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1. Molarity = concentration and it expressed as mol per dm^3 Change the mass to mole and the 851 ml to dm^3. Then find the ratio of the number of moles to volume in dm^3 2. Soluble ionic compounds breaks up the water molecules in H^+ ions and OH^- ions. Sodium carbonate give Na^+ ions and CO3^2- ions......Other ions that are present are HCO3^- ions called hydrogen carbonate ions. I have found that a solution of sodium carbonate is alkaline: implying that there are more OH^- ions to H^+ ions, due to the formation of HCO^- ions. 3. OH^- (aq) + H^+(aq) ----> H2O (l) 4. H^+ ions (ie. hydrogen ions), OH^- ions (ie. hydroxyl ions), Ca^2+ calcium ions and Cl^- chloride ions 5. Hydrogen ions, hydroxyl ions. ammonium ions and sulphate ions. 6. Actually the calcium metal reacts with water to give hydrogen gas and calcium hydroxide. At the same time, calcium reacts with nitric acid to give calcium nitrate, water and brown nitrogen dioxide gas. Hence there is no simple ionic equation for your question You may want to consider this one: Ca^0 + 2H^+ + 2OH^- -----> H2(g) + Ca^2+ + 2OH^- 7. Hydrogen ions, hydroxyl ions, iron (III) ions Fe^3+ and sulphate ion SO4^2-
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1. heat lost by the metal = heat gained by the water metal's m C dT = water's m C dT (136.9854 g)(C)(101.1C- 35.15) = (46.90g)(4.184J/g-C)(35.15 - 29.25) (136.9854 g)(C)(65.95C) = (46.90g)(4.184J/g-C)(5.9) 9034.18 C = 1157.75 C = 0.128 a) The Specific Heat of the metal: your answer(2 sigfigs): 0.13 J/g-C (to have water's change in temp only 2 sigfigs @ 5.9C, may have been an oversight, but it really rounds off your answer) ======================================... Calculated using the rule of Dulong and Petit http://www.stetson.edu/~wgrubbs/datadriv... (specific heat) (Molar mass)= 25 J/mol-K Molar mass = 25 / 0.128 = your answer: Atomic Mass = 195 g/mole(again it is only good for 2 sigfigs though) ======================================... ======================================... I. To a calorimeter containing 44.60 g dH = m C dT dH = 44.60g (4.184 J/g-C)(25.90 - 21.00 ) dH = 44.60g (4.184 J/g-C)(4.90) dH = 914.37 J lost by the water , gained by the dissolution your answer: Heat of solution = + 94.3 j/g-C ======================== find the Heat of solution per mole: 88.44grams @ 914.37 J/g-C =80,867 joules/g-C .....(aka 80.87kJ/g-C) but given this: "Given that the Heat of Formation of the solid salt is.... -110.30 Kj/mol Calculate: d) The Heat of Formation of the Solution ______________ Kj/mol" it is a strange question... it sounds like they don't want the heat of solution per mole... but rather someting like dHreaction = dHf prod - reactnats dH rxn = (Heat of Formation of the Solution) - (Heat of Formation, solid) 80.87kJ/g-C = (Heat of Formation of the Solution) - (-110.30 Kj/mol) 80.87kJ/g-C = (Heat of Formation of the Solution) + 110.30 Kj/mol (Heat of Formation of the Solution) = 80.87kJ/g-C - 110.30 Kj/mol (Heat of Formation of the Solution) = - 29.43 Kj/mol i haven't seen this asked before, I have only seen them ask for the "Heat of solution" whatever
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Based on the reaction 2Mg(s)+O2(g)=2MgO(s) Write the three chemical equations (including phase subscripts) that demonstrate how nitrogen is eliminated from your product. These include the reaction between magnesium and nitrogen, treatment of that product with water, and heating of that new product. Use this series of reactions to explain why the side reaction of magnesium with nitrogen does not interfere with the results
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1.) 520.0 cm3 = 520.0 mL 520.0mL x 1dL/10mL = 52.00 dL to four significant figures 2.) 1 in. = 2.54 cm; 1 in3 = 16.4 cm3 0.025lb/1in3 x 454g/1lb x 1in3/16.4cm3 = 0.70g/cm3 to two significant figures 3.) Let the soda be called S. 55mgNa/355mLS x 1gNa/1000Namg x 1lbNa/454gNa x 1000mLS/1LS = 0.00034 lb Na to two significant figures ( = 3.4 x 10^-4) Before pulling out your calculator, you would cancel the two 1000's top and bottom 4.) 3.00x10^8m/1sec x 1km/1x10^3m x 0.63mi/1km x 60sec/1min x 60min/1hr = 6.80 x 10^8 mi/hr The rest go on pretty much the same as the ones I've done. 7.) There is a neat way to interconvert F and C temperatures based on the fact that -40C = -40F. 210K = 210 - 273 = -63C. First add 40 to the temp to be converted: -63 + 40 = -23 Going from F to C, multiply by 5/9. Going from C to F, multiply by 9/5: -23 x 9/5 = -41.4 Then subtract the 40 again: -41.4 - 40 = -81.4
Would chemistry help with cooking?
