# Tag Info

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First of all, in chemistry two types of formulas exist: structural and empirical. A structural formula shows the way atoms are connected. An empirical formula only summarizes atoms and their ratios. While often an empirical formula references a molecule, there are cases when the compound in question is not molecular, or it is unknown if it is molecular. In ...

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Here's how to solve the system in python: >>> import numpy as np >>> stoich_mat = np.array([[-1, 0, 1, 0, 0], # copper balance [0, -1, 0, 0, 2], # hydrogen [0, -1, 2, 1, 0], # nitrogen [0, -3, 6, 1, 1], # oxygen [1, 0, 0, 0, 0]]) ...

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This confusion pops up often around here. The problem usually comes down to insufficient consideration of units. When you say "rate", what do you mean? It has units of mol per volume per second, e.g. $\mathrm{\frac{mol}{L\cdot s}}$. But that isn't enough information. You need to know moles of what. First example: if the rate is defined as loss of ...

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Sure, most of the equation-balancing Javascript calculators on the web use that method. It's overkill to balance a single equation that way, but if you're automating the process or if you have a lot of equations to balance and check it can be convenient. If you search on "balancing equations matrix" you'll find many simple matrix inversion approaches for ...

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The reaction as you state it is correct only if there will react only one molecule of oxygen. But the reaction describes burning of methane which is supposed to be in the presence of excess of oxygen. Then not only methane is burnt, but also the arised hydrogen. So in "first" step: $\ce{CH_4 + O_2 -> CO_2 + 2H_2}$ but then the hydrogen will be also ...

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Given the reaction $\ce{a Cu + b HNO3 -> c Cu(NO3)2 + d NO + e H2O}$, we can set up four equations: \begin{align} a &= c & &\text{(for Cu)} \tag{1} \\ b &= 2e & &\text{(for H)} \tag{2} \\ b &= 2c+d & &\text{(for N)} \tag{3} \\ 3b &= 6c+d+e & &\text{(for O)} \tag{4} \end{align} Substituting equation $(1)... 19 I think we are over-interpreting the poor$+$sign. In older books you will also see an$=$sign instead of an arrow just like what Lavoisier did. We have to get hold of La Traité Élémentaire de la Chimie (1789). We should look at the historical perspective and I always want to see the earliest usage in chemistry just to see what was the original person ... 16 The term you are looking for is formula unit, I think. Wikipedia doesn't really describe it super well, but just to give an example, you could write the sentence$\pu{58.44 g}$of$\ce{NaCl}$contains$6.022 \times 10^{23}$formula units of sodium chloride and it would be pretty well understood. See also: What is the difference between a Chemical ... 15 You can do this in the 'guided trial-and-error' method that LordStryker showed which is probably quickest for simple reactions, or approach it in a purely mathematical fashion which is the method I will explain. This method works well for arbitrarily difficult reactions. Your chemical equation contains 3 atomic species:$\ce{C}$,$\ce{H}$and$\ce{O}$. This ... 15 You can expand the second equation so it has the same amount of educt on the left side as the first one: $$\ce{6XeF4 + 12H2O → 3Xe + 3XeO3 + 24HF + 1.5 O2}$$ Difference on the product side, compared to first equation, is$\ce{-1 Xe + 1 XeO3 - 1.5 O2}$. This excess of one$\ce{XeO3}$can go $$\ce{XeO3 → Xe + 1.5 O2}$$ , exactly the amount of xenon and oxygen ... 14 Hint : n-factor of a molecule/compound is defined as the change in oxidation state per molecule. You have correctly calculated the change of one carbon atom as 4. But how many carbon atoms are there in the glucose molecule? Note: The average oxidation state of carbon in glucose is zero while in reality the different carbons have different OS. (... 14 When using matrices, you are really just solving a system of equations. You can use any technique you know to solve the system of equations that a chemical equation represents. Using matrices is only one technique that may be useful for complicated equations or if you have a calculator/MATLAB on hand. In the given example, it would be easier to solve it ... 14 Suppose that you want to make a cake. You would first need to go your fridge to check whether you have the necessary ingredients to make one. So you go to your fridge. Luckily, you have all the ingredients: milk, eggs, flour, and frosting. So you decide to make one. You also decide to make the cake as big as possible to avoid the hassle of cleaning/preparing ... 13 The picture you showed does have an unequal number of sodium cations and chloride anions. However, the picture shows only part of a crystal. Every atom that is on a boundary of the shown cube, whether on a face, edge, or vertex of the cube, is shared with other "cubes" in the crystal that aren't shown in the picture. Each of the 8 corner Cl atoms in your ... 13 The rate of production of$\ce{B}$is given by: $$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = k[\ce{A}]^2$$ The rate of consumption of$\ce{A}$is given by: $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -2k[\ce{A}]^2$$ The reason for having the coefficients this way is because of the way the rate of the reaction,$r$, is defined (source: IUPAC Gold Book): $$r ... 13 It is probably supposed to be the 'corresponds to' sign[ref, p 105]; in this case simply representing that for every two moles of \ce{NaOH} one mole of \ce{H2SO4} is converted by reaction stoichiometry. This sign is difficult to typeset in LaTeX which is why I presume the authors used something else instead. Creating the sign is even more annoying in ... 12 The mole is a base unit as specified in the Système international d’unités (SI) by the bureau international des poids et mesures. Its decisive definition is that published in French: La mole est la quantité de matière d’un système contenant autant d’entités élémentaires qu’il y a d’atomes dans 0,012 kilogramme de carbone 12 ; son symbole est « mol ». ... 12 The sign ≏ is read "is chemically equivalent to". The term chemically equivalent refers to a specific chemical reaction. For the reaction$$\ce{N2(g) + 3 H2(g) -> 2NH3(g)}\pu{1 mol } \ce{N2} \, ≏ \, \pu{3 mol } \ce{H2} \qquad \pu{1 mol } \ce{N2} \, ≏ \pu{2 mol } \ce{NH3}$$Reference: Atkins, P. W.; Jones, L. L.; Laverman, L. E. Chemical ... 11 This happens when you have a sum of two (or more) independent reactions. You can balance each one, and then add them together in an arbitrary proportion. People usually run into this when trying to come up with an equation for combustion of gunpowder, where the oxidation of carbon and oxidation of sulfur are pretty much independent. Now, in your case the ... 11 Here is an interpretation I came up for it, so suppose you have the molecule \ce{CH4}, then we can think of this as a vector as \langle 1,4 \rangle where the 1 stands for number of carbons and 4 as the number of hydrogen. Generally for each 'nice' chemical reaction(*), we can associate a vector length 'n' to the reactant side, where n is the ... 10 The mass of the sulfate is m_{\ce{SO4}} = 0.610~\mathrm{g}, the mass of the metal is m_\text{M} = 1.250~\mathrm{g}. With the molar mass of the sulfate anion M_{\ce{SO4}} = 96.061~\mathrm{g\, mol^{-1}} we can calculate the amount of sulfate present in the final compound:$$ n_{\ce{SO4}} = \frac{m_{\ce{SO4}}}{M_{\ce{SO4}}} = 6.35~\mathrm{mmol} $$Since ... 10 \ce{S6}, \ce{S8}, \ce{S12} – does it make a difference or is it just a trick to make the question more complicated than it is? What you know for sure is: M(\ce{S}) = 32.065\ \mathrm{g}\cdot \mathrm{mol}^{-1} M(\ce{H2SO4}) = 98.079\ \mathrm{g}\cdot \mathrm{mol}^{-1} \frac{5000\,\mathrm{g}}{32.065\,\mathrm{g}\cdot \mathrm{mol}^{-1}} = 155.93\,\... 10 Below is a basic example. If percent yield is based on a mass, then a >100% yield can result from a product contains materials that should not be there. For instance, a product that should be anhydrous may not be completely dry and water would add mass to the product increasing your yield. Other present contaminants can also do this and can even appear ... 10 What is the difference between one egg and one dozen eggs? A dozen is simply a certain count, in this case 12, that is rendered as one unit for our convenience. Avogadro's number s just the same thing as a dozen, except we made the unit count larger than 12 in order to match it up with our conventionally used units. Your assumptions about a "mole" are ... 9 You should use the the same product species in both half -reactions. In the example you gave, this is \ce{N2}.$$\ce{NH4NO3-> N2}$The half-reactions are constructed from the ammonium and nitrate ions separately, since these feature nitrogen in different oxidation states: \begin{align} \ce{NH4+ &-> N2} \tag{1} \\ \ce{NO3- &-> N2} \tag{2} \... 9 You can make certain assumptions in these and similar problems because concentration of uniform substances (whether it's expressed as molarity, molality, mass density, or something else entirely) is an intensive property, i.e., it has no dependence on the total amount of material in the object or sample being measured. For example, if you have a homogeneous ... 9 Preface: The problem is that you have overstated your percentage yield, and the symptom is that it is above 100%. Example: You and your friend each do an experiment where the literature states the yield should be ~65%. You both use poor technique and fail to dry your product, causing you to attribute water mass as product mass. Your friend's yield is 101%, ... 9 The equation is a little dubious, as I mentioned in the comments. How do we come up with coefficients of this sort? Let's go from scratch. We don't know the coefficients at this point so we'll just call thema$,$b$, and$c$. Note that the coefficient of$\ce{S$_x$H$_y$O$_z$}$is taken to be 1. It doesn't have to be 1; however, for the sake of ... 9 This is a semi-standard example of why 'naive' balancing does not work. Ozone is a source of atomic oxygen, producing free oxygen molecules. So, in 'mild' conditions only one oxygen per ozone molecule would react as a strong oxidizer, and the remaining molecular oxygen would require elevated temperatures to react. Thus the equation would be$\ce{PbS + 4O3 -...

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You seem to have had the right idea of fixing the scale by arbitrarily choosing the value of one coefficient, and then solving for the rest. Apparently, you just got stuck at some point, presumably either because you couldn't solve for $b$ just with simple substitutions, or because your initial choice of $a = 1$ gave you fractional values for the other ...

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