O2=32 g/ mol
1.15/32=0.035
N2=28 g/mol
1.55/28=0.055
in STP every 22.4 litters is 1 mol
Answer:
and 
Explanation:
The equation for the reaction is AgNO3(aq) + KCl(aq) ==> AgCl(s) + KNO3(aq)
With all the ions, it is
(aq) + 
(aq) + (aq) + 
(aq) ==> AgCl(s) +
and do not change, so they are the spectator ions and are removed
The ionic equation is:
(aq) + (aq) ==> AgCl(s)
Answer:
a) the minimun of acetic anhydride required for the reaction is 2.175 g (CH3CO)2O
b) V acetic anhydride = 2.010 mL
Explanation:
C6H4OHCOOH + (CH3CO)2O ↔ C9H8O4 + C2H4O2
⇒ mol salicylic acid = 2.94 g C6H4OHCOOH * ( mol C6H4OHCOOH / 138.121 g ) = 0.0213 mol C6H4OHCOOH
⇒ mol acetic anhydride = 0.0213 mol C6H4OHCOOH * ( mol (CH3CO)2O / mol C6H4OHCOOH ) = 0.0213 mol (CHECO)2O
⇒ g acetic anhydride = 0.0213 mol * ( 102.1 g/mol ) = 2.175 g CH3CO)2O
b) V = 2.175 g (CH3CO)2 * ( mL / 1.082 g ) = 2.010 mL (CH3CO)2O
Answer:
Hello attached below is the data found in Aleks Data tab
answer :
i) N0
ii) N0
iii) YES , pH of highest solubility = 5
Explanation:
i) For CuBr
solubility does not change with pH hence answer = NO
ii) For MgCl2
solubility does not change with pH hence the answer = NO
iii) For Ba(OH) 2
Solubility does change with pH hence the answer = YES
and the pH at which the highest solubility will occur is = 5
attached below is the reason for the answers given
Answer:
Q.1
Given-
Volume of solution-1 L
Molarity of solution -6M
to find gms of AgNO3-?
Molarity = number of moles of solute/volume of solution in litre
number of moles of solute = 6×1= 6moles
one moles of AgNO3 weighs 169.87 g
so mass of 6 moles of AgNO3 = 169.87×6=1019.22
so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution
