Tutor profile: Sheetal K.
Ba(OH)2 + HCl --> BaCl2 + H2O Balance the equation above. What are the stoichiometric coefficients? How many grams of BaCl2 will be produced from 6.0 g of Ba(OH)2 and 12.0 g HCl?
The chemical equation needs to be balanced before doing any calculations. The stoichiometric coefficients are Ba(OH)2 + 2HCl --> BaCl2 + 2H2O. Oxygens and hydrogens always should be balanced last. The barium is already balanced on both sides, but chlorine is not balanced since there is one on the left side and 2 on the right side, therefore adding a 2 in front of HCl balances the chlorine in the reaction. Then there are 4 hydrogens on the left side, while there are 2 on the right side, therefore adding 2 in front of H2O would balance the hydrogens and oxygens in the chemical reaction. Now, we can begin the second part of the question. First, we need to find the limiting reagent, which can be found using dimensional analysis, shown below, 1. Find the number of mols of each reactant 6.0 g Ba(OH)2 * (1 mol Ba(OH)2)/(171.34 g Ba(OH)2) = .035 mol Ba(OH)2 12.0 g HCl * (1 mol HCl)/(36.34 g HCl) = .329 mol HCl 2. Using the stoichiometric ratios from the chemical reaction find the number of mols of BaCl2 to see which reactant is the limiting reagent .035 mol Ba(OH)2 * (1 mol BaCl2)/(1 mol Ba(OH)2) = .035 mol BaCl2 .329 mol HCl * (1 mol BaCl2)/(2 mol HCl) = .165 mol BaCl2 The smaller amount is the limiting reagent, which is Ba(OH)2 meaning the reaction is going to reach completion with HCl in excess while Ba(OH)2 is used completely 3. Find the amount of BaCl2 is needed for the reaction .035 mol BaCl2 * (208.23 g BaCl2)/(1 mol BaCl2) = .00017 g BaCl2 (which is pretty small)
Subject: Basic Chemistry
How many valence electrons does oxygen have and what is its charge?
Oxygen is in the second row of the periodic table and has 2 shells, 2s and 2p. The 2s shell has 2 electrons but according to Hund's rule, the 2p orbital is not full and has 4 electrons in it. Adding the electrons in the 2s and 2p shell gives us the numbers of valence electrons, therefore there are 6 valence electrons in oxygen. Its charge is 8- # valence electrons, and since it is on the right side of the periodic table, the charge is negative. So oxygen has 6 valence electrons and its charge is -2.
Find the derivative of the function f(x) = (x^2)*(sin(x))
Here the product rule will need to be applied, which states, d(f(x)*g(x))/dx = f'(x)*g(x) + g'(x)*f(x) Here the derivative of x^2 is 2x, and the derivative of sin(x) is cos(x) Therefore the derivative of the function f(x) = (x^2)*(sin(x)) is 2x*(sin(x)) + x^2(cos(x))
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