My degree is in Food Chemistry, and while it did give me a great understanding of food manufacturing, the components of food and how they interact as well as many other things, I would say it only marginally improved my cooking. In your case, you would learn about the chemical mechanisms of baking, what reacts to make bread turn brown and things like that. That wouldn't necessarily translate to you being an excellent baker though. If baking is your passion, you might consider being trained as a pastry chef later on. Edit: To address your concern about whether you should take biology instead, I think if you prefer that then you should go with it. Neither subject is going to have a huge effect on whether you become a great cook and you will need special training if you want to improve, irrespective of the one you choose. One more thing: before selecting one, check out the requirements to be trained as a baker or pastry chef to make sure that a background in chemistry is not one of them (I highly doubt that is the case though).
I need help with my chemistry project. I need an item i can find at home or at a store.?
1. 6.02 x 1023 of anything. Calculations required. 2. A substance with a density less than 1. Calculations required. (no work, no credit, no kidding) 3. A chemistry cartoon. Summarize what it means or what it a satire of. 4. An acid base indicator from nature other than one you have previously made for this class. The indicator can not be a substance sold as an indicator. State the source of the indicator, what color the indicator is with an acid and what color it is with a base. 5. A weak Arrhenius acid. Define Arrhenius acid. State the common and chemical names and chemical formula. 6. A weak Arrhenius base. Define Arrhenius base. State the common and chemical names and chemical formula. 7. A physical equilibrium system containing two or more phases. Define equilibrium. Identify the common name phase of the substances at equilibrium. 8. A chemical equilibrium system containing at least two phases. Define equilibrium. Write a balanced equation representing the equilibrium. 9. A covalently bonded substance. Chemical formulas, common names and chemical names required. 10. A non-polar molecule. Common name, chemical name, chemical formula required.
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1. The balanced equation is: 2 H2 (g) + O2 (g) → 2 H2O (g) The number of moles of H2 that is consumed by the reaction is equal to the number of moles of H2O that is produced (2:2 stoichiometry). The molecular weight of H2O is 18 g/mole and H2 is 2 g/mole. The number of moles of H2 needed is: (60 g) (2 g/mole H2) / (18 g/mole H2O) = 6.7 g of H2 2. Na2CO3 (aq) + CaSO4 (aq) → CaCO3 (s) + Na2SO4 (aq) Molecular weight of CaSO4 = 136.1 g/mole Molecular weight of Na2CO3 = 106.0 g/mole The mass of Na2CO3 needed is (1:1 stoichiometry) (0.68 g) (106.0 g Na2CO3) / (136.1 g CaSO4) = 0.53 g of Na2CO3 3. Cu (s) + 2 AgNO3 (aq) → 2 Ag (s) + Cu(NO3)2 (aq) Atomic mass Ag = 107.9 g/mole; Atomic mass Cu = 63.5 g/mole. The stoichiometry of the reaction is 2:1, so the mass of Cu needed is: (72 g Ag) (63.5 g/mole Cu) / (2) (107.9 g/mole Ag) = 21.2 g Cu.
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Rb+, Na+, Mg2+
I need free chemistry help online, is that possible?
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I need chemistry help ASAP on Molecular Mass!?
The roman numerals show the charge or electronic state of the ions. when you do mass calculations, just ignore this. The mass of an atom is given by the protons and neutrons which are in the nucleus. The charge refers to the number of electrons in the outermost shell, the electron mass is really really small, that is why we don't really care if it is Fe or Fe3+ or Fe2+, their masses will all be the same.