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diamong [38]
2 years ago
10

What would be the observation when pineapple juice is dipped in a red litmus paper​

Chemistry
2 answers:
Elena L [17]2 years ago
7 0

Answer:

Red to blue as it has boelomin which is acidic but alkalanic(basic)in nature.

Explanation:

Pineapple contains bromelain. This naturally occurring substance is most concentrated in fresh pineapple.

Bromelain has anti-inflammatory and anticancer properties. Although it’s acidic, some experts believe that it has an alkalizing effect as you digest it. This may be beneficial to people with acid reflux. People believe the bromelain enzyme reduces swelling, bruising, and other injury-related pains.

Sloan [31]2 years ago
6 0

Answer:

red litmus paper change into blue when it is dipped into Pineapple juice.

hope it helps

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What is the total number of orbitals found in the second energy level? openstudy?
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On the second shell there are two individual subshells:
The "s" subshell has only 1 orbital with max. two electrons spinning around; and the so-called "p" subshell has 3 orbitals with max. 6 electrons (2 on each!)
In total, there are four orbitals with 8 revolving electrons on the second shell.
Hope could help :)
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3 years ago
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Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
               = 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
       = -㏒ 0.088 = 1.06 
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