# Tutor profile: Megan S.

## Questions

### Subject: Chemistry

Show the net reaction of the following balanced equations. Balanced equations: $$\textrm{IO}_3^-$$ (aq) + $$6\textrm{H}^+$$ (aq) + $$8\textrm{I}^-$$ (aq) $$\rightarrow$$ $$3\textrm{I}_3^-$$ (aq) + $$3\textrm{H}_2\textrm{O}$$ (l) $$\textrm{I}_3^-$$ (aq) + $$2\textrm{S}_2\textrm{O}_3^{2-}$$ (aq) $$\rightarrow$$ $$3\textrm{I}^-$$ (aq) + $$\textrm{S}_4\textrm{O}_6^{2-}$$ (aq)

The first step in this problem is to solve for the net reaction of the balanced equations above. In order to solve for a net reaction, you first cancel out the spectator ions, which are those that appear on both sides of the equations. $$\textrm{IO}_3^-$$ (aq) + $$6\textrm{H}^+$$ (aq) + $$8\textrm{I}^-$$ (aq) $$\rightarrow$$ $$3\textrm{I}_3^-$$ (aq) + $$3\textrm{H}_2\textrm{O}$$ (l) $$\textrm{I}_3^-$$ (aq) + $$2\textrm{S}_2\textrm{O}_3^{2-}$$ (aq) $$\rightarrow$$ $$3\textrm{I}^-$$ (aq) + $$\textrm{S}_4\textrm{O}_6^{2-}$$ (aq) There are $$\textrm{I}^-$$ and $$\textrm{I}_3^-$$ on both sides of the equation. The way to cancel out the spectator ions is to "subtract" the alike ions from one another. The alike ions with the larger coefficient (the number in front of the ion) is the greater one and the one that gets to "keep" that ion on their side. So, for our equation we would "subtract" alike ions: $$8\textrm{I}^- - 3\textrm{I}^-$$ and $$3\textrm{I}_3^- - \textrm{I}_3^-$$. Next step in solving for a net reaction is putting all terms in both equations on the left side of the arrow into one equation (still on the left side) and all terms in both equations on the right side of the arrow into one equation (still on the right side). This leaves us with our net reaction: $$\textrm{IO}_3^-$$ (aq) + $$6\textrm{H}^+$$ (aq) + $$5\textrm{I}^-$$ (aq) + $$2\textrm{S}_2\textrm{O}_3^{2-}$$ (aq) $$\rightarrow$$ $$2\textrm{I}_3^-$$ (aq) + $$\textrm{S}_4\textrm{O}_6^{2-}$$ (aq) + $$3\textrm{H}_2\textrm{O}$$ (l)

### Subject: Calculus

Let $$f(x) = \sqrt{x}$$. Find $$c$$ if the rate of change of $$f$$ at $$x = c$$ is twice the rate of change at $$x = 1$$.

We first start with figuring out the rate of change of $$f$$, which is the derivate of $$f$$. rate of change = $$\frac{d}{dx} (\sqrt{x})$$ rate of change = $$\frac{1}{2}(x)^{-\frac{1}{2}}$$ rate of change = $$\frac{1}{2\sqrt{x}}$$ The next step is solving the rate of change at $$x=1$$. rate of change = $$\frac{1}{2\sqrt{1}}$$ The square root of 1 is 1. So that leaves us with: rate of change = $$\frac{1}{2}$$. We know from the information provided in the problem that the rate of change at $$x=c$$ is twice the rate of change at $$x=1$$. We know that the rate of change at $$x=1$$ is $$\frac{1}{2}$$. Therefore, we multiply the rate of change at $$x=1$$ by 2 to get the rate of change at $$x=c$$. $$\frac{1}{2}*2 = \frac{2}{2} = 1$$ Lastly we have to solve for $$c$$. The rate of change of $$f(x)$$ at $$c$$ is $$f'(c) = \frac{1}{2\sqrt{c}}$$. We know from the previous step that the rate of change of $$f(x)$$ at $$c$$ is 1. Therefore, we set these two equations equal to each other: $$\frac{1}{2\sqrt{c}} = 1$$. To solve for $$c$$, we need to get $$c$$ on one side of the equation and the numbers on the other side of the equation. In order to do this, my approach is to first cross multiply the equations: $$\frac{1}{2\sqrt{c}} = \frac{1}{1}$$. This gives us: $$1=2\sqrt{c}$$. Then we divide both sides of the equation by 2 (in order to get $$c$$ by itself): $$\frac{1}{2}=\frac{2\sqrt{c}}{2}$$. This gives us: $$\frac{1}{2}=\sqrt{c}$$. The last step is to square (or raise to the power of two) to both sides. This is done because squaring and taking the square root are opposites, so when you square a square root, your answer will be the number/variable/expression under the square root. $$(\frac{1}{2})^2=(\sqrt{c})^2$$. The answer is: $$c=\frac{1}{4}$$.

### Subject: Algebra

Simplify the expression below: 3(a-2)+5b-4(a+b-1)+3

The first step in simplifying an expression is to multiple out the factors. This gives us the below expression: 3a-6+5b-4a-4b+4+3 The second step is to group together "like" terms: 3a-4a+5b-4b-6+4+3 The third step is to add or to subtract "like" terms: -1a+1b+1