At a minimum, the identity and mass of each atomic element present must be known. If the total mass of the compound is known, then it is necessary to know all but one mass of the elements that compose the compound. When balancing a chemical equation, changing the subscripts changes the identity of the substance. The subscripts in a formula may not be changed unless one is determining the molecular formula from the empirical formula.
To convert grams of a substance to molecules of the same substance, the molecular mass of the substance, and Avagadro's number are needed.
Student B is correct. Student A wrote a properly balanced equation. However, by changing the subscript for the product of the reaction from an implied one, NaCl, to a two, NaCl2, this student has changed the identity of the product.
When balancing chemical equations, never change the values of the subscripts given in the unbalanced equation. Convert moles of B to moles of compound, A5B2; then using the stoichiometric ratio of moles of A to moles of A5B2, determine the moles of A; and finally convert the moles of A to grams of A using the molecular mass of A.
To determine the number of grams of sulfur that would react with a gram of arsenic, the stoichiometric ratio of the arsenic to the sulfur in the compound is needed, as well as the atomic masses of sulfur and arsenic. The scale of the reaction is determined by the number of moles used as reactants in the experiment. Then calculate the number of moles of NH4NO3 is in 1. The formula CaC2 indicates that there is 1 mole of Ca for every 2 moles of C.
Therefore, if there are 0. The molecular formula is some integer multiple of the empirical formula. This means that we can divide the molecular formula by the largest possible whole number that gives an integer ratio among the atoms in the empirical formula.
We begin by realizing that the mass of oxygen in the compound may be determined by difference: 0. Next we can convert each mass of an element into the corresponding number of moles of that element as follows:.
Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the simplest mole ratio among the three elements in the compound: for Na, 4.
To solve this problem we will assume that we have a g sample. This implies we have This implies that we have The amount of oxygen was determined by subtracting the total amounts of the other three elements from the total assumed mass of g.
Convert each of these masses into a number of moles:. The empirical formula weight is This means that the molecular formula is twice the empirical formula, or Hg2C4H6O4. The grams of O are determined knowing that the total mass of all elements in the compound must add to g. These relative mole amounts give us the empirical formula C4. The relative mole ratios are: for C, 0. This type of combustion analysis takes advantage of the fact that the entire amount of carbon in the original sample appears as CO2 among the products.
Hence the mass of carbon in the original sample must be equal to the mass of carbon that is found in the CO2. Similarly, the entire mass of hydrogen that was present in the original sample ends up in the products as H2O:. Now, convert these masses to a number of moles:. The relative mole amounts are: for C, 4. However, we see that if we double the relative number of moles of each compound, there are approximately 19 moles of C, 30 moles of H and 2 moles of O. In most problems where we attempt to determine an empirical formula, the relative mole amounts should work out to give a nice set of values for the formula.
Rarely will a problem be designed that gives very odd coefficients. With experience and practice, you will recognize when a set of values is reasonable. The mass of oxygen is determined by subtracting the mass due to C and H from the total mass: 7. Thus, the empirical and molecular formulas are equivalent. By inspection, though, we can see there are the same number of moles of Hg and Br.
Consequently, the simplest mole ratio is and the empirical formula is HgBr. To determine the molecular formula, recall that the ratio of the molecular mass to the empirical mass is equivalent to the ratio of the molecular formula to the empirical formula.
Thus, we need to calculate an empirical mass: 1 mole Hg The ratio of these is:. From the information provided, the mass of sulfur is the difference between the total mass and the mass of antimony:. Now, divide each of these values by the smaller quantity to determine the simplest mole ratio between the two elements:. For Sb: 3. Since the molecular mass reported in the problem is the same as the calculated empirical mass, the empirical formula is the same as the molecular formula.
First, determine the amount of oxygen in the sample by subtracting the masses of the other elements from the total mass: 0. Now, convert these masses into a number of moles for each element:. In order to obtain whole numbers, each of these values is multiplied by 2 and we determine the empirical formula is C21H22N2O2.
This means that the molecular formula is the same as the empirical formula, or C21H22N2O2. There are 9 moles of Ca, 6 moles of P, and 24 moles of O present. The picture shows 9 molecules of O2 and 3 molecules of C2H6S. The balanced reaction shows that 2 molecules of C2H6S react with 9 moles of O2. Therefore, O2 is the limiting reagent.
Since you have an unreacted oxygen molecule in the mixture, the reaction mixture had an excess of O2 and C2H6S was the limiting reagent. First determine the amount of Fe2O3 that would be required to react completely with the given amount of Al:. This can be confirmed by calculating the amount of C2H4 that would be required to react completely with all of the available H2O:. We know that only 5. Therefore, the amount left unused is: In that case we would have seen that AgNO3 was in excess.
