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Chemistry: Post your doubts here!

Student12

Student12

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I'm just too worried i don't know how to find the slop of the line by using the construction liness HELP ! with example please!
 
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MaxStudentALevel

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fatemakhan said:
i think yours is right =/
even i dont know how to explain what i did there... i think to say i just took out % dec. in solubility and considered it to be the % dec in dissolved solid would be the right explanation...but then i shouldve stopped here ... and 14.8g wouldve been the ans according to me....
does that make any sense ? =/
Click to expand...

Yes it does! Your calculation was right, but in your mind you were calculating the wrong thing!
 
Student12

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Soldier313

Soldier313

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MaxStudentALevel said:
FOR B) The values calculated should not be less than the values used to calculate! So for this, all readings are to 2 dp so the values used to calculate the mass of iodine are all to 2dp ours should be as well! The trick is always to do equal dp as readings UNLESS stated otherwise (like sometimes they say (3sf) specifically so follow that in that case! :)

-AS for the second part I'm slightly confused with that also! Can't seep to find a logical reason! xhizors ?

-I think the logical way here, rather the Intuitive way was to calculate I:Zn as we need an answer of 2 for that! Because I:Zn ratio would be 2 if exp showed its validity. That's the only logical reason I can think of!

-ANOMALIES: IF I:Zn ratio differs from the rest by a lot it is an anomalous result!
-If I:Zn is too high (meaning mass ofZn was too low) then means Zn must've been blown out of tube while heating!
-If I:Zn too low (meaning mass if Zn too high) maybe not dried enough!

Any other confusions?
Click to expand...

hey sis,
thanx for the explanation!
-About the siginificant figures issue, the reason why i am confused is, the masses i calculated for part a, were to 3sgf and then suddenly as i move to part bi, i am required to put the moles to 2 sgf! and moving further to part bii, the mole ratio, i am supposed to use 2 dp! I mean why isn't the required accuracy consistent??

about the anomalies, i thought exactly the way you have put it down, but the ms has switched the resons:
it says if the I:Zn ratio is too high, then the Zn has been oxidised
and if the I:Zn ratio is too low, then the Zn must have been blown out during heating!
 
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hellangel1

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Two moles of compound P were placed in a vessel. The vessel was heated and compound P was partly decomposed to produce Q and R. A dynamic equilibrium between chemicals P , Q and R was established.

At equilibrium x moles of R were present and the total number of moles present was (2+ (x/2))

what is the equation for this equilibrium reaction ?

A. P ↔ 2Q + R
B. 2P ↔ 2Q + R
C. 2P ↔ Q+ R
D. 2P ↔ Q + 2R
 
X

xhizors

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MaxStudentALevel said:
FOR B) The values calculated should not be less than the values used to calculate! So for this, all readings are to 2 dp so the values used to calculate the mass of iodine are all to 2dp ours should be as well! The trick is always to do equal dp as readings UNLESS stated otherwise (like sometimes they say (3sf) specifically so follow that in that case! :)

-AS for the second part I'm slightly confused with that also! Can't seep to find a logical reason! xhizors ?

-I think the logical way here, rather the Intuitive way was to calculate I:Zn as we need an answer of 2 for that! Because I:Zn ratio would be 2 if exp showed its validity. That's the only logical reason I can think of!

-ANOMALIES: IF I:Zn ratio differs from the rest by a lot it is an anomalous result!
-If I:Zn is too high (meaning mass ofZn was too low) then means Zn must've been blown out of tube while heating!
-If I:Zn too low (meaning mass if Zn too high) maybe not dried enough!

Any other confusions?
Click to expand...
i did answer to that part, b/c examiner stated ratio of I:Zn
 
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