Sorasyn

I've corrected it now, thank you. :) I took the wrong approach, and didn't fully complete the question. I gave him bounds by accident instead of following through.

H₂SO₄ is classified as a "strong" acid, which means it dissociates completely. Knowing this, we can surmise that the molarity of hydrogen ions will be the same as the molarity of our given acidic solution (H₂SO₄), at 0.0005M. 

$$[H_2SO_4] = 0.0005M$$

Dissociates to equal concentrations of the ions H⁺, and HSO₄^-:

$$[H^+] = 0.0005M$$

$$[HSO_4^-] = 0.0005M$$

Thus the molarity of hydrogen atoms up to this point is 0.0005M, giving a pH = -log([H⁺]) = -log(0.0005) = 3.

Now, we need to calculate the molarity of hydrogen dissociated from the decidedly "weak" acidic ion HSO₄^-. Since HSO₄^- is a weak acid, we need to find and use the dissociation constant of HSO₄^- to calculate the molarity of H⁺ from the parent compound. 

$$Ka = [H^+][SO_4^2^-]/[HSO_4^-] = 1.1 x 10^-^2$$

Solving for Ka will give us our desired hydrogen molarity, so consider the following:

$$([H^+]*[SO_4^2^-]) / [HSO_4^-] = 1.1 x 10^-^2$$

We know that [H⁺] = [SO₄^2-] = 0.0005M so we can consider those, respectively, as a variable, "x" for example. We also know that the concentration of [HSO₄^-] will be it's curent concentration MINUS the concentration of those that dissociated further to the ions [H⁺] and [SO₄^2-]; giving us:

$${\frac{\left({{\mathtt{x}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{0.000\: \!5}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}} = {\mathtt{1.1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{2}}}$$

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.011}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{0.000\: \!005\: \!5}} = {\mathtt{0}}$$

Solving for "x", we'll get:

$$x = [H^+] = 0.00047913$$

Which indicates our [H⁺] concentration for the second hand dissociation of H₂SO₄, or the first hand dissociation of HSO₄^-. To get the total concentration of [H⁺] ions in the solution, you simply add them together and run the resulting number through the pH formula.

 $$0.0005M + 0.00047913M = 0.00097913M$$

$$pH = -log([H^+]) = -log(0.00097913) = 3.009 \approx 3.01$$

Hope this helps. Thank you Melody, and CPhil, for pointing out my mistake. I've corrected it now. :)

Get a common denominator, and then solve the equation like so:

$$\left({\frac{{\mathtt{4}}}{{\mathtt{5}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{7}}}}\right) = {\mathtt{6}}$$

$$\left({\frac{{\mathtt{28}}}{{\mathtt{35}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{35}}}}\right) = {\mathtt{6}}$$

$${\frac{\left({\mathtt{28}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}{{\mathtt{35}}}} = {\mathtt{6}}$$

$${\mathtt{28}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{210}}$$

$${\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{182}}$$

$${\mathtt{x}} = {\frac{{\mathtt{182}}}{{\mathtt{20}}}}$$

Distributive Property is exactly what it sounds. When you distribute something, you're giving equal quantities to all parties, right? The same applies to mathematics. You're multiplying, or "distributing," one factor amongst another factor, or set of factors.

$$4(x + 2) \Rightarrow 4x + 8$$

$$3(x + 3) \Rightarrow 3x + 9$$

$$(x + 2)(x + 2) \Rightarrow x^2 + 2x + 2x + 4 \Rightarrow x^2 + 4x + 4$$

Think of it as a product rather than a quotient. 

$${\frac{\left({\frac{{\mathtt{53}}}{{\mathtt{552}}}}\right)}{{\mathtt{100}}}} = \left({\frac{{\mathtt{53}}}{{\mathtt{552}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{100}}}}\right)$$

Which gives us a final fraction of:  $$\left({\frac{{\mathtt{53}}}{{\mathtt{55\,200}}}}\right)$$

Scientific Notation is just moving decimals around while maintaining the consistency of the number. It's manipulating large or small numbers with powers of ten. Multiplying a number by ten shifts a decimal place to the right, similarly dividing shifts a decimal place to the left. 

$${\mathtt{24.91}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{1}}} = {\mathtt{249.1}}$$ 

$${\mathtt{24.91}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{0}}} = {\mathtt{24.91}}$$

$${\mathtt{24.91}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{1}}} = {\mathtt{2.491}}$$

To know which way to shift the decimal is solely reliant on the power of ten. 

$${\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{X}}}$$

Should X be positive then the decimal will shift X places to the right, otherwise X is negative and should be shifted X places to the left. 

Hope this helps. 

Nauseated : Did you happen to miss where I stated I was fuzzy about the question? Congratulations on establishing yourself as an incredibly moronic and condescending idiot. (Moronic Condescending Idiot disorder, MCI for short.) Pull your head out of your a*s.

Not sure I understand what your question is. To balance a chemical equation the sides must be equal to one another. 

When Sodium Chlorate, NaClO₃, decomposes it produces Sodium Chloride, NaCl. Using this we can surmise that the missing compound in our balanced equation is Ozone, O₃:

$$NaClO3 - NaCl \Rightarrow O3$$

Now, we piece the balanced equation back together as:

$$NaClO3 \Rightarrow NaCl + O3$$

Note: Ozone, O₃, will further decompose to O₂ as Ozone is too unstable under higher atmospheric pressure.

$$O3 \Rightarrow O2 + O$$

0.333333333 ... = 1/3

So you can think of 21.33333 as:

$${\mathtt{21}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}$$

Multiply 21 by 3/3 to allow them to be added:

$${\frac{{\mathtt{63}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}} = {\frac{{\mathtt{64}}}{{\mathtt{3}}}}$$

Converting between units in Chemistry is as follows:

$$Grams\Leftrightarrow Mols\Leftrightarrow Atoms$$

So, firstly, you need to convert from grams to mols of Fe (gFe is grams of Fe -- Note that they cancel):

Fe = 55.845g per mol

$$\left({\mathtt{500}}{\mathtt{\,\times\,}}{\mathtt{gFe}}\right){\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molFe}}\right)}{\left({\mathtt{55.845}}{\mathtt{\,\times\,}}{\mathtt{gFe}}\right)}}\right) = {\mathtt{8.953}}{\mathtt{\,\times\,}}{\mathtt{molFe}}$$ 

Lastly you'll need to convert from mols to atoms (molFe is mols of Fe -- Again note that they cancel):

The number of atoms per mol is represented by Avogadro's Constant = 6.022 x 10^23.

$$\left({\mathtt{8.953}}{\mathtt{\,\times\,}}{\mathtt{molFe}}\right){\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{6.022}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{23}}}{\mathtt{\,\times\,}}{\mathtt{atomsFe}}\right)}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molFe}}\right)}}\right) = {\mathtt{5.39}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{24}}}{\mathtt{\,\times\,}}{\mathtt{atomsFe}}$$

5.39 x 10^24 atoms of Fe in 500g of Fe.