steel alloys for a wide range of uses. \nThe waste slag is used for road construction or filling in quarries which can then be landscaped. \n \n\n\nThe Extraction of Aluminium \n \n\n\n The purified bauxite ore of aluminium oxide is continuously fed in. Cryolite is added to lower the melting point and dissolve the ore. \nIons must be free to move to the electrode connections called the cathode (-, negative), attracting positive ions e.g. Al3+, and the anode (+, positive) which attracts negative ions e.g. O2-. \nWhen the d.c. current is passed through aluminium forms at the negative cathode (metal*) and sinks to the bottom of the tank. \nAt the positive anode, oxygen gas is formed (non-metal*). This is quite a problem. At the high temperature of the electrolysis cell it burns and oxidises away the carbon electrodes to form toxic carbon monoxide or carbon dioxide. So the electrode is regularly replaced and the waste gases dealt with!  \nIt is a costly process (6x more than Fe!) due to the large quantities of expensive electrical energy needed for the process. \n* Two general rules: \nMetals and hydrogen (from positive ions), form at the negative cathode electrode. \nNon-metals (from negative ions), form at the positive anode electrode. \n\n\nThe original extraction of copper from copper ores\n \nfrom copper carbonate ores* ... \nThe ore can be roasted to concentrate the copper as its oxide. \nWater is driven off and the carbonate thermally decomposed. \ncopper(II) carbonate ==> copper oxide + carbon dioxide \nCuCO3(s) ==> CuO(s) + CO2(g) \nThe oxide can be smelted by heating with carbon (coke, charcoal) to reduce the oxid
747941
To provide 3.0 moles of KOH we need 2 L of a 1.5M KOH solution
221298
2.5% copper, 97.5% zinc
664298
molecular: \nHCl + K2CrO4 + SnCl2 --> CrCl3 + SnCl4 + KCl + H2O\n\nnet ionic: H+ + CrO42- + Sn2+ --> Cr3+ + H2O + Sn4+\n\nreduction half reaction: \n3 e- + 3 H+ + 5 HCl + K2CrO4 --> CrCl3 + 2 KCl + 4 H2O\noxidation half reaction:\n2 HCl + SnCl2 --> SnCl4 + 2 H+ + 2 e-\n\nBalancing them:\n6 e- + 6 H+ + 10 HCl + 2 K2CrO4 --> 2 CrCl3 + 4 KCl + 8 H2O\n6 HCl + 3 SnCl2 --> 3 SnCl4 + 6 H+ + 6 e-\n-----------------------------------------------------------\n6 H+ + 16 HCl + 2 K2CrO4 + 3 SnCl2 --> 2 CrCl3 + 4 KCl + 8 H2O + 3 SnCl4
844628
a) 53.4\n34gCO  x 1mol CO/28gCO x 2molCO2/2molCO x44gCO2/1molCO2 = 53.4gCO2\nThe 2 moles CO cancel the 2 moles CO2 (from the balanced equation).
145956
MY ANSWER IS :\n\n18.2g  of Na2S is produced.
507301
5cc in a tspn and 3tspn in a Tbsp so 15cc total.\n30cc in an oz. in case you need to know.