conjugate base conjugate acid 1:1 There is a ✓stochiometric ratio between NaOH and H₂CO3. Therefore, if we have 5 moles of NaOH, we need 5 moles of not neutralized H₂CO3 for neutralization. If we have 5 mL of 1 M NaOH and add 2.5 mL of 2 M H₂CO3, the acid is in excess and is therefore therefore neutralized If we have 10 mL of 0.1 M NaOH and add 16 mL of 0.05 M equal moles to the base H₂CO3, the acid is V the acid is and H₂O is the ✓ [Select] neutralized If we have 2 mL of 5 M NaOH and add 5 mL of 3 M H₂CO3, [Select] ✓and is therefore not neutralized V and is
conjugate base conjugate acid 1:1 There is a ✓stochiometric ratio between NaOH and H₂CO3. Therefore, if we have 5 moles of NaOH, we need 5 moles of not neutralized H₂CO3 for neutralization. If we have 5 mL of 1 M NaOH and add 2.5 mL of 2 M H₂CO3, the acid is in excess and is therefore therefore neutralized If we have 10 mL of 0.1 M NaOH and add 16 mL of 0.05 M equal moles to the base H₂CO3, the acid is V the acid is and H₂O is the ✓ [Select] neutralized If we have 2 mL of 5 M NaOH and add 5 mL of 3 M H₂CO3, [Select] ✓and is therefore not neutralized V and is
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
Both are the same question, first image for context.
The second part is asking about whether the acid will be in: excess, limiting reagent, or equal moles to base.
The therefore part is referring to whether it will be neutralized or not.
Can someone help with this? Thanks
![0
There is a 1:1
between NaOH and H₂CO3. Therefore, if we have 5 moles of
NaOH, we need
5
moles of
H₂CO3 for neutralization.
If we have 5 mL of 1 M NaOH and add 2.5 mL of 2 M H₂CO3,
LIKE
and is therefore
the acid is
not neutralized
in excess
therefore
If we have 10 mL of 0.1 M NaOH and add 16 mL of 0.05 M
and is
H₂CO3, the acid is equal moles to the base
[Select]
neutralized
If we have 2 mL of 5 M NaOH and add 5 mL of 3 M H₂CO3,
the acid is [Select]
and is therefore
stochiometric ratio
Overall
N₂.CO
HICL
NHCO.
N₂CL
L
*1
F
m
lab
Ma](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F84577f47-6855-4e4f-91d0-241ebc051b86%2Fafe9fbdf-5090-46b1-8176-511fcb7bcb27%2Fkase2bx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:0
There is a 1:1
between NaOH and H₂CO3. Therefore, if we have 5 moles of
NaOH, we need
5
moles of
H₂CO3 for neutralization.
If we have 5 mL of 1 M NaOH and add 2.5 mL of 2 M H₂CO3,
LIKE
and is therefore
the acid is
not neutralized
in excess
therefore
If we have 10 mL of 0.1 M NaOH and add 16 mL of 0.05 M
and is
H₂CO3, the acid is equal moles to the base
[Select]
neutralized
If we have 2 mL of 5 M NaOH and add 5 mL of 3 M H₂CO3,
the acid is [Select]
and is therefore
stochiometric ratio
Overall
N₂.CO
HICL
NHCO.
N₂CL
L
*1
F
m
lab
Ma
![conjugate base
conjugate acid
There is a
stochiometric ratio
between NaOH and H₂CO3. Therefore, if we have 5 moles of
NaOH, we need 5
moles of
1:1
the acid is
H₂CO3 for neutralization.
If we have 5 mL of 1 M NaOH and add 2.5 mL of 2 M H₂CO3,
in excess
and is therefore
0
not neutralized
If we have 10 mL of 0.1 M NaOH and add 16 mL of 0.05 M
H₂CO3, the acid is equal moles to the base
therefore neutralized
and H₂O is the
✓ [Select]
If we have 2 mL of 5 M NaOH and add 5 mL of 3 M H₂CO3,
the acid is [Select]
and is therefore
neutralized
not neutralized
Overall
Na CO
HC
and is
NaHCO.
NaCl](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F84577f47-6855-4e4f-91d0-241ebc051b86%2Fafe9fbdf-5090-46b1-8176-511fcb7bcb27%2Ffpw2gbo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:conjugate base
conjugate acid
There is a
stochiometric ratio
between NaOH and H₂CO3. Therefore, if we have 5 moles of
NaOH, we need 5
moles of
1:1
the acid is
H₂CO3 for neutralization.
If we have 5 mL of 1 M NaOH and add 2.5 mL of 2 M H₂CO3,
in excess
and is therefore
0
not neutralized
If we have 10 mL of 0.1 M NaOH and add 16 mL of 0.05 M
H₂CO3, the acid is equal moles to the base
therefore neutralized
and H₂O is the
✓ [Select]
If we have 2 mL of 5 M NaOH and add 5 mL of 3 M H₂CO3,
the acid is [Select]
and is therefore
neutralized
not neutralized
Overall
Na CO
HC
and is
NaHCO.
NaCl
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