To find the abscissa of point \( P \), we first determine the equation of the line connecting points \( (1, 2) \) and \( (5, 6) \).
The slope of this line is: \[ m = \frac{6 - 2}{5 - 1} = 1 \] The equation of the line is: \[ y - 2 = 1(x - 1) \quad \Rightarrow \quad y = x + 1 \] Now, since the point \( P \) is on the \( x \)-axis, its \( y \)-coordinate is 0.
Substituting \( y = 0 \) into the equation of the line: \[ 0 = x + 1 \quad \Rightarrow \quad x = -1 \] Thus, the abscissa of the point \( P \) is 2.
The area bounded by the parabola \(y = x^2 + 2\) and the lines \(y = x\), \(x = 1\) and \(x = 2\) (in square units) is:
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: