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svp [43]
3 years ago
6

Consider the reaction: HCl(aq) + NaOH(aq) → NaOH(aq) + H2O(l) How much 0.113 M NaOH solution will completely neutralize 1.25 L o

f 0.228 M HCl solution?
Chemistry
2 answers:
k0ka [10]3 years ago
3 0
1.25L Hcl(.228 mol HCl/1 L)=0.285 mol HCl
0.285 mol HCl(1 mol NaOH/1 mol HCl)=0.285 mol NaOH
0.285 mol NaOH(1 L/.113 mol NaOH)=2.52212 L of .113M NaOH solution
lisabon 2012 [21]3 years ago
3 0

The  much of 0.113 M NaOH  solution that  is required  to  neutralize  is  2.522 L of NaOH


    <u><em>calculation</em></u>

HCl (aq) + NaOH (aq) → NaOH (aq)  + H2O  (l)

Step 1:  calculate the   moles of  HCl

  moles= molarity  x volume in liters

volume in liters =  1.25 L

molarity =  0.228 M  =  0.228 mol/ L

moles is therefore = 0.228 mol/L  x 1.25 L = 0.285  mols


Step  2:  use the mole ratio to  calculate  the   moles of NaOH

The  mole ratio   of  HCl : NaOH  is 1:1  therefore  the moles  of NaOH= 0.285 moles


Step  3:  calculate the volume  of  NaOH

volume=moles/molarity

molarity  0.113 M  = 0.113 mol/l

= 0.285 moles /0.113 Mol/  L = 2.522 L  of NaOH




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The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-
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Answer:

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Explanation:

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According to first order kinematics

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initial           4.70                         0             0

change        -2x                          +2x           +x

final             4.70 -2x                     2x           x

pressure ofO_2 after first half life  = 2.35 = 4.70 - 2x

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pressure of N_2 after first half life  =  2x = 2(1.175) = 2.35 ATM

Total pressure  = 2.35 + 2.35 + 1.175

                          = 5.875 atm

5 0
3 years ago
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