So, there is excess H2O present. The amount that remains is First calculate the number of moles of water that are needed to react completely with the given amount of NO Since this is less than the amount of water that is supplied, the limiting reactant must be NO2. Therefore, to calculate the amount of HNO First calculate the number of moles of water that are needed to react completely with the given amount of PCl Determine the theoretical yield:.
This is determined by dividing the actual yield by the percent yield. Recall actual yield Substituting the values from this problem gives the There is an excess of F2 present. The mass of the wet paint is:. The mass of Cr in the original sample is thus 0. Converting to moles, we have:. The relative mole amounts are: for Cl: 3. First determine the percentage by weight of each element in the respective original samples. This is done by determining the mass of the element in question present in each of the original samples.
The percentage by weight of each element in the unknown will be the same as the values we calculate. Next, we assume g of the compound, and convert these weight percentages into mole amounts:. Since the molecular mass is the same as the empirical mass, the molecular formula is CaC2S2N2. Next calculate the amount of H2 needed to react completely with all of the available O2.
Since only , lb. Since 6. Note that the drawing below is an anion. The negative charge has not been included in the drawing. Formic acid, a weak acid will form. Insoluble nickel hydroxide will precipitate. You want to use a metathesis reaction that produces CoS, which is insoluble, and a second product that is soluble.
You may want the reactants to be soluble. Determine the moles of HCl in 1. This is the number of moles of HCl in the 0. If we were working with a full liter of this solution, it would contain 0. The molar mass of the salt is However, we are working with just 50 mL, so the amount of Sr NO3 2 needed is slightly more than a twentieth of 40 g, or 2 g.
The answer, 2. Review Questions 5. Concentration the ratio of the quantity of solute to the quantity of solution or quantity of solvent. Unsaturated Any solution with a concentration less than that of a saturated solution of the same solute and solvent. Supersaturated a solution whose concentration of solute exceeds the equilibrium concentration. Solubility the ratio of the quantity of solute to the quantity of solvent in a saturated solution.
Chemical reactions are often carried out using solutions because this allows the reactants to move about and come in contact with each other. Furthermore, solutions can be made with a high enough concentration to allow the reaction to proceed at a reasonable rate. When a sugar crystal is added to a a saturated sugar solution, the sugar crystal will not dissolve.
Precipitate a solid that separates from a solution usually as the result of a chemical reaction For a precipitate to form spontaneously in a solution, the equilibrium must be disrupted. A supersaturated solution may form a precipitate spontaneously, or if the temperature changes in the direction that will cause a precipitate to form.
Electrolytes are soluble, ionic compounds. An ion is hydrated when it is surrounded by water molecules. Dissociation the dissolving of an ionic compound in water such that the individual ions that compose the ionic compound become separated from one another via hydration , and move about freely in solution, acting more or less independently of one another.
In a balanced ionic equation, both the mass and the electrical charge must be balanced. It must have the correct formulas of reactants and products. The product is not correct in the equation. Acid sour taste, turns litmus red, corrode some metals, etc Base bitter taste, turns litmus blue, soapy feel, etc If a solution is believed to be basic, red litmus paper should be used so that it would turn blue. The blue litmus paper may not change color if the solution is neutral.
Dynamic equilibrium is a condition in which two opposing processes are occurring at equal rates. Double arrows are not used for the reaction of a strong acid with water because the reaction is not in equilibrium. These are not reversible reactions, i. The student removed a hydrogen attached to the methyl group, CH3. Hydrogen attached to carbon atoms are not acidic protons and will not be removed in water.
The correct structure of the ion is one where the hydrogen attached to the oxygen atom is removed. The structure should be:. The molecules is diethylamine, as base. The structure of the resulting ion is:. Since this solution is immediately supersaturated in the moment of mixing, a precipitate of AgBr forms spontaneously.
Any solution containing ammonium ion will react with a strong base to yield ammonia. The presence of ammonia is easily detected by its odor. Molarity is the number of moles of solute per liter of solution, also known as molar concentration. The number of moles of HNO3 in the solution has not changed because none of the original sample was removed. Instead, the concentration has decreased since more water was added. The number of moles of CaCl2 is the same in both solutions, but A is 0.
The volume of solution A is 50 mL, therefore the volume of B is 25 mL:. Qualitative analysis is the use of experimental procedures to determine what elements are present in a substance. Quantitative analysis determines the percentage composition of a compound or the percentage of a component in a mixture. Qualitative analysis answers the question, "what is in the sample?
