bohr theory explains that the emission spectral line is due to Page Practice mid
ID: 548855 • Letter: B
Question
bohr theory explains that the emission spectral line is due to Page Practice mid-ter 1. In the tollow ing reaction 2 KClO3(s)- 2 KCK s) + 3O2(g) 14.0 g KCIO3 yielded 1.40g KCL What is the percent yield? 2. An 8.07 g sample of impure Ag,O decomposes into solid silver and O, gas. If 395 ml of Ag) is collected over water at 25 C and 749.2 mmHg barometric pressure, then wh is the percent by mass of Ag O in the sample? The vapor pressure of water at 25 C is 23.8 mmHg, What is the volume of 02 collected STP? 3. Suppose that we have a mixture of N2ig) and O2(g) of unknown composition whose total pressure is 385 torr. If all the 02g) is removed from the mixture by reaction with phosphorus, which does not react directly with nitrogen, then the new pres sure is 250 torr. Calculate the mole fraction of each gas in the original mixture 4. Explain why gases tend to behave idea#y at high1emperatures and low, presaares and nonideally at low temperatures and high presswres 5. The Bohr theory explains that an emission spectral line is: A) due to an electron losing energy but keeping the same values of its four quantum numbers B) due to an electron losing energy and changing shells C) due to an interaction between electrons in two different principal shells D) due to an electron gaining energy and changing shells E) due to an increase in the principal quantum number n of an electron 6. When an electron in an atom goes from a high energy state to a low one, what occurs? A) Another electron goes from a low energy state to a high one. B) The aom moves faster C)Light is given off. D) This process is not possible. E) Light is absorbed 7. The comect electron configuration for Br- is B) [Ar 3d104s24p6 D)IArl 3dl04s24p5 E) [Nel 3d104s24p6Explanation / Answer
Q1
% yield = actual / theoretical * 100%
actual = 1.4 g of KCl
get theoreitcal KCl from
mol of KClO3 = mass/MW = 14/122.55 = 0.1142 mol of KClO3
mol of KCl = 0.1142
mass of KCl = mol*MW = 0.1142*74.5513 = 8.513 g of KCl
% yield = 1.40/8.513 *100
% yield = 16.44 %
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