Dec 6, 2015 · 2H_2+O_2rarr2H_2O H_2+O_2rarrH_2O Notice the imbalance of atoms. On the left side, the reactant side, there are 2 hydrogen atoms and 2 oxygen atoms. On the right side, the product side, there are 2 hydrogen atoms and only 1 oxygen atom. An important thing to remember when balancing equations is that the molecules themselves may not be …

Dec 24, 2013 · Bond polarity refers to the separation of charge within a bond. Bonds between atoms of different electronegativity are polar. The more electronegative atom has a higher density of bonding electrons around it, giving it a partial negative charge (δ⁻). The less electronegative atom has some of its electron density taken away, giving it a partial positive charge (δ⁺). In …

Nov 16, 2015 · 2Na + 2H_2O -> 2NaOH +H_2 You have to count the number of atoms in both sides of the reaction to be equal, to make it a little easier you can make a list. So in the equation Na+H_2O->NaOH+H_2, there are 1-Na-1 2-H-3 1-O-1 So you multiply NaOH and H2O by 2, so you have 4 hydrogens in both sides, and then multiply Na by 2 to be equal to the NaOH you …

Oct 23, 2015 · From the original equation, you first need to write each component separately. #H_2 + O_2 = H_2O# Left side: H = 2; O = 2 (number based on the subscript)

Jul 15, 2014 · You can find a procedure for drawing Lewis structures at this location. For H₂O, O must be the central atom The skeleton structure is H-O-H. O has 6 valence electrons, and each H has one. You must arrange 8 electrons in pairs so that O has 8 and each H has two electrons in its valence shell. The trial structure is You have eight valence electrons in your trial structure, …

Mar 5, 2017 · 2SO_2 + O_2 + 2H_2O -> 2H_2SO_4 Start with the skeleton equation: SO_2 + O_2 + H_2O -> H_2SO_4 Atoms on the left: 1-S, 5-O, 2-H Atoms on the right: 1-S, 4-O, 2-H To make the number of oxygen on the left to be even, add 2 molecules to H_2O and a 2 to the right. SO_2 + O_2 + 2H_2O -> 2H_2SO_4 Atoms on the left: 1-S, 6-O, 4-H Atoms on the right: 2-S, …

May 9, 2018 · Hydrogen is formally reduced, and is thus the "oxidizing agent.." And so ....underbrace(H_2O(l) +e^(-) rarr 1/2H_2(g)uarr + HO^(-))_"reduction half equation" And natrium metal is oxidized....i.e. the metal is the "REDUCING agent..." underbrace(Na(s) rarr Na^+ + e^(-))_"oxidation half equation" And we simply add one t'other to remove the electrons.... Na(s) + …

Jun 18, 2018 · Well the simplest answer is..."because God wanted in that way.." As you know, the normal boiling points of water, 100 ""^@C, and ammonia, -33.3 ""^@C..are EXCESSIVELY HIGH....especially given these are molecules of paltry molecular mass, the which gives rise to negligible dispersion force. So why so? Clearly, there is an intermolecular force operating …

Apr 17, 2018 · SO3 + H2O=H2SO4 How do I balance this equation? Chemistry. 1 Answer Junaid Mirza Apr 17, 2018 Equation is ...

Jan 11, 2016 · For the conjugate base, simply remove a proton from the parent molecule/species and CONSERVE mass and charge. If I remove H^+ from H_2O, clearly I get the hydroxide ion, OH^-. Mass and charge have been conserved. If I remove a proton from sulfuric acid, I get bisulfate anion, HSO_4^-; thus bisulfate is the conjugate base of sulfuric acid. What is the …