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The emission spectrum of an unknown element contains two lines-one in the visible portion of spectrum and the other, ultraviolet. Based what you have learned about Niels Bohr's model of atom, account for the difference in energy between these lines.
55.0 mL of a 1.70 M solution is diluted to a volume of 288 mL. A 144 mL portion of that solution is diluted using 181 mL of water. What is the final concentration?
The mass spectrum of an organic compound shows the relative abundances of M to be 58.09% and M 1 to be 15.08%. Assuming the peaks are caused by 12C and 13C isotopes, determine the number of carbon atoms in the compound
If 3.50 g of H2S are used in the above reaction, what will be the theoretical yield of water in grams?
The standard heat of formation of WO3(s) is -843 KJ/mol, and its standard molar entropy
A sample of He gas (V = 3.0 L) at 5.6 atm and 25 °C was combined with 4.5 L of Ne gas at 3.6 atm and 25 °C in a 9.0 L flask (which was initially evacuated).
What technology or combination of technologies would you recommend for treating the following contaminated air streams?
Calculate [OH] for a solution formed by adding 5.00 mL of 0.140M KOH to 15.0 mL of 9.0×10-2M Ca(OH)2.
determine the structures of Compounds A, B, C, D, and E. Show calculations. Support all of your structure determinations.
This question deals with the reaction between benzene and 1-chloro-2-methylpropane in the presence of AlCl3. Knowing that the arenium ion intermediate has positive charge on the ortho and para carbons
The black silver sulfide discoloration of silverware can easily be removed by heating the silver article in a sodium carbonate solution in an aluminum pan.
Consider the voltaic cell , which is based on the following cell reaction. Zn(s) + Cu2+(aq) Zn2+(aq) +Cu(s) Under standard conditions
Write a balanced combustion reaction for natural gas composed of 90 mol% methane and 10 mol% ethane. [Fractional moles are convenient for the subsequents calculations. Your reaction should have .90 mol methane and .10 mol ethane on the left side. ..
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