# Tutor profile: Melanie T.

## Questions

### Subject: Pre-Calculus

Let $$f(x)=\sqrt{9-x}$$ and let $$g(x)=\sqrt{\frac{1}{9-x^2}}$$. What is the domain of each function?

The domain of $$f(x)$$ is the interval $$(-\infty,9]$$ and the domain of $$g(x)$$ is the interval $$(-3, 3)$$.

### Subject: Chemistry

What is the molarity of 5.30 g of Na$$_{2}$$CO$$_{3}$$ dissolved in 400.0 mL solution?

Answer: 0.125 M Solution: Molarity is equal to number moles divided by volume: $$M=\frac{n}{V}$$ where M=molarity, n=number of moles, and V=volume (in liters) In order to find the number of moles of Na$$_{2}$$CO$$_{3}$$, we must first calculate the molar mass of this compound. Looking at the periodic table, we see that the molar mass of Na is 22.99 g/mol, the molar mass of C is 12.01 g/mol, and the molar mass of O is 16.00 g/mol. The molecular formula of the compound tells us there are two Na atoms, one C atom, and three O atoms, hence: $$(22.99\frac{g}{mol})(2)=45.98 \frac{g}{mol} Na$$ $$(12.01\frac{g}{mol})(1)=12.01 \frac{g}{mol} C$$ $$(16.00\frac{g}{mol})(3)=48.00 \frac{g}{mol} O$$ Therefore the molar mass of the entire compound, Na$$_{2}$$CO$$_{3}$$, is: $$45.98 \frac{g}{mol}+12.01 \frac{g}{mol}+48.00 \frac{g}{mol}=105.99\frac{g}{mol}$$ Now that we have the molar mass of Na$$_{2}$$CO$$_{3}$$, we divide the given amount of the compound (5.30 g) by the molar mass (105.99 g/mol). The mass units (g) will cancel and we will be left with a number of moles: $$(5.30g)(\frac{1mol}{105.99g})=0.050$$ mol Na$$_{2}$$CO$$_{3}$$ Next, we must convert the given 400 mL of solution into liters: $$(400mL)(\frac{1L}{1000mL})=0.4L$$ Now that we have number of moles of Na$$_{2}$$CO$$_{3}$$ (0.050 mol) and liters of solution (0.4 L), we can calculate molarity: $$M=\frac{0.050mol}{0.4L}=0.125M$$ Therefore, the solution is 0.125 M.

### Subject: Calculus

Evaluate the given integral: $$ \int(4x^6-2x^3+7x-4)\mathrm{d}x $$

$$\frac{4}{7}x^7-\frac{1}{2}x^4+\frac{7}{2}x^2-4x+c$$ where $$c$$ is an arbitrary constant.