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Chemistry: Post your doubts here!
- Thread starter XPFMember
- Start date
Yes.
Molecular mass: sum of atomic masses of atoms in a molecule
Atomic mass: weighted average of isotopic masses
Isotopic mass: mass of isotope
(All relative to 1/12th of C-12)
The sample is going to have different stable carbon isotopes with different masses. The atomic mass derived from the sample is the weighted average of these different masses.
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THANKS
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the ideal gas equation: pV=nRT.
now if you rearrange the equation, you'll get pv/RT=n. So n will always have a constant value for a fixed mass of gas. Hence the expected graph for pV/RT against p should be a horizontal line.
EXTRA: If you want to plot a graph having the same axes but with a different mass of gas, the graph would be a different horizontal line...nothing much. Think about plotting a graph of mass of gas trapped inside a cylinder as you push in the piston to change the pressure. The mass never changes
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32 Use of the Data Booklet is relevant to this question.
In which pairs do both species have the same number of unpaired p electrons?
1 O and Cl +
2 F+ and Ga–
3 P and Ne+
I don't get the part 2. The answer is 1 and 2 but I gave 1 now I don't understand how 2 is also included
In which pairs do both species have the same number of unpaired p electrons?
1 O and Cl +
2 F+ and Ga–
3 P and Ne+
I don't get the part 2. The answer is 1 and 2 but I gave 1 now I don't understand how 2 is also included
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Thank you
You have to remember that in an equilibrium reaction there is no maximum yield,
H+ ions are in a higher concentration compared to the other products since its formed twice, when sulfuric acid dissociated and when HSO4- dissociated.
Try setting your own values like volume of H2SO4 and you'll get what I mean.
Hope it helped!
O AND CL+ - There electronic configuration is as follows:
[He]+2s2 2p4 , [He]+2s2 2p4 (since chorine lost one e )
F+ AND Ga- - There electronic configuration is as follows:
[He]+2s2 2p4 , [He]+2s2 2p6, 3s2 3p6 , 4s2 , 3d10 , 4p2
To actually understand this, you want to draw the orbitals and fill them. You'll realise for part one regards 2p4 lets say () is the orbital. And we know that a p subshell has 3 orbitals. there fore
(11)(1)(1) [the paired e are in opposite direction] [you can see there are 2 orbitals that are half filled]
In the second example,
for 2p4 (11)(1)(1) and for 4p2 , (1)(1)() [You can see here that also both have two orbitals
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Last edited:
- Messages
- 500
- Reaction score
- 425
- Points
- 73
- Messages
- 8
- Reaction score
- 4
- Points
- 3
In these types of questions you really need to work on just two atoms. See the answer will depend on balancing number of nitrogen atoms and oxygen atoms.
I'll explain each options and why they are possible/not possible.
A- for the given value of x and z, the number of nitrogen atoms on the L.H.S is 7 and on the R.H.S is 2.5 so they are not balanced hence this can't be the answer.
B- for the given value of x and y, no. of oxygen atoms on the L.H.S is 10 and on the R.H.S is 8 so they are not balanced hence this can't be the answer.
C- The number of nitrogen atoms on the L.H.S and the R.H.S is 7. Also the oxygen atoms in L.H.S and R.H.S is 8. no. of H atoms (8) are also equal. SO this can be the ans.
D - no. of nitrogen atoms on the L.H.S is 7 and on the R.H.S is 6 so this can't be the answer.
so you see it's really simple to just consider two atoms rather than trying to balance the whole equation taking each possible options. You might notice i didn't take Hydrogen atoms for balancing at first but it is automatically balance in option C because the no. of other two atoms are balanced. It is better to start by thinking about whole numbers provided in the options.
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what is meant by H+ is high or SO42- is low?
I think I can explain this! So if you see the periodic table, potassium is below the sodium.
- Potassium is in period 4 and Sodium is in period 3 but they belong to the same group.
- Potassium's atomic number is 19 while the Sodium's is 11. Potassium > Sodium (in terms of number of electrons).
- Using Aufbau principle
K= 1s2, 2s2, 2p6, 3s2, 3p6, 4s1
Na= 1s2, 2s2, 2p6, 3s1
- Electron configuration
K= 2,8,8,1
Na= 2,8,1
Hence, we can say that potassium occupies one more shell than sodium.
K+ and Na+. Both mean that they lose one electron from their outermost electron shell.
Losing one electron will make potassium to become full 3rd shell and sodium to become full 2nd shell.
So basically it's like,
K+= 1s2, 2s2, 2p6, 3s2, 3p6
Na+= 1s2, 2s2, 2p6
Therefore K+ has a greater ionic radius than Na+
Hope you get my point my friend!
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Reaction with 2,4-dinitrophenylhydrazine shows the presence of either an aldehyde or a ketone.
Check the options a positive result will be shown with option A,B,D
Decolourising of pottasium magnate shows oxidation took place option A and D
And the emperical formula shown of C2H3O is only consistent with option A i.e C4H6O2-------->C2H3O
Thus,correct option is A.
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Potassium magnate is an oxidizing agent it is originally purple in colour and on oxidizing it becomes colourless.
In option A the aldehyde will be oxidized to carboxylic acid thus Oxidation took place.
In option D on the right side of the compound there is also an aldehyde which oxidizes.