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1) Which of these compounds is a week electrolyte? a) HC1 b) CI-1.5COOH (acetic acid) c) CcHtz06 (glucose) d) e) NaC1 2) The oxidation number of Mn in ICMn04 is: a) +8 b) +7 c) +5 d) -7 e) -8 3) What...

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1) Which of these compounds is a week electrolyte?
a) HC1 b) CI-1.5COOH (acetic acid) c) CcHtz06 (glucose) d) e) NaC1
2) The oxidation number of Mn in ICMn04 is:
a) +8 b) +7 c) +5 d) -7 e) -8
3) What mass of K, 0; is needed to prepare 200. rnL of a solution ha v nig a potassium ion concentration of 0.150 M?
a XXXXXXXXXXg b) 10.4 g c) 13.8 g d) •2_074 e) 1.49g
3/Po I AI? 38-'0 i1V9Z:0? 010 it'12. 4?
4) Lithium metal disailves in water to yield hydrogen gas and aqueous lithium hydroxide_ What is the final concentration of hydroxide ions when XXXXXXXXXXg of lithium metal is dropped into 750, mL of venter'? 0,) 111 M ' ill 1)--2-' .. b XXXXXXXXXXM 5- ti. 11'17111 o [ )91-0 =- c) 1.06 l d XXXXXXXXXXM XXXXXXXXXX), ,ITe ' ..I, e XXXXXXXXXXM
5) What is the chemical formula of the salt produced by the neutralization of sodium hydroxide with sulfuric acid?
a) Na2S h) Na2004)3 Nu(SO4)2 d) Na503 e) NalSO4
(C) Vhc. initial state of a as is P=0.80 atm, t=45 'DC, V.= 1.20 L. The conditions were clanged to P-A..76 atm, V= 1.4 L the new temperature is:
a) 45.5 cc 1,) 79.5 c) 49.8 °C d) 238 °C. e) 337 DC
1. .; XXXXXXXXXXJe47 r —
(iiihe volume in liters eor 3..4 mol of CHI gas al I rC kind 54 ton' is:
a) 87L b XXXXXXXXXXc) 6.7 x 10'1.. d) 4_7 L. e) 111 L.
Answered Same Day Dec 21, 2021

Solution

Robert answered on Dec 21 2021
130 Votes
1) Among the given options O2 is a weak electrolyte because it hasn’t nonpolar bonds to
eak the
atoms easily. Hence the answer is d)
2) From the given information –
1molLi
6.94g/mol
 
 
 
    4KMnO K +1 Mn 4 2 for O 0
Therefore
Mn = +7
    

Hence answer is b)
3)      22 3 3K CO 2K COaq aq aq
  
Every 1 mol potassium ca
onate gives 2 mol K+, hence –(0.150 M)(0.2 L) = 0.03 mol K+ obtained
y - =  + 2 3 2 3+
2 3
1molK CO 138.21g K CO
= 0.03molK
2molK 1molK CO
  
  
  
= 2.07 g (answer d)
4) The balanced equation for the given reaction is –
       2 22Li H O 2LiOH Hs l aq g  
The final concentration of hydroxide ion is –
 
-
-1molLi 2molLiOH 1molOH 1= 5.500g Li = 1.05MOH
6.94g/mol 2molLi 1molLiOH 0.750L
     
     
     

Thus answer is c).
5) The balanced equation for the given reaction is –
       2 4 2 4 22NaOH H SO Na SO 2H Oaq aq aq l  
Thus 2 4Na SO is the answer.(answer a))
6) From the given information –
  
 
  
1 1 2 2
1 2
0
Since-
0.80atm 1.20L 0.76atm 1.4LP V P V
= =
T T 45+273 K T K
Bysolving -
T =352.45K = 79.5 C


Thus answer is b)
7) From the given information –
   
Since-
PV = nRT
3.4mol 0.0821L.atm/mol.K 12+273 KnRT
V = =
543P
atm
760
V = 111 L
 
 
 

Thus answer is e)
8) From the given information –
  
 
  
 
1 1 2 2
1 2
Since-
234mmHg 0.250L 645mmHg VP V P V
= =
T T 46+273 K 323 273 K
Bysolving -
P = 0.167 L


Thus answer is d)
9) From the given information –
  2
2
1molH1mol Zn 22400mL
= 43.2g Zn 14800mL
65.39g 1mol Zn 1molH
   
   
   

Thus answer is – d).
10) From the given information –
  
 
  
 
1 1 2 2
1...
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