Question

116 minutes remaining | 13 of 26 questions remaining
calculate the energy (in j) of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49×10^14 hz. express numerical answer only in scientific notation
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integer, decimal, or e notation allowed
question 15
the orbital shape below is an example of which atomic orbital?
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question 16
a block of metal has a mass of 29.28 g and displaces 58.79 ml of water. calculate the density of the metal in g/ml
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Explanation:

Question 13

Step1: Recall energy - frequency formula

The formula for the energy of a photon is $E = h
u$, where $h$ is Planck's constant ($h=6.626\times 10^{- 34}\ J\cdot s$) and $
u$ is the frequency.

Step2: Substitute values

Given $
u = 5.49\times 10^{14}\ Hz$, then $E=(6.626\times 10^{-34}\ J\cdot s)\times(5.49\times 10^{14}\ Hz)$.
Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $E=(6.626\times5.49)\times10^{-34 + 14}\ J$.
$6.626\times5.49 = 36.37674$, so $E = 3.637674\times 10^{-19}\ J$.
Rounding to three - significant figures, $E=3.64\times 10^{-19}\ J$.

The dumb - bell shaped orbital is a p - orbital. Atomic orbitals are regions in an atom where electrons are likely to be found. The p - orbitals have a dumb - bell shape.

Step1: Recall density formula

The density formula is $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass, and $V$ is volume.

Step2: Substitute given values

Given $m = 29.28\ g$ and $V = 58.79\ ml$.
$
ho=\frac{29.28\ g}{58.79\ ml}\approx0.498\ g/ml$.

Answer:

$3.64\times 10^{-19}$

Question 15