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ALL CIE CHEMISTRY DOUBTS HERE !!

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Master_Key:

--- Quote from: Fidato on May 05, 2011, 10:46:42 am ---Question 1 please.

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w06_qp_1.pdf


--- End quote ---

Answer is A.

required 25 cm3 of 1.0 × 10–2 mol dm–3 potassium hydroxide for complete neutralisation.

moles = .025*1.0 × 10–2

=2.5 × 10-4 moles.


Ca2+ = 2H+

= 2.5 × 10-4 * 0.5 moles of Ca2+.

= 1.25 × 10-4 moles of Ca2+ in .05 dm3.

C = m/v

         1.25 × 10-4
C  =  ---------------
                0.05

C = 2.5 × 10-3

Vin:
^Yep I got that too

Arthur Bon Zavi:

--- Quote from: Master_Key on May 05, 2011, 12:15:49 pm ---Answer is A.

required 25 cm3 of 1.0 × 10–2 mol dm–3 potassium hydroxide for complete neutralisation.

moles = .025*1.0 × 10–2

=2.5 × 10-4 moles.


Ca2+ = 2H+

= 2.5 × 10-4 * 0.5 moles of Ca2+.

= 1.25 × 10-4 moles of Ca2+ in .05 dm3.

C = m/v

         1.25 × 10-4
C  =  ---------------
                0.05

C = 2.5 × 10-3

--- End quote ---

I didn't see the ratio as 2 : 1 and so didn't divide. Thanks.

EMO123:
i got one fantastic doubt
M/J 2002 P1 Q32 :)

Arthur Bon Zavi:

--- Quote from: EMO123 on May 06, 2011, 05:35:15 am ---i got one fantastic doubt
M/J 2002 P1 Q32 :)

--- End quote ---

H2O (2 bonding pairs, 2 lone pairs) = 104.5o
CH4 (4 bonding pairs) = 109.5o
NH3 (3 bonding pairs, 1 lone pair) = 107o
SF6 (6 bonding pairs. S has 12 electrons now, after bonding) = 90o
CO2 (linear) = 180o
BF3 (3 bonding pairs. Look at the number of electrons in the outer shell of B after bonding !) = 120o

Do the rest yourself, you may find answer as D.

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