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Ask me for chemistry help! (1 Viewer)

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Depends on what you mean. If they've recorded the heat as only 70% of the heat of combustion, then divide it by 0.7 to get all the heat.
why wud u divide? u wud times it by 0.7 to find 70% of the theoretical value
 

olipierce

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is there an easier way to calculate structural variants of isomers than just drawing them out... like some kind of formula
 

nazfiz

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What volume of this 0.200 mol/L lead nitrate solution is just sufficient to react with all the iodide in 25mL of 0.247 mol/L potassium iodide solution?
 

nightweaver066

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There are 2 moles of PbI(subscript2)

Find the no of moles of I(subscript2)

Do you need to find the molar ratio molarmass(Isubscript2)/molarmass(PbI(subscript2)
??
2 moles of PbI(subscript2) means 2 moles of Pb2+ ions and 2 moles of 2I- ions.
 

nightweaver066

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What volume of this 0.200 mol/L lead nitrate solution is just sufficient to react with all the iodide in 25mL of 0.247 mol/L potassium iodide solution?
First, write out the equation. (going to be ignoring states as it would be annoying)

Pb(NO3)2 + 2KI -> 2KNO3 + PbI2

Number of moles of potassium iodide solution you have:
0.247 x 0.025 = 6.175 x 10^-3 moles

Mole ratio of lead nitrate to potassium iodide is 1:1

So you have the same amount of moles, now calculate volume.
Volume = (6.175 x 10^-3)/0.200
= 0.031L (2sf)
 
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Alkanes

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First, write out the equation. (going to be ignoring states as it would be annoying)

Pb(NO3)2 + 2KI -> 2KNO3 + PbI2

Number of moles of potassium iodide solution you have:
0.247 x 0.025 = 6.175 x 10^-3 moles

Mole ratio of lead nitrate to potassium iodide is 1:1

So you have the same amount of moles, now calculate volume.
Volume = (6.175 x 10^-3)/0.200
= 0.03L (2sf)
That's 1 sig fig lol.
 

nazfiz

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First, write out the equation. (going to be ignoring states as it would be annoying)

Pb(NO3)2 + 2KI -> 2KNO3 + PbI2

Number of moles of potassium iodide solution you have:
0.247 x 0.025 = 6.175 x 10^-3 moles

Mole ratio of lead nitrate to potassium iodide is 1:1

So you have the same amount of moles, now calculate volume.
Volume = (6.175 x 10^-3)/0.200
= 0.031L (2sf)
oh k thanks, but the answers say 154 mL? do you think the answers are wrong?
 

FCB

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oh k thanks, but the answers say 154 mL? do you think the answers are wrong?
154 mls is right as the stoichiometric ratio of Pb(NO3)^2 : 2KI is 1:2 respectively.

And thus you would only require half as its the limiting reagent.
 

_deloso

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First, write out the equation. (going to be ignoring states as it would be annoying)

Pb(NO3)2 + 2KI -> 2KNO3 + PbI2

Number of moles of potassium iodide solution you have:
0.247 x 0.025 = 6.175 x 10^-3 moles

Mole ratio of lead nitrate to potassium iodide is 1:1


So you have the same amount of moles, now calculate volume.
Volume = (6.175 x 10^-3)/0.200
= 0.031L (2sf)
no, you actually look at the stoichiometry of lead nitrate to potassium iodide which is 1:2 because you're reacting it with the lead nitrate(given concentration) not the lead iodide (this is the product)
Therefore moles of lead nitrate is half of moles of potassium iodide which is 0.006175/2 = 0.0030875 mol
now using c=n/v --> v=n/c --> v= 0.0030875/0.200 v=0.0154L which is 15.4 ml

oh k thanks, but the answers say 154 mL? do you think the answers are wrong?
hmm the answer should be 15.4 ml
 

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