Stupid question

[scotties123]

can we use calculators on the mcat? i know theres tricks to predicting things like pH and logs and stuff like that, but some physics questions seem impossible to calculate without a calculator.

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[fisko82]

can we use calculators on the mcat? i know theres tricks to predicting things like pH and logs and stuff like that, but some physics questions seem impossible to calculate without a calculator.

no. Also, like EK says, to paraphrase, if you're spending more than 2 minutes or so on any one problem in the physical science, it's likely you're missing something or making things harder than they appear to be.

 

[scotties123]

no. Also, like EK says, to paraphrase, if you're spending more than 2 minutes or so on any one problem in the physical science, it's likely you're missing something or making things harder than they appear to be.

thanks, but for example: ka = 4.3x10^-7, and u gotta find kb then use that to find pH. how wud u do this type of math without a calcultor?

 

[DrBowtie]

Intense calculations are rare on the MCAT. The saying goes "If you're doing rigorous calculations, you're probably doing the problem wrong"

 

[swim2006]

thanks, but for example: ka = 4.3x10^-7, and u gotta find kb then use that to find pH. how wud u do this type of math without a calcultor?

You can round the number and then do the math...like round to 4x10^-7 because the answer you get will be close to the correct answer

 

[fisko82]

thanks, but for example: ka = 4.3x10^-7, and u gotta find kb then use that to find pH. how wud u do this type of math without a calcultor?

is this is a polyprotic acid? In any event, I'd round 4.3 to 4, then say kw=ka*kb, then deal with an easy fraction (1/4th) or .25e-7 = kb. Then use kb to find OH concentration, assuming this is the typical kb=x^2 type problem, I'd prob convert .25e-7 to 2.5e-8 then take the square root of that and know that sq.root of e-8 is e-4 and I know that sq.root of 4 is 2 and sq.root of 1 is 1, so sq.root of 2 has to be between those to, so I'd guess its between that and I'd estimate it's around 1.5e-2 = OH concentration.

From that, pOH=-log OH, approximate this as being something slightly less than 2 and then use ph+poh=14 and get ph around 10. Something like that. This type of problem could be solved in under 2 minutes if you round and are good with estimations. In any event, I don't think I've seen any types of problems this drawn out. Is this a kaplan question?

 

[scotties123]

is this is a polyprotic acid? In any event, I'd round 4.3 to 4, then say kw=ka*kb, then deal with an easy fraction (1/4th) or .25e-7 = kb. Then use kb to find OH concentration, assuming this is the typical kb=x^2 type problem, I'd prob convert .25e-7 to 2.5e-8 then take the square root of that and know that sq.root of e-8 is e-4 and I know that sq.root of 4 is 2 and sq.root of 1 is 1, so sq.root of 2 has to be between those to, so I'd guess its between that and I'd estimate it's around 1.5e-2 = OH concentration.

From that, pOH=-log OH, approximate this as being something slightly less than 2 and then use ph+poh=14 and get ph around 10. Something like that. This type of problem could be solved in under 2 minutes if you round and are good with estimations. In any event, I don't think I've seen any types of problems this drawn out. Is this a kaplan question?

thanks for the help, my problem was not realizing to right them as fractions using 1x10^-14 and not just 10^-14. and its exam krackers