Upon reaction of 1.274 grams of copper sulfate with excess zinc metal, 0.392 grams of copper metal was obtained according to the following equation: CuSO 4(aq)+Zn(s)⟶Cu(s)+ZnSO 4(aq). The use of stoichiometric factors to generate the desired values is explained in detail. Step-by-step solutions are available for computing the amount of reactants needed and the theoretical yield in addition to the percent yield. To solve this, just enter “ convert 5 mL of 1.5 mM magnesium hydroxide to atoms”.Īfter running a chemical reaction, one often wants to know how the reaction went by computing the reaction yields. How many atoms are in five milliliters of a 1.5 mM magnesium hydroxide solution? Step-by-Step Solution Unit conversions and dimensional analysis details are provided. As such, step-by-step solutions are available for converting among moles, mass, volume, molecules and atoms. What is its percent composition? Step-by-Step Solutionįor the answer, just enter “ antihemophilic factor elemental composition”.Ĭhemical conversions crop up in nearly every chemistry homework or research problem. Example ProblemĪntihemophilic factor is a coagulant with the formula C 11794H 18314N 3220O 355S 83. Ways in which you can check your work during the calculations are also available via the “Show intermediate steps” buttons.
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Details of the relevant equations, as well as how to compute the necessary intermediate values, are provided. The step-by-step solution provides a general framework for solving this class of problem in the Plan step.
One way to analyze individual chemicals is to compute and compare the mass and atom percentages. In this case, just enter “ molar mass silver sulfate”. Example ProblemĬalculate the molar mass of silver sulfate, Ag 2SO 4. Details of which formula to use and how to gather the necessary information are provided. In all cases, a general framework for solving these types of problems is provided via the Plan step. Step-by-step solutions are available for the molecular mass and relative molecular mass in addition to the molar mass. To do this, one needs the molar mass for each reactant. Step-by-Step Solutionįor this class of problem, just enter “ balance copper + nitric acid -> copper nitrate + nitrogen dioxide + water”.Īfter balancing the related chemical equations, the next step in planning a laboratory experiment is computing how much of each reactant must be measured out. Write the balanced equation for the reaction of copper with nitric acid to produce copper nitrate, nitrogen oxide and water. The step-by-step solution walks you through a robust algebraic approach to identifying the stoichiometric coefficients.
If chemical equations are the language in which chemical processes are expressed, then balancing chemical equations is the corresponding grammar. Read on for example problems in chemical reactions and their step-by-step solutions! Balancing Chemical EquationsĪ fundamental aspect of chemistry is balancing chemical equations. Over the next few weeks, we’ll be exploring some of the popular topics that middle-school, high-school and college students encounter in their chemistry courses and final exams: chemical reactions, structure and bonding, chemical solutions, and finally, quantum chemistry. The guides not only hone efficient problem solving, but also facilitate digging deeper into concepts that might still be murky. The step-by-step solutions provide stepwise solution guides that can be viewed one step at a time or all at once. Luckily, the Step-by-Step Solutions feature of Wolfram|Alpha has got your back! Whether you’re studying remotely or collaborating via video conferencing, Wolfram|Alpha helps you learn and apply the problem-solving frameworks for chemical word problems.
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However, although free textbooks keep cash in your wallet, they don’t include solution guides for all the homework problems. To combat this, the chemical education community has developed open educational resources to provide free chemistry textbooks. If you’re studying chemistry or are in a discipline requiring chemistry prerequisite courses, then you know how expensive the required textbooks can be.
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