Chemistry AS paper 1 questions- URGENT, exam tomorrow!

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jazz_xox_
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Please if anyone could explain any of following questions step by step, I would appreciate it so much. My exam, is tomorrow, and my chemistry teacher has been off sick for a while so can't help me !
Thank you
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5h3bi
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(Original post by jazz_xox_)
Please if anyone could explain any of following questions step by step, I would appreciate it so much. My exam, is tomorrow, and my chemistry teacher has been off sick for a while so can't help me !
Thank you
What paper is this btw
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5h3bi
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(Original post by 5h3bi)
What paper is this btw
Ok for the Kinetic energy one. You rearrange the KE formula to get v=squarerootof2KE/M
You know that one mole of mgis equal to 24.3g as its equal to the Mr. To find the mass of one ion you divide 24.3 by Avogadro constant and divide that by 1000 to get it into Kg. Once this has been done you substitute your values into the velocity equation and rearrange the v=d/t to get V*T=D
Should give you 2.16M or 2.13 depends on which Ar you use eg 25.
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5h3bi
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(Original post by jazz_xox_)
Please if anyone could explain any of following questions step by step, I would appreciate it so much. My exam, is tomorrow, and my chemistry teacher has been off sick for a while so can't help me !
Thank you
Chemical properties are determined by the electron configuration. So you know that isotopes are different atomic forms of the same element with the same number of protons(= number of electrons) but different number of neutrons. So if the electrons are equal then chemical properties will also be the same.
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Jay.vyas
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For the kinetic energy question:

You have:
Ke = 4.52x10^-16 J
T = 1.44x10^-5 s

You can work out the mass of the Mg+ ion using the periodic table and Avogadros constant. The mass of one mole of Mg (from the Periodic table) is 24.3 g. We can divide this by Avogadros constant get the mass of a single Mg atom, which would be the same as an Mg+ as the removal of one electron is negligible in terms of mass. We'll need to convert the mass of Mg to Kg first. So 24.3/1000 = 24.3x10^-3

So Mass of Mg+ ion = 24.3x10^-3 Kg / 6.022x10^23 = 4.04x10^-26 (Rounded to 2 dp). This is our m.

Rearrange the equation for Kinetic energy to get: 2Ke/m=V^2 and input the values..... [2 x (4.52x10^-16 J)] / 4.04x10^-26 = 2.24x10^10 (2 dp)

This would give you V^2, so square root it to get the velocity = 1.50x10^5 (2 dp)

So we can use the Time they gave us before and the Velocity we worked out to calculate the distance.

Distance = Speed x Time

Distance = 1.50x10^5 m/s x 1.44x10^-5 s = 2.15 metres (2 dp)

This question is more physics/maths. But I hope this helped. Sorry for the extremely long post. If you spot any mistakes I'll try to correct them.
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epsilon9012
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Remember the main topics that come up tomorrow and the big mark q's and post on this tomorrow please
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epsilon9012
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Actually never mind this is aqa i thought it was ocr
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rexs2000
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rearrange by after simultaneous equation
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IDONTW1N
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The last one:

Theres 0.01 moles of NH3, to find the number of molecules you times by Avogadros constant: (0.01x6.022x10^23) as its that many molecules per mole. This gives 6.022x10^21 molecules. Each NH3 molecule is made up of 4 atoms (1N 3H's) so if theres 6.022x10^21 molecules times that by 4 it gives the amount of atoms which gives 2.4088x10^22> 2.41x10^22
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IDONTW1N
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(Original post by IDONTW1N)
The last one:

Theres 0.01 moles of NH3, to find the number of molecules you times by Avogadros constant: (0.01x6.022x10^23) as its that many molecules per mole. This gives 6.022x10^21 molecules. Each NH3 molecule is made up of 4 atoms (1N 3H's) so if theres 6.022x10^21 molecules times that by 4 it gives the amount of atoms which gives 2.4088x10^22> 2.41x10^22
Forgot to add the second last one, its:

40.8 grams of ZnSO4.7H20 so divide by 0.7 to get the mass if all of the reactants had reacted which gives 58.286... now find the number of moles (Ms/Mr) so 58.286.../(287.50)=0.2 moles > its a 1:1 ratio of zinc so times 0.2 by Mr of zinc to get the original mass = 0.2x65.4=13.25>13.3g
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