<p>im having trouble with a couple of CB qs i would really appreciate the feedback 
*<em>How many mL of .200-molar sodium hydroxide must be added to a 100-mL solution of .100-molar nitric acid to obtain a solution with a pH of 7 ?
A-10mL
B-25mL
C-50mL </em> ( correct ans. WHY ?? )
D-100mL
E-200ml</p>
<p>**The ground state electron configuration of the silicon atoms is characterized by which of the following ??</p>
<p>I. Partially filled 3p
II.The presence of unpaired electrons
III.Six valence electrons
A.I
B.II ONLY
C.I , II ONLY **** ( WHY ? )
D.I, III ONLY
E.I,II,III</p>
<p>**Boron trifluoride BF3, is a nonpolar molecule whereas ammonia NH3 is a polar molecule the difference in polarities is related to the fact that </p>
<p>ANSWER–>…BF3 is a trigonal planar and NH3 is pyramidal<br>
how is BF3 a trigonal planar dont both of then have a pair of lone electrons ? i dont get that !!</p>
<li>Neutralization of 500 mL of 2-molar NaOH requires the smallest vlume of which of the following?
A. 1 M H2SO4 *** why??
B. 1 M CH3COOH
C. 1 M HCl
D. 1 M NH3
E. 0.1 M H2SO4</li>
</ol>
<p>THATS IT THANKS</p>
<p>
[quote]
The ground state electron configuration of the silicon atoms is characterized by which of the following ??
[/quote]
The electron configuration of Si is 1s2, 2s2, 2p6, 3s2, 3p2
So:
I is true (there's 2 electrons in 3p)
II is also true (the 2 electrons in 3p are unpaired)
III is untrue: there's only 4 valence electrons</p>
<p>
[quote]
how is BF3 a trigonal planar dont both of then have a pair of lone electrons ? i dont get that !!
[/quote]
Boron is one of the exceptions to the octet rule, and only bonds 3 times.
Beryllium is another exception, and bonds 2 times.</p>
<p>
[quote]
How many mL of .200-molar sodium hydroxide must be added to a 100-mL solution of .100-molar nitric acid to obtain a solution with a pH of 7 ?
[/quote]
100mL of 0.1mol/L HNO3 means there are 0.01moles of H+ ions (0.100L X 0.1mol H+/L)
You need 0.01 moles of OH- to neutralize the 0.01 moles of H+ ions.
0.01molOH- X L/0.2mol OH- X 1000mL/L = 50mL</p>
<p>
[quote]
5. Neutralization of 500 mL of 2-molar NaOH requires the smallest vlume of which of the following?
[/quote]
0.500L X 2mol OH-/L = 1 mol OH-</p>
<p>A. 1 M H2SO4
H2SO4 dissociates into 2H+ and SO4(2-)
So, to get 1 mol of H+, you need 500mL
1mol H2SO4/L X 2mol H+/1mol H2SO4 X 0.5L</p>
<p>B. 1 M CH3COOH
Ignore; it's an extremely weak acid.</p>
<p>C. 1 M HCl
Ignore; there is only 1mol H+ for every 1mol HCl. You've already seen in answer A that there's 2mol H+ for every 1 mol H2SO4.</p>
<p>D. 1 M NH3
Ignore; very weak acid again.</p>
<p>E. 0.1 M H2SO4
Ignore; weaker concentration of the same acid in A, therefore it would require more of this solution than the solution in A to achieve the same result.</p>
<p>Thanks A LOT CDN-dancer !!</p>
<p>A simpler explanation for the first one.</p>
<p>(.2)(x/1000) = (.1)(100/1000)</p>
<p>.2x = 10
(10)/.2 = x, x = 50 mL</p>