Correct Answer: C) 3.76

Explanation

To answer this question, we first need to understand the combustion reaction of ethanol. The balanced equation for the combustion of ethanol (C?H?OH) is:

\(C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O\)

From the equation, we can see that 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of CO?. The question states that 3g of pure oxygen is used in the reaction. Oxygen has a molar mass of 32g/mol, so we can calculate the moles of oxygen:

\(\text{moles of } O_2 = \frac{3}{32}\)

Now, we can use the stoichiometry of the reaction to find out how many moles of CO? are produced. For every 3 moles of oxygen, 2 moles of CO? are produced. Hence, we can write:

\(\frac{\text{moles of } CO_2}{\text{moles of } O_2} = \frac{2}{3}\)

Substituting the moles of oxygen, we get:

\(\text{moles of } CO_2 = \frac{2}{3} \times \frac{3}{32}\)

After canceling out the 3's, we are left with:

\(\text{moles of } CO_2 = \frac{1}{16}\)

To find the number of CO? molecules, we multiply the moles of CO? by Avogadro's number (N = 6.02

Correct Answer: B) 100N

Correct Answer: B. 3.4

Explanation

Let the linear transformation T be represented by the following:

\(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} -1 & 7 \\ 4 & 11 \end{pmatrix}\).

\(3a + 2b = -1 ... (1)\)

\(-a + 4b = 7 ... (2)\)

\(3c + 2d = 4 ... (3)\)

\(-c + 4d = 11 ... (4)\)

\((2) \times 3 : -3a + 12b = 21... (5)\)

\((1) + (5) : 14b = 20 \implies b = \frac{10}{7}\)

Substitute for b in (1),

\(3a + 2(\frac{10}{7}) = -1 \implies 3a = \frac{-27}{7}\)

\(a = \frac{-9}{7}\)

Solving for c and d,

\((4) \times 3 : -3c + 12d = 33 ... (6)\)

\((3) + (6) : 14d = 37 \implies d = \frac{37}{14}\)

\(-c + 4(\frac{37}{14}) = 11 \implies -c = 11 - \frac{74}{7} = \frac{3}{7}\)

\(c = \frac{-3}{7}\)

\(\therefore T = \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix}\)

\(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)

From \(P : (x, y) \to (x + y, x + 2y)\)

\(P : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} x + y \\ x + 2y \end{pmatrix}\)

\(P : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)

\(\therefore P = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\end{pmatrix}\)

(b) \(TP = \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\)

= \(\begin{pmatrix} \frac{-9}{7} + \frac{10}{7} & \frac{-9}{7} + \frac{20}{7} \\ \frac{-3}{7} + \frac{37}{14} & \frac{-3}{7} + \frac{74}{14} \end{pmatrix}\)

= \(\begin{pmatrix} \frac{1}{7} & \frac{11}{7} \\ \frac{31}{14} & \frac{34}{7} \end{pmatrix}\)

(c) The image of X(4, 3) under TP:

\(TP : \begin{pmatrix} 4 \\ 3 \end{pmatrix} \to \begin{pmatrix} \frac{1}{7} & \frac{11}{7} \\ \frac{31}{17} & \frac{34}{7} \endd{pmatrix} \begin{pmatrix} 4 \\ 3 \end{pmatrix}\)

= \(\begin{pmatrix} \frac{4}{7} + \frac{33}{7} \\ \frac{62}{7} + \frac{102}{7} \end{pmatrix}\)

= \(\begin{pmatrix} \frac{37}{7} \\ \frac{164}{7} \end{pmatrix}\)

The image of X(4, 3) under TP is \((\frac{37}{7}, \frac{164}{7})\).

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