Chemistry, Yuck....

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dotty4

New member
Joined
Oct 8, 2009
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12
I think I posted it in the wrong section initially,
Hi Everyone,
This is my first time posting. I am taking a beginning Chemistry course after not taking math for 23 years.
I need major help with Expodentials, I think that is what they are.

I totally get how to set up the questions, but do not know how the answers came to be.
#cs = 10u x 10^-6/1u x 1cs/10^-2s = 2.5 x 10^-3

My answer was 10^8

If the exponents are -6 and -2, two negs equal a positive? And how did 2.5 come into it?
 


I do not understand the meaning of #cs, u, cs, and s. Are these some combination of unit abbreviations, variables, and constants?

Without typing grouping symbols to clearly show what belongs in numerators versus denominators, expressions like this are difficult to decipher. Here's my guess.

\(\displaystyle \frac{10 \times 10^{-6}}{1} \cdot \frac{1}{10^{-2}}\)

With a ratio of powers when the bases are the same, we subtract the lower exponent away from the upper exponent, in order to eliminate the ratio.

Symbolically, the rule looks like this:

b^m/b^n = b^(m - n)

\(\displaystyle 10 \times 10^{-6 \;-\; (-2)}\)

\(\displaystyle 10 \times 10^{-6 \;+\; 2}\)

\(\displaystyle 10 \times 10^{-4}\)

\(\displaystyle 10^{-3}\)

I have no idea from where the 2.5 comes.

 
Sorry I think I typed the last one wrong.
Okay here is another example. They were unit measurements.

If 1mg = 10^-3 g and 1Mg = 10^6g, how many milligrams are there in .25 Mg

So I write it like this

#mg= .25Mg x 10^6/1Mg x 1mg/10^-3 g =

What is the step by step for this?
 
We're trying to convert 0.25 Mg to mg, and we're given information to create conversion factors to do this.

\(\displaystyle \frac{0.25 \; \mbox{Mg}}{1} \cdot \frac{10^6 \; g}{1 \; \mbox{Mg}} \cdot \frac{1 \; \mbox{mg}}{10^{-3} \; g}\)

Notice how I set up the two conversions ratios.

In the first one, I put Mg on bottom and g on top because we want the Mg units on 0.25 to cancel. This can only happen if Mg appears in the bottom.

The first conversion ratio has g on top, so g has to go in the bottom of the second conversion factor, in order to get the g units to cancel (leaving mg as the only remaining unit, which is what we want).

Does this make sense, so far?

What's left is just the arithmetic.

Of course, the rule for multiplying these ratios is (numerator times numerator times numerator) over (denominator times denominator times denominator).

Simplify, using the rule posted in my first response.
 
Can you post the rule like the first one you posted that has to do with this last question?

Then it will make sense.


Oh, thanks so much for helping me!!!
 
Okay so I go

.25 x 10 x 1
1x1x10

=

2.5
10

Where do the exponents come in to it?
 
I told you that it's the same rule; could you scroll up to read it?

I'll finish the math for you, instead.

After we cancel units Mg with Mg and units g with g, we have the following product.

\(\displaystyle \frac{0.25 \cdot 10^6 \cdot 1}{1 \cdot 1 \cdot 10^{-3}} \; \mbox{mg}\)

\(\displaystyle 0.25 \cdot 10^{6 \;-\; (-3)} \; \mbox{mg}\)

\(\displaystyle 0.25 \cdot 10^{6 + 3} \; \mbox{mg}\)

\(\displaystyle 0.25 \cdot 10^9 \; \mbox{mg}\)

Scientific notation requires moving the decimal point on 0.25

\(\displaystyle 2.5 \times 10^8 \; \mbox{mg}\)
 
OMG,
I did the second question and it was right!!!

thank you!!! :mrgreen:
 
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