3.5- Calculating an average atomic mass is like doing a weighted average to find your grade. You take the relative abundance (%) times the atomic masses of the respective isotopes, and add. So, 34.968 x 0.7553 + 36.956 x 0.2447 = answer. The answer should match the number on the periodic table.
3.65- Multiply by the mole ratio using train tracks. 3.60 mol CO x 2 mol CO2 /2 mol CO. The question was asking for moles of CO2 so you're done.
3.77- grams -> moles -> moles-> grams. Convert 1E5 to grams, then to moles of (NH4)2SO4. Then multiply by the mole ratio (moles NH3/ moles (NH4)2SO4). Then covert moles of NH3 to grams, then to kg. It can all be done on train tracks. See example 3:13. It is similar but doesn't have the gram<-> kg conversions.
3.83- Do the stoichiometry problem twice. Multiply each amount by the respective mole ratio. So, 0.886 mol NO x 2 NO2/2NO. gives 0.886 moles IF NO is limiting. Now, 0.503 mol O2 x 2NO2/1O2 makes 1.006 moles IF O2 is limiting. Thus, the limiting reactant is NO and the maximum amount of NO2 is 0.886 moles. See the top half of pg 104 for an example.
3.89- To calculate percent yield you need to a stoichiometry problem to find out the theoretical number of kg of HF. So, grams ->moles -> moles -> grams again. Then, once you have a theoretical number of kg of HF, find percent yield. Take the actual yield (2.86 kg) and divide by theoretical. Lastly, turn it into a percent by multiplying by 100. Example 3:16 can help.
2:57b- The ion HPO4(-2) is the hydrogen phosphate ion. See pg 61. Then, put it together with potassium. Potassium hydrogen phosphate.
1:47- use the train tracks. Start with 62 m on top and 1 s on bottom. There are 1000 m in a km and 0.621 mi in a km. Then, convert seconds to minutes and then to hours.
1:49- use train tracks again. a) start with the speed of light 3.00e8 m/s with 3.00e8 m on top and 1 sec on bottom. Then, covert meters to km and then to miles like in 47. Then convert seconds to minutes to hours to days to years. NOW you should have a number with units miles/year. Multiply this by 1.4 years to get miles
So, the key here is that not all the oxygen that is inhaled can be used. (oxygen is breathed out) 20% of incoming air was oxygen and 16 percent of outgoing air was oxygen. Thus, only 4% of the incoming air gets to count as "oxygen". This means that 4% of the total breath volume is the oxygen that can be used (the 20 mL per breath). So, 0.04 x (breath volume) = 20 mL.
On the list of assigned problems, the second 39 should be 49.
I'll answer 58m and 59 and then get back to 58f.
58m- This one is sodium peroxide. O2 (2-) is the peroxide ion, each oxygen essentially has a -1 charge but they always come in a pair- O2 (-2). That's why it can't be NaO. The peroxide ion is a tricky one and usually you'll only see it show up in hydrogen peroxide- H2O2.
59c,e,f- with these, it's just like adding a hydrgen (or 2 in the case of di or 3 in the case of tri) to the anion which decreases the charge of the anion. For example-sodium hydrogen sulfide is Na (+1) and HS-1 so you get NaHS. Calcium hydrogen phosphate would be Ca+2 and HPO4(-2) so it makes CaHPO4. In potassium dihydrogen phosphate, you add 2 hydrogens to the phosphate (decreasing the charge by 2 this time). so K+1 and H2PO4-1 makes KH2PO4. If it had been potassium hydrogen phosphate, you would only add one hydrogen so you'd get K+1 and HPO4-2 so that would make K2HPO4.
So now for the oxoacids...
not so bad with a few rules...They always start with hydrogen and have a polyatomic ion with oxygen. Once you recognize it as an acid, follow these rules-
Look at the polyatomic anion. Name it. Then change the ending as follows:
So, for example... HClO4. Ion name is perchlorate so it becomes perchloric acid.
HClO3 Ion name is chlorate so the acid is chloric acid.
HClO2 Ion name is chlorite so the acid is chlorous acid.
HClO Ion name is hypochlorite so the acid is hypochlorous acid.
SO, for HIO, the ion is hypoiodite so the acid is hypoiodous acid. (a weird acid that we don't really use)