Qualification > Sciences
ALL CIE CHEMISTRY DOUBTS HERE !!
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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