Hello!
Find the Energy of the Photon by Planck's Equation, given:
E (photon energy) =? (in Joule)
h (Planck's constant) = ![6.626*10^{-34}\:J * s](https://tex.z-dn.net/?f=%206.626%2A10%5E%7B-34%7D%5C%3AJ%20%2A%20s%20)
f (radiation frequency) =
Therefore, we have:
![E = h*f](https://tex.z-dn.net/?f=%20E%20%3D%20h%2Af%20)
![E = 6.626*10^{-34}*8*10^{12}](https://tex.z-dn.net/?f=%20E%20%3D%206.626%2A10%5E%7B-34%7D%2A8%2A10%5E%7B12%7D%20)
![E = 53.008*10^{-34+12}](https://tex.z-dn.net/?f=%20E%20%3D%2053.008%2A10%5E%7B-34%2B12%7D%20)
![E = 53.008*10^{-22}](https://tex.z-dn.net/?f=%20E%20%3D%2053.008%2A10%5E%7B-22%7D%20)
![\boxed{\boxed{E = 5.3008*10^{-21}\:Joule}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Cboxed%7BE%20%3D%205.3008%2A10%5E%7B-21%7D%5C%3AJoule%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark%20)
I Hope this helps, greetings ... DexteR! =)
Answer:
The empirical formula is CH2O, and the molecular formula is some multiple of this
Explanation:
In 100 g of the unknown, there are 40.0⋅g12.011⋅g⋅mol−1 C; 6.7⋅g1.00794⋅g⋅mol−1 H; and 53.5⋅g16.00⋅g⋅mol−1 O.
We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers. Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.Clearly n=2, and the molecular formula is 2×(CH2O) = CxHyOz.
Answer:
Are you in flvs, if so im prettyb sure if yo look on page 3 of lesson 1.04 it tells you the answer.
Explanation:
![\LARGE{ \boxed{ \purple{ \rm{Answer}}}}](https://tex.z-dn.net/?f=%20%5CLARGE%7B%20%5Cboxed%7B%20%5Cpurple%7B%20%5Crm%7BAnswer%7D%7D%7D%7D)
☃️ Chemical formulae ➝ ![\sf{I_2}](https://tex.z-dn.net/?f=%5Csf%7BI_2%7D)
How to find?
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
![\boxed{ \sf{No. \: of \: moles = \frac{Given \: weight}{Molecular \: weight} }}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Csf%7BNo.%20%5C%3A%20of%20%5C%3A%20moles%20%3D%20%20%5Cfrac%7BGiven%20%5C%3A%20weight%7D%7BMolecular%20%5C%3A%20weight%7D%20%7D%7D)
Solution:
❍ Molecular weight of ![\sf{I_2}](https://tex.z-dn.net/?f=%5Csf%7BI_2%7D)
= 2 × 126.90
= 253.80
= 254 (approx.)
❍ Given weight: 12.7
Then, no. of moles,
⇛ No. of moles = 12.7 / 254
⇛ No. of moles = 0.05 moles
⚘ No. of moles of Iodine molecule in the given weight = <u>0.05</u><u> </u><u>moles </u>
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