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# Tutor profile: Sudibyo G.

Sudibyo G.
Specialist Tutor for Maths, Physics and Chemistry

## Questions

### Subject:Organic Chemistry

TutorMe
Question:

Why $$SOCl_2$$ is used as a preferred reagent to convert alcohol $$R-OH$$ to chloroalkane $$R-Cl$$ rather than $$PCl_5$$?

Sudibyo G.
Answer:

Below is the chemical equation when alcohol reacts with the respective reagents: $$R-OH\ +\ PCl_5 \to \ R-Cl\ +\ HCl\ +\ POCl_3$$ $$R-OH\ +\ SOCl_2 \to \ R-Cl\ +\ HCl +\ SO_2$$ When using $$SOCl_2$$, the products are gaseous $$\ HCl$$ and $$SO_2$$, which can easily be cleaned up using pyridine (a base) to obtain a pure sample of $$R-Cl$$ which is the desired end product. On the other hand, the purification of $$R-Cl$$ is more tedious when using $$PCl_5$$.

### Subject:Pre-Calculus

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Question:

How do we know if a function is continuous over a given interval?

Sudibyo G.
Answer:

Basically, there are 3 conditions to check for continuity: 1) If $$f(x)$$ is continuous at a point $$a$$, then $$\lim_{x\to a} f(x)=f(a)$$ 2)if $$f(x)$$ is defined for $$x\geq a$$, then $$\lim_{x\to a^+} f(x)=f(a)$$ 3) 2)if $$f(x)$$ is defined for $$x\leq a$$, then $$\lim_{x\to a^-} f(x)=f(a)$$ This means $$\lim_{x\to a} f(x)=\lim_{x\to a^+} f(x)=\lim_{x\to a^-}f(x) =f(a)$$. In simple words, this implies that the value of the limit of the function will be equal to the value of the function at $$a$$ (i.e $$f(a))$$, even when approaching from the left side of $$a$$ (i.e $$x \to a^-$$) or from the right side of $$a$$ (i.e $$x \to a^+$$) And the function is continuous on an interval ($$x_1,x_2)$$ if it's continuous at every point $$a$$, where $$a\in (x_1,x_2)$$.

### Subject:Calculus

TutorMe
Question:

Evaluate the following indefinite integral as a power series, $$\int \frac{x^{2}}{1-x^{3}}\ {d}x$$.

Sudibyo G.
Answer:

Using the standard expression for Maclaurin series for $$\frac{1}{1-x}= \sum_{n=0}^{\infty} x^n , \mid x \mid <1.$$ We can express $$\frac{x^{2}}{1-x^{3}} = x^2.\frac{1}{1-x^3}=x^2\sum_{n=0}^{\infty} (x^3)^n= \sum_{n=0}^{\infty} x^{3n+2}$$. (Notice we replace $$x$$ with $$x^3$$ in the standard expansion) Thus, $$\int \frac{x^{2}}{1-x^{3}}\ {d}x = \int \sum_{n=0}^{\infty} x^{3n+2}\ {d}x = C+\sum_{n=0}^{\infty} \frac{x^{3n+3}}{3n+3}$$ for some constant $$C$$.

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