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Stella [2.4K]
3 years ago
8

jin has four colorful glass bottles on his windowsill. one is red, one is violet, one is green, and one is yellow. the light tha

t passes through which bottle has the lowest frequency? red violet green yellow

Chemistry
2 answers:
Over [174]3 years ago
4 0
The red bottle would have the lowest frequency because red light has the longest wavelengths. The light passing through the violet would have the highest frequency because its wavelengths are the shortest.
Sloan [31]3 years ago
3 0

Answer: The light passing through Red bottle will have the lowest frequency.

Explanation: Relation between wavelength and frequency follows:

\nu =\frac{c}{\lambda }

As, from the above relation, it is clear that the wavelength and frequency follows inverse relation, so the light having highest wavelength will have the lowest frequency.

From the spectra of visible region, the wavelength of Red light is maximum and hence, it will have the lowest frequency.

Therefore, the light passing through the bottle which is red in color will have the lowest frequency.

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Answer:

contain hereditary information

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Metabolism occurs in the cells. The cells use oxygen to break down food materials to produce energy.

5 0
3 years ago
An ideal liquid mixture of benzene and toluene (50% by volume each) is used in a plant at 27°C and 1 atm. Determine (a) the mixt
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See explaination

Explanation:

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7 0
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Is bioluminescence an endothermic process bioluminescence an endothermic process or exothermic process?\?
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6 0
3 years ago
would a cloud or a rock be more dense? PLSSSSSS HELP PLSSS I WILL MARK BRAINLIEST AND ANSWER SOME OF YOUR QUESTIONS
QveST [7]
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6 0
3 years ago
Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne
777dan777 [17]

Answer:

Explanation:

From the information given:

CaF_2 \to Ca^{2+} + 2F^-

Ksp = 3.2 \times 10^{-11}

no of moles of Ca^{2+} = 0.01 L × 0.0010 mol/L

no of moles of Ca^{2+} = 1 \times 10^{-5} \ mol

no of moles of F^- = 0.01 L × 0.00010 mol/L

no of moles of F^- = 1 \times 10^{-6}\ mol

Total volume = 0.02 L

[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\  \\  \[[Ca^{2+}}] = 0.0005 \ mol/L

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[F^{-}] = 5 \times 10^{-5}  \ mol/L

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Since Q<ksp, then there will no be any precipitation of CaF2

3 0
3 years ago
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