I need to balance a chemistry equation.

Let m represent the subscription rate for a man, and w the subscription rate for a woman.
m/w=4/3, or m=(4/3)w.
2 men and 5 women paid 460. Thus
2m+5w=460.
Substitute (4/3)w for m, in 2m+5w=460.
This gives an equation in the unknown w. Solve for w
2(4/3)w +5w=460 or (23/3)w=460. Answer is w=60
Use this value in the relation m=(4/3)w to get m=80.
Check m/w=80/60=4/3

Use the rules for exponents and you will calculate it by hand
(1*10^-14)/(5*10^-12)=(1/5)*10^(-14+12)= 0.2*10^-2=2*10-3
To enter powers of 10 use the short-cut key [EE]
For example to enter 6.02*10^23 type in 6.02* [EE] 23. When you use the EE key, the base (10) should not appear.

You'll find lots of applications here.

Surviving Chemistry AP Exam One Day at a Time.
at www.e3chemistry.com

Something new different cleaner clearer more simplified and less overwhelming for preparing for your AP exam.

This problem is the type often encountered in a first semester chemistry course, whether in high school or college.

It is very easy to solve if you already understand the following concepts, and if you also have had lots of practice applying them!

• PROPORTIONS
• THE MOLE
• The LAW OF CONSERVATION
• CHEMICAL EQUATIONS
• BALANCED CHEMICAL EQUATIONS
• STOICHIOMETRIC CALCULATIONS
• SIGNIFICANT FIGURES, PRECISION, & ROUNDING OFF CALCULATED QUANTITIES

Initially, I will assume you already understand the above concepts, except for "stoichiometric calculations." I'd be happy to focus more on the other topics at another time, depending on interest, as indicated by posted questions.

To solve this problem you must be aware of the balanced chemical equation, as follows:
Notice that there is one C, one O2, and one CO2.
(For convenience in my post, I am writing the subscripts on the same line,)

"One what?" You might ask. It can be one "molecule" for each the O2 and CO2; and one "atom" of C. For the method of stoichiometry in this problem, we should use one "mole." So, there is one mole of C, one mole of O2, and one mole of CO2. In other words, the PROPORTION of these reactants (C and O2) and product (CO2), is one to one to one, also written one:one:one (or 1:1:1).

As I am assuming you know, these numbers are called the "coefficients" of the balanced chemical equation.

To calculate the mass in grams of O2 (the unknown quantity), we will need only the moles of O2 and the moles of one other substance from the balanced chemical equation. Since we are given the mass of C, we can use the moles of C.

TO WORK A STOICHIOMETRIC problem using a balanced chemical equation, YOU MUST INCLUDE THE MOLES OF TWO SUBSTANCES from the chemical equation in your calcuation, in this case, moles of C and moles of O2. As you know, you can calculate moles of C from the mass (in grams) of C that was given in this problem. You must calculate the moles of O2 by use of the following math expression:

mol O2 = (MOLE RATIO of O2 to C) x mol C.

(Notice that I have used the official abbreviation of moles, which is "mol".)

I will show you how to calculate the MOLE RATIO a moment.

Once you have caculated the moles of O2, you will have only one more calculation to do to calculate the number of grams of O2. Since I am assuming you already know how to calculate grams from moles, I will skip this step until my closing summary.

The MOLE RATIO can be calculated as follows: The calculated mole ratio will always equal the ratio of corresponding coefficients.

Notice that the mole ratio actually represents the ratio of the coefficients that are in the balanced chemical equation, and is written in the same order as are the calculated (calc'd) moles.

Therefore,

So, you can see that the number of moles of O2 = calculated moles of C.

The moles of C were calculated as follows:

The calculated moles of C = [grams C] / [MW of C] = [188 g C / 12.0 g] = 15.7 mol C

Therefore, there are 15.7 moles of O2, as calculated from use of the math expression (with boxes) shown above.

The wanted GRAMS of O2 are calculated from MW of O2 times its number of moles, as follows:

Grams of O2 = 32.0 g O2 x 15.7 mol O2 = 502 grams O2.

For your convenience as you make conversions among grams and moles, you may use the following memory aid:

To go from moles to grams, just notice that moles are beside MW, so multiplying them will give grams (moles x MW = grams). If you want to go from grams to moles, notice that grams are over MW, so (grams/MW) = moles. If you should want to calculate MW, you would divide grams by moles, (grams/moles) = MW.

One more tip: When doing calculations, use at least the same number of significant figures (sig figs) in the molecular weights (and atomic weights) as is in the given numerical data. That is, since there are three sig figs in 188 grams of C, use 12.0 as the atomic weight of C, and 44.0 as the MW of CO2. You would obtain the same answer using 12 and 44, but as you may know, following this rule would be more important in other problems in which the digit past the decimal is not zero. This allows proper rounding off of the final calculated quantity.

In Summary

• Write the chemical equation, and balance it.
• Use MOLES to do the necessary stoichiometric calculations. Convert grams to moles to make this possible.
• Multiply the calculated moles of the given amount of chemical compound (or element) by the COEFFICIENT RATIO of the two compounds involved in the problem.
• Convert moles to grams if grams are asked for by the problem.
• Round off correctly.
• Show the answer quantity, which should always show both numerical value AND unit.

Suggestion: Try out this method to other problems involving a chemical reaction. Good luck!

***

Faraday's Law is actually more accurately referred to as the plural of that, Faraday's LawS.
The reason for this plurality is indicated by the existence of three different laws or equations, as shown below:

• Faraday's Law of Induction - related to principles of transformers, inductors, electric motors, generators
• The Maxwell-Faraday Equation - states that a changing magnetic field creates an electric field in a closed loop, or circuit; appears in a modern set of Maxwell's Equations, and is commonly called "The Faraday Law."
• Faraday's Laws of Electrolysis - refers to the relationships between change in mass of substances (at electrodes) and amount of electricity (Coulombs) flowing through an electrolytic cell.

I know that the last one listed above is taught in the second semester of general chemistry courses offered at community colleges. The first two, and maybe the last one, are taught in physics courses.

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Hello. 1. Select the characters that you would like to be on subscript. Go to Format menu and Click Font, from the Font tab you have to select subscript, and click OK; or alternatively you may use the shortcut key CTRL + +(plus sign) for subscript; 2. From the tools menu, click on "track changes". you will see a vertical line on the left margin of your document. Just scroll on the document and you will see how and when the text was edited by whom. You can use the options on the "reviewing toolbar" to either accept or reject the changes made on the document.

chemistry diagram...???? give me the equations, let me try...

the way i enter my equations is a bit different from what you're doing, i use the fraction button to do division problems, i just feel it makes it easier to read.
anyway, the fraction button is on the far left, 3rd button down (it looks like a white box on top of a line, and below the line is another empty box). press that, and you can type 1.0x10^-14 on the numerator.
it seems that you haven't discovered the EXPONENT button. that button is on the same row as the fraction button, it looks like an x with a white box next to it. press that once you've typed out "1.0x10", and then type -14. the negative button btw is below the fraction button, and looks like this: (-)
press the right arrow twice on the circle thingy at the top of all the keys, and that'll advance you to the denominator. type 2.0x10^-5, using that nifty exponent key. press the equal sign and it'll solve it for you, all in scientific notation (the answer is 5x10^-10). the problem with this calculator is, it won't expand that answer for you with all the decimals and everything, but if its chemistry that shouldn't be a problem.
also, you can certainly use the "x10^x" button on the very bottom of the keypad (its next to the decimal button). by using this key, you won't have to press the exponent button since you can just type in the exponent directly next to the 10.

as for that negative log of 5.6, just input the negative, the log button is right next to the exponent button, and your number and yeah....it's that easy :]

i hope you like this caculator, it has so many functions (have you seen the size of the manual??)
good luck on your chemistry test!

6.02 X 10y*23

y* is y raise to X