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mr_godi [17]
3 years ago
6

ASAP: The main difference between the gravitational force and electrical force is that

Chemistry
1 answer:
Dahasolnce [82]3 years ago
5 0

Answer:

Electrical force can pull and push

Explanation:

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Please help fast<br>I WILL MARK YOU AS BRAINLIEST ​
Elan Coil [88]

Answer:

c

Explanation:

all the atoms must be balanced.

3 0
3 years ago
Write the structure of a D-aldohexose that gives a meso compound on