Buret a long glass tube fitted with a stopcock, graduated in mL, and used for the controlled, measured addition of a volume of a solution to a receiving flask. Titration a procedure for obtaining quantitative information about a reactant by a controlled addition of one substance to another until a signal usually a color change of an indicator shows that equivalent quantities have reacted.
Titrant the solution delivered from a buret during a titration. End point that point during a titration when the indicator changes color, the titration is stopped, and the total added volume of the titrant is recorded. The indicator provides a visible signal that the solution has changed from an acid to a base.
Phenolphthalein is colorless in acid solution. Phenolphthalein is pink in base solution. The electrical conductivity would decrease as the solution is neutralized because there are half the amount of ions as products of this reaction than there were to start with.
The electrical conductivity would increase since HC2C3O2 is a weak acid and is only partially dissociated, and as the NH3 is added two ions are formed, increasing the concentration of ions in solution. These reactions have the following "driving forces": a formation of a gas, CO2. There are numerous possible answers. One of many possible sets of answers would be: a b c d e. We need to choose a set of reactants that are both soluble and that react to yield only one solid product. Choose b.
The solution needs to be kept acidic enough to prevent the formation of Cu OH 2. The volume of water to be added is: This produces 0. First, calculate the number of moles HCl based on the titration according to the following equation:.
First convert the density of vinegar to a value appropriate for one liter of solution: 1. We know that one liter of this vinegar contains 0. Since lactic acid is monoprotic, it reacts with sodium hydroxide on a one to one mole basis: 0. The equation for the reaction indicates that the two materials react in equimolar amounts, i. Because this reaction is , we can see by inspection that the AgNO3 is the limiting reagent.
Since we start with equal volumes, there are fewer moles of the AgNO3. Assuming that AgCl is essentially insoluble, the concentration of silver ion can be said to be zero since all of the AgNO3 reacted. The number of moles of chloride ion would be reduced by the precipitation of 9. If we assume that the Ca3 PO4 2 is completely insoluble, then its concentration may be said to be essentially zero. The concentrations of the other ions are determined as follows: For nitrate:.
Because we know the amounts of both starting materials this is a limiting reactant problem. So start by assuming that the barium hydroxide is the limiting reactant. Therefore the barium hydroxide is the limiting reactant. Now we can calculate the mass of aluminum hydroxide that is produced. All of the barium ion and hydroxide ion are reacted so the concentration of each is 0. We started with the following:. Sodium is oxidized since it loses electrons.
Chlorine is reduced and is, therefore, the oxidizing agent. Al is oxidized and is, therefore, the reducing agent. First the oxidation numbers of all atoms must be found. The oxidation numbers for K and Na do not change.
However, the oxidation numbers for the chlorine atom decreases. The oxidation numbers for nitrogen increase. This reaction is the redox reaction. In the other reaction, the oxidation numbers of the atoms do not change. The oxidation numbers for O and Na do not change.
However, the oxidation numbers for all chlorine atoms change. There is no simple way to tell which chlorines are reduced and which are oxidized in this reaction. In this case Cl2 is reduced and is the oxidizing agent, while NaClO2 is oxidized and is the reducing agent.
If H2O2 acts as an oxidizing agent, it gets reduced itself in the process. The product is therefore water. The oxidation number for C increases and the oxidation number for O decreases.
Therefore, CH4 undergoes oxidation and O2 undergoes reduction. This means that CH4 is the reducing agent and O2 is the oxidizing agent. Then, to balance the reaction in base use the above balanced reaction in acid and add 12OH- to each side.
Oxidation is the loss of one or more electrons. Reduction is the gain of one or more electrons. The oxidation number decreases in a reduction and increases in an oxidation. The number of electrons involved in both the reduction and the oxidation must be the same; only those electrons that come from the reductant and go to the oxidant are involved.
No electrons from external or uninvolved sources are allowed to enter the process, and there cannot be any electrons left unreacted at the end of the process.
An oxidizing agent is the species that is reduced or gains electrons in an oxidation reduction reaction. A reducing agent is the species that is oxidized or loses electrons in an oxidation reduction reaction. The oxygen atoms are -2 in the reactants and products. Thus, the reaction is not a redox reaction. No other element changes oxidation state.
Therefore this is not a redox reaction. This change in oxidation number represents a reduction of nitrogen by 5 units, and it requires that nitrogen gain 5 electrons. The equation is not balanced since the charge is different on either side of the arrow.
The best reducing agents are those that are most easily oxidized and are found at the bottom of the activity series.
The best oxidizing agents are those that are most easily reduced and are found at the top of the activity series. See Table 6. This would be any metal higher less reactive than hydrogen, i. Combustion is the rapid reaction of a substance with oxygen, which is accompanied by the evolution of light and heat.
Historically, the reaction of a substance with oxygen was termed oxidation. Now we realize that reaction with oxygen most typically means that oxygen acquires electrons from the substance with which it reacts. The oxidation of a substance is, therefore, taken to represent the loss of electrons by a substance, whether the substance has reacted with oxygen or with another oxidizing agent.
The reaction of sulfur with air produced sulfur dioxide, SO2 gas. The limiting reagent is sulfur. Review Problems 6. The free radical OH is a molecular species. Thus, oxidations states of H and O must add up to zero. Rule 5 assigns an oxidation number of -2 to O. Since the oxidations numbers do not add to aero we have a conflict. That means rule 4 takes precedence and O must then have an oxidation number of -1 in OH.
Therefore Cl2 is reduced. One atom is reduced, the other is oxidized. In the reverse direction: The Cl ion begins with an oxidation number of 1 and ends with an oxidation number of 0.
Therefore the Cl ion is oxidized: This means Cl is the reducing agent. N is both reduced and oxidized. N is reduced in the conversion of NO2 to NO, a two electron reduction. Thus, NO2 is both the oxidizing agent and the reducing agent. In each case, the reaction should proceed to give the less reactive of the two metals, together with the ion of the more reactive of the two metals.
The reactivity is taken from the reactivity series table 6. The equation given shows that Cd is more active than Ru. Coupled with the information in Review Problem 6. The observation shows that Mg is more active than Ni, however, the information is not sufficient to determine which is easier to oxidize, Mg or Mo. Therefore it cannot be determined which reaction will occur spontaneously.
The density of the wine was 0. Since there is one mole of Cr in each mole of CrO42, then the above number of moles of CrO42 is also equal to the number of moles of Cr that were present: 0. Bromine is both oxidized and reduced. This is an example of an esterification reaction, the reaction of an organic acid with an organic base, an alcohol.
The first reaction demonstrates that Al is more readily oxidized than Cu. The second reaction demonstrates that Al is more readily oxidized than Fe. Reaction 3 demonstrates that Fe is more readily oxidized than Pb. Reaction 4 demonstrates that Fe is more readily oxidized than Cu.
The fifth reaction demonstrates that Al is more readily oxidized than Pb. The last reaction demonstrates that Pb is more readily oxidized than Cu. Any metal that is lower than hydrogen in the activity series shown in Table 6. We choose the metal that is lower more reactive in the activity series shown in Table 6.
The number of moles of this ion in the reactant solution is: 0. The mass of copper consumed is 0. The amount of unreacted copper is thus: The mass of Ag that is formed is: 0. The final mass of the bar is: If we assume that there is excess copper available, we need to determine the number of moles of Ag that will be produced. Convert this nmber of moles to a number of grams: 0. The final mass of the bar will include the unreacted copper and the silver that is formed: The activity series only lists metals in order of reactivity, not ions, non-metals, or molecules.
This number of moles of tin ion remaining is subtracted from the total that was available in the Multiply this number by 10 to get the moles of MnO4 that had not reacted with the SO2 in the original mL of 0. The balanced equation is the place to start. This amount is subtracted from the initial number of moles of hydrogen ion, and the amount of titrant is calculated by dividing moles by molarity of NaOH solution. The sample of oxalic acid to be titrated with NaOH is We can solve for x and y from these and calculate the initial concentration of SO32 and S2O This should then be converted to a value representing kJ per mole of reactant, remembering that the sign of H is to be negative, since the process releases heat energy to surroundings.
This should then be converted to a value representing kJ per mole of reactant, remembering that the sign of H is to be positive, since the process absorbs heat energy from the surroundings. We can proceed by multiplying both the equation and the thermochemical value of Example 7.
Energy is something that matter possesses by virtue of an ability to do work. Work is the energy expended in moving an opposing force through some particular distance. The energy of the child on a swing is all potential energy when she is at the top of the arc. As she descends, the energy is converted to kinetic energy. At the bottom at the arc, all the energy is kinetic energy. The potential energy increases as she rises and is completely potential energy at the top of the arc.
Chemical energy is the potential energy in substances, which changes into other forms of energy when substances undergo chemical reactions. Heat is a form of energy that is transferred between objects, and is the molecular kinetic energy possessed by molecules as a result of the temperature of the sample. Heat is related to the total kinetic energy of the molecules and temperature is related to the average kinetic energy. Thermal equilibrium is when two objects in contact with each other are at the same temperature.
The molecules in the hot object are moving with more kinetic energy than the colder object. As heat is transferred from the hot object to the cold one, the atoms in the hot object slow down and the atoms in the cold speed up until they have the same average kinetic energy.
The SI unit of energy is a joule kg m2 s The heat produced by combustion of gasoline does no useful work. It is expended into the surroundings. The internal energy is the sum of the molecular kinetic energy and the potential energy.
The change in internal energy is defined as the difference in internal energy between the energy of the products and the energy of the reactants. The maximum of the curve will be lower and shifted to the right. The quart of boiling water has more heat and will cause a more severe burn because the quart has more water, so that the quart has more kinetic energy.
The first diagram represents an isolated system. Isolated systems cannot transfer mass or energy across its boundary. The second diagram represents a closed system.
Closed systems cannot transfer mass but can transfer energy across its boundary. The third diagram represents an open system. Open systems transfer mass and energy across boundaries. The state of a system in chemistry is usually specified by its current conditions such as its chemical composition, its pressure, its temperature and its volume. A state function is a quantity whose value depends only on the initial and final states of the system and not on the path taken by the system to get from the initial to the final state.
The system is that part of the universe under study and separated from the surroundings by a real or an imaginary boundary. The surroundings are that part of the universe other than the system being studied and separated from the system by a real or an imaginary boundary. An isolated system does not allow matter or energy to be transferred between the system and the surroundings. The closed system can absorb or release energy but not mass across the boundary between the system and its surrounding.
The energy depends directly on the specific heat, so the material with the large specific heat requires the higher energy input for the 5 C rise in temperature.
Heat capacity is an extensive property and is proportional to the mass of the sample. Specific heat is an intensive property. If object A has twice the specific heat and twice the mass of object B, and the same amount of heat is applied to both objects, the temperature change of A will be one-fourth the temperature change in B. The potential energy of gasoline and oxygen is higher than the potential energy of carbon dioxide and water vapor since energy is released as heat in the reaction.
Kinetic energy must increase. The temperature of the system must increase because kinetic energy increased.
Since the heats of reaction will in general depend on temperature and pressure, we need some standard set of values for temperature and pressure so that comparisons of various heats of reaction are made under identical conditions. The standard temperature is 25 C, slightly above room temperature, and the standard pressure is 1 atmosphere. Fractional coefficients are permitted because thermochemical properties are extensive they depend on the amount of material present.
The coefficients of thermochemical equations are moles of substance. The reaction must 1 produce one mole of a compound at 25 C and 1 atm, and 2 the reactants must be elements in their standard states. The heat gained by the water must equal the heat lost by the copper. To get the correct signs for the heat transfer, remember heat lost is negative and heat gained is positive.
Next determine the number of moles of reactant involved in the reaction: 0. Divide the given equation by 2. Since NO2 does not appear in the desired overall reaction, the two steps are to be manipulated in such a manner so as to remove it by cancellation.
If we label the four known thermochemical equations consecutively, 1, 2, 3, and 4, then the sum is made in the following way: Divide equation 3 by two, and reverse all of the other equations 1, 2, and 4 , while also dividing each by two: Multiply all of the equations by 1 2 1 2 1 2 1 2.
We need to eliminate the NO2 from the two equations. To do this, multiply the first reaction by 3 and the second reaction by two and add them together.
The heat of formation is defined as the enthalpy change when one mole of a compound is produced from its elements in their standard states. Only c satisfies this requirement. The state for C is not given as graphite or as a solid. For choice d , the reaction is not balanced. Only b satisfies this requirement.
For choices a and c the reactants are not elements in their standard state. Choice d might look okay but it does not fit the definition that one mole of product is formed. Water has the highest specific heat and therefore will take the greatest amount of heat to raise its temperature as given amount. Lead has the smallest specific heat, though gold is on 0. Now divide this equation by two to give the thermochemical equation for the formation of 1 mol of HBr g : 1 2.
Comparing this value to the Hf value listed in Appendix C. Thus, for It is difficult to carry out this reaction as it is a non-spontaneous reaction requiring an input of energy. However, when energy is input to this mixture it is almost impossible to control what product will be formed.
We must assume the density of the solution is 1. The heat generated by the reaction of the salts with water is given by:. Also, it would be useful to have the enthalpy in units of kJ g Cloud Computing Tutorial - Tutorialspoint. Getting Started with Arduino, 2nd Edition. Beginning Arduino. Android Programming Cookbook.
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The overwhelming majority of bugs and crashes in computer programming stem from problems of memory access, allocation, or deallocation.
Such memory related errors are also notoriously difficult to debug.
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