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How did you do that?
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How did you do that?
use 30 percent .You sure?
But I have one doubt.
In ques it is said Calculate the density of the
KOH solution?
So do we have keep mass for whole solution i.e. 100 g ( I took it) or only mass of KOH i.e. 30 grams? [ In finding density, the last step]
Can you show your solution?use 30 percent .
I did it orally .Can you show your solution?
#LOLI did it orally .
I need some definitions for Chemistry as.. Anyone?
M means mol per dm cubeRoOkaYya G you were right
the answer is 1.3
Here's the solution -->
Given, molar mass of KOH = 56 g mol−1
30% by mass KOH solution means that 30 g of KOH is present in 100 g of the solution.
6.90 M solution contains 6.90 mol of KOH in 1000 cm3 of the solution.
1 mol of KOH = 56 g
6.90 mol of KOH = (56 × 6.90) g
= 386.4 g
∴ 386.4 g of KOH is present =
= 1288 g of solution
Density of KOH solution =
= 1.288 g cm−3
Hence, the density of the given solution is 1.288 g cm−3
now explain this all plz
I didn't get anything
this is in syllabus, and it's not difficult 120 and 90 is the answerHello everyone. How do we predict bond angles and shapes? Advice, notes or anything will be appreciated. I remember in M/J 14, PCl5 bond angle and shape was asked and I had no clue.
each equatorial P–Cl bond makes two 90° and two 120° bond angles with the other bonds in the moleculeHello everyone. How do we predict bond angles and shapes? Advice, notes or anything will be appreciated. I remember in M/J 14, PCl5 bond angle and shape was asked and I had no clue.
don't confuse it with hybridization P-Cl makes 90 degree with axial and 120 degree with the equatorialeach equatorial P–Cl bond makes two 90° and two 120° bond angles with the other bonds in the molecule
each axial P–Cl bond makes three 90° and one 180° bond angles with the other bonds in the molecule.
it is sp3 hybridised n there is 5 charge clouds (5 bond pair) hence its trigonal bipyramidal
don't confuse it with hybridization P-Cl makes 90 degree with axial and 120 degree with the equatorial
hybridisation helps to determine SHAPE of molecule im talking about SHAPE here
i already explained about bond angle in 1st two lines
Bipyrimidal shape, is it the answer?http://hschemsolutions.com/files/Download/4.7 VSEPR II(SLN).pdf
To all those having doubts with determining shapes of molecules
♣♠ Magnanimous ♣♠
yes trigonal bipyramidalBipyrimidal shape, is it the answer?
just don't go for other shapes, consult syllabus it's written there. I think it's not in syllabussomebody please explain how is the shape of I3^- linear?
doesn't it have 2 bonding and 3 lone?
this is written is syllabus "somebody please explain how is the shape of I3^- linear?
doesn't it have 2 bonding and 3 lone?
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