A particular color of light has a wavelength of 663 nm. What is the energy possessed by the light?
(Planck’s constant $h = 6.63 \times 10^{-34}$ J·s; Velocity of light $c = 3 \times 10^8$ m/s)
$3.0 \times 10^{-19}$ J
Step 1: The energy of a photon is given by: \[ E = \frac{hc}{\lambda} \] where: - $h = 6.63 \times 10^{-34}$ J·s - $c = 3 \times 10^8$ m/s - $\lambda = 663$ nm = $663 \times 10^{-9}$ m
Step 2: Substituting values: \[ E = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{663 \times 10^{-9}} \] Step 3: Calculating the numerator: \[ (6.63 \times 10^{-34}) \times (3 \times 10^8) = 1.989 \times 10^{-25} \] Step 4: Dividing by the denominator: \[ E = \frac{1.989 \times 10^{-25}}{663 \times 10^{-9}} \] \[ E = 3.0 \times 10^{-19} { J} \] Step 5: Therefore, the correct answer is (E).
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:
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
Then \( f'(-4) \) is equal to:If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to: