Question

1 multiple choice 1 point
what color of light would have a frequency of 4.4 x 10¹⁴ hz? youll most likely need to convert the frequency to a wavelength first.
green
red
yellow
orange
2 multiple choice 1 point
a compound with the empirical formula ch₃ has a molecular mass of 60 amu. what is the molecular formula for this molecule?
ch₃
c₄h₁₂
c₃h₉
c₂h₆

Explanation:

Question 1

Step1: Recall the formula relating frequency (\(f\)) and wavelength (\(\lambda\))

The formula is \(c = f\lambda\), where \(c = 3.0\times10^{8}\, \text{m/s}\) (speed of light in vacuum). We need to solve for \(\lambda\): \(\lambda=\frac{c}{f}\)

Step2: Substitute the values

Given \(f = 4.4\times10^{14}\, \text{Hz}\), \(c = 3.0\times10^{8}\, \text{m/s}\)
\(\lambda=\frac{3.0\times10^{8}\, \text{m/s}}{4.4\times10^{14}\, \text{Hz}}\approx6.82\times10^{-7}\, \text{m}=682\, \text{nm}\)

Step3: Relate wavelength to color

Visible light wavelengths: Red is around 620 - 750 nm, green ~495 - 570 nm, yellow ~570 - 590 nm, orange ~590 - 620 nm. 682 nm is in the red range. Wait, no—wait, maybe miscalculation? Wait, \(3e8 / 4.4e14 = (3/4.4)e-6 ≈ 0.6818e-6 m = 681.8 nm\), which is red? Wait, but maybe I mixed up. Wait, green is lower frequency (higher wavelength? No, frequency and wavelength are inversely proportional. Higher frequency means lower wavelength. Wait, 4.4e14 Hz: let's check standard wavelengths. Red: ~4.3e14 to 4.8e14? Wait no, actually, frequency of red is lower (longer wavelength), violet is higher (shorter wavelength). Wait, the frequency of green is around 5.4e14 Hz, red is around 4.6e14 Hz? Wait, maybe my initial wavelength - color mapping was reversed. Let's correct: wavelength (nm) to frequency (Hz): \(f = c/\lambda\), so \(\lambda = c/f\). So for \(\lambda = 680 nm\) (red), \(f = 3e8 / 680e-9 ≈ 4.41e14 Hz\), which matches 4.4e14 Hz. So the color is red? Wait, but the options: green, red, yellow, orange. Wait, 682 nm is red. Wait, but maybe I made a mistake. Wait, let's recalculate: \(3\times10^{8}\, \text{m/s} = 3\times10^{17}\, \text{nm/s}\) (since 1 m = 1e9 nm). Then \(\lambda=\frac{3\times10^{17}\, \text{nm/s}}{4.4\times10^{14}\, \text{Hz}}≈681.8\, \text{nm}\). Red light is typically in the range of 620 - 750 nm, so 682 nm is red. Wait, but the options: the first option is green, then red. Wait, maybe I messed up. Wait, no—wait, maybe the frequency of green is higher. Let's check standard values:

• Red: ~620 - 750 nm, frequency ~4.0 - 4.8e14 Hz
• Green: ~495 - 570 nm, frequency ~5.3 - 6.1e14 Hz
• Yellow: ~570 - 590 nm, frequency ~5.1 - 5.3e14 Hz
• Orange: ~590 - 620 nm, frequency ~4.8 - 5.1e14 Hz

So 4.4e14 Hz is in the red range (since red's frequency is ~4.0 - 4.8e14 Hz). So the color is red.

Step1: Calculate the molar mass of the empirical formula

Empirical formula \(CH_3\). Molar mass of \(C = 12\) amu, \(H = 1\) amu. So molar mass of \(CH_3 = 12 + 3(1)=15\) amu.

Step2: Find the ratio of molecular mass to empirical formula mass

Molecular mass is 60 amu. Ratio \(n=\frac{60}{15}=4\)

Step3: Multiply the empirical formula by \(n\)

Molecular formula = \((CH_3)_n=(CH_3)_4 = C_4H_{12}\) (since 14 C, 34 H)

Answer:

red (the option with "red")

Question 2