Because the solid is soluble in water, it cannot be calcium fluoride.
This leaves us with either silver fluoride or potassium bromide, as both are soluble in water.
Upon addition of sodium sulfate, a white precipitate forms, confirming that the substance was silver fluoride as the precipitate are of silver sulfate.
Observe a blue light be wavelength flame test during
I'm pretty sure its the weight of the book
Sorry it’s messy, hope this helps :]
<h3>
Answer:</h3>
4.73 × 10^4 m
<h3>
Explanation:</h3>
From the question;
Frequency of the photon = 634 × 10^12 Hz
We are required to calculate the wavelength of the photon.
We need to know the relationship between wavelength and frequency of a wave.
The relationship between f and λ is given by;
c = fλ
Where c, is the speed of light, 2.998 × 10^8 m/s
Therefore, to get the wavelength we rearrange the formula such that;
λ = c ÷ f
= 2.998 × 10^8 m/s ÷ 634 × 10^12 Hz
= 4.73 × 10^-5 m
But we require wavelength in nm
1 M = 10^9 nm
Therefore;
Wavelength = 4.73 × 10^-5 m × 10^9 nm/m
= 4.73 × 10^4 m
Hence, the photon's wavelength is 4.73 × 10^4 m
