anonymous
  • anonymous
can someone please help me find these values?
Chemistry
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
@phi
anonymous
  • anonymous
@robtobey

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NoelGreco
  • NoelGreco
For someone to comment, he's have to know more about the lab.
aaronq
  • aaronq
the moles you can simply find by dividing the masses by their molar mass... for the volume of H2 formed, you need to write the equation of the reaction in your exp and work with that.
anonymous
  • anonymous
@aaronq so for magnesium it would be 24.31/0.035?
aaronq
  • aaronq
the other way around
anonymous
  • anonymous
oh ok and for h2? @aaronq
aaronq
  • aaronq
you have to find the volume first
anonymous
  • anonymous
is the volume of hydrogen gas produced the same as teh volume of liquid displaced? @aaronq
aaronq
  • aaronq
unless you made other gases in your experiment
aaronq
  • aaronq
yes
anonymous
  • anonymous
no i didnt
anonymous
  • anonymous
so the volume divided by the mass of h2 will give me the moles of h2?
aaronq
  • aaronq
well for ideal gases you can just use the fact that 1 mole = 22.414 L at STP
aaronq
  • aaronq
otherwise you have to know the density of the gas
anonymous
  • anonymous
therefore, how would the equation look?
aaronq
  • aaronq
so for ideal gases, moles = volume of gas/(22.414 L/mol)
aaronq
  • aaronq
make sure you have the same units
anonymous
  • anonymous
oooh
anonymous
  • anonymous
@aaronq for the moles of magnesium igot 0.001 for all the trials, but my friend wrote it in dimensional analysis . whats betteR?
aaronq
  • aaronq
it's the same thing, since it's a gas, the volume is proportional to the amount of moles
anonymous
  • anonymous
ok and how would i find the ratio of volume displaced in mL to moles h2 gas produced? @aaronq
aaronq
  • aaronq
a ratio is just a division volume of gas/moles of gas
anonymous
  • anonymous
so volume of h2/moles of h2?
anonymous
  • anonymous
@aaronq
aaronq
  • aaronq
yeah
anonymous
  • anonymous
@aaronq i get a huge number
anonymous
  • anonymous
37.00ml/0.002
aaronq
  • aaronq
yep, thats what the want
anonymous
  • anonymous
18500?
aaronq
  • aaronq
they actually want the volume displaced/molesof H2 which is the same thing
aaronq
  • aaronq
mL/mol
anonymous
  • anonymous
so its right?
anonymous
  • anonymous
@aaronq
aaronq
  • aaronq
to what they're asking you, yes
anonymous
  • anonymous
but its such a big number? @aaronq
aaronq
  • aaronq
yep thats not that big of a number, remember it's in mL ..
anonymous
  • anonymous
18.5 L DAMM
aaronq
  • aaronq
lol that's not a lot, it really should be around 22.414 L
anonymous
  • anonymous
ARE YOU POSITIVE?
aaronq
  • aaronq
well unless they want you to calculate using Mg (s) + HCl -> H2 + MgCl (aq)
anonymous
  • anonymous
how would that be?
aaronq
  • aaronq
this would be the theoretical way you balance the equation and use diminutional analysis, but really your answer is gonna be like off by very little, also, because of rounding
aaronq
  • aaronq
did they give you the concentration of the acid?
anonymous
  • anonymous
no
anonymous
  • anonymous
all we had to do was place a piece of magnesium inside a buret that contained hcl and this was then converted and all the hcl migrated down to the bottom of the buret
anonymous
  • anonymous
basically the magnesium was reacting with the hcl
anonymous
  • anonymous
and after that we had to find the values in that table
aaronq
  • aaronq
hm well they really wouldn't need to tell you, do you know what a limiting reagent is?
anonymous
  • anonymous
limiting reactant?
aaronq
  • aaronq
yep, it's the same thing. you could use the number of moles of that and translate it to moles of H2 produced 2Mg (s) + 2HCl -> H2 + 2MgCl (aq) 0.001439 moles of Mg/2 = x moles of H2/1 moles of H2 = 0.0028794734677088 thats the theoretical yield
anonymous
  • anonymous
well that makes more sense because thats what we are currently learning in class haha
aaronq
  • aaronq
oh wait sorry i multiplied, it should be moles of H2 = 0.0007198683669272
aaronq
  • aaronq
haha oh true, well maybe let me know next time? haha
anonymous
  • anonymous
sorry!!!
anonymous
  • anonymous
so in other words, to find lets say trial #1
anonymous
  • anonymous
where the volume of hydrgen gas produced was 37.00mL and the moles of h2 gas produced were 0.002 mol
aaronq
  • aaronq
you would convert the mass of Mg to moles then using the balanced equation, build a relationship between H2 and Mg using their respective coefficients.. this just means: moles of Mg/coefficient=moles of H2/coefficient
anonymous
  • anonymous
oh my god
anonymous
  • anonymous
moles of mg that reacted?
anonymous
  • anonymous
coefficient is 2?
aaronq
  • aaronq
yeah because thats your limiting reactant..
aaronq
  • aaronq
for Mg is 2 for H2 is 1
aaronq
  • aaronq
wait no i balanced it wrong because it's Mg + 2HCl -> MgCl2 + H2
aaronq
  • aaronq
so it's coefficient is 1 for both
aaronq
  • aaronq
sorry, i'm like lost today idk why
anonymous
  • anonymous
i dont get it
aaronq
  • aaronq
Mg (s) + 2HCl -> H2 + MgCl2 0.001439 moles of Mg/1 = x moles of H2/1 moles of H2 = 0.001439 thats the theoretical yield which is close to what you would get if you just divided the volume of gas produced by 22.414 L
anonymous
  • anonymous
thats the ratio?
aaronq
  • aaronq
yep the ratio is 1:1
anonymous
  • anonymous
but what value would i put?
aaronq
  • aaronq
i would go with the theoretical since thats what you're doing in class
anonymous
  • anonymous
0.001439?
aaronq
  • aaronq
yeah, even round up to 0.0015
aaronq
  • aaronq
or down to 0.0014
anonymous
  • anonymous
ok wait where did the 0.001439 come from>?
aaronq
  • aaronq
its the number of moles of magnesium
anonymous
  • anonymous
ooooh yes thats my 0.001
aaronq
  • aaronq
haha yep, i mean round to whatever you think it's best, or whatever they're teaching you at school
anonymous
  • anonymous
but for the ratio isnt moles of h2 divided by moles of magnesium? so 0.002h2/0.001mg
aaronq
  • aaronq
which ratio?
anonymous
  • anonymous
ratio of volume displaced to moles h2 gas prodced
aaronq
  • aaronq
well the amount of moles of H2 produced is equal to the moles of Mg that reacted according to the stoichiometric ratios of the equation... so to answer your question it would be 37 mL/0.001 moles.. which is what you had before, if i remember correctly
anonymous
  • anonymous
37 being the volume of hydrogen gas produced and 0.001 moles of magnesium reacted?
aaronq
  • aaronq
well, technically yes, but the question is asking for the volume displaced divided by the moles of H2 produced..which are the same values
anonymous
  • anonymous
oh so volume of liquid displaced divided by the moles of h2
anonymous
  • anonymous
and thats the ratio?
anonymous
  • anonymous
wow i feel stupid
aaronq
  • aaronq
that they're asking you for in the question yes. They're basically seeing how close that value is to the ideal gas constant volume of 1 mole =22.414 L, since 37mL/0.0014 mol = 26428.5714285714285714 mL/mol 26.428 L/mol so it's close if you use more than 0.001, which i think you should do lol
anonymous
  • anonymous
oh my i get the big number again hahaha
aaronq
  • aaronq
haha don't feel stupid
aaronq
  • aaronq
your friend in the other thread used 0.00144 moles
anonymous
  • anonymous
0.0014 is in the correct significant figures? i keep using 0.001
aaronq
  • aaronq
hm idk though if you don't use that your error is pretty large
anonymous
  • anonymous
exactly
anonymous
  • anonymous
but look when i divide 0.035/ 24.31
aaronq
  • aaronq
0.0014397367338544
anonymous
  • anonymous
and the right sig figures are four so you round it to 0.001?
aaronq
  • aaronq
zeros dont count though
aaronq
  • aaronq
well zeros on the left side
anonymous
  • anonymous
oh..
anonymous
  • anonymous
still..
anonymous
  • anonymous
so now its 3 sig figures?
aaronq
  • aaronq
i would use 2 sig digits b/c 0.035 .. and have 0.0014
anonymous
  • anonymous
so only 14 are significant
aaronq
  • aaronq
yes
anonymous
  • anonymous
what about 0.016/24.31
anonymous
  • anonymous
0.0066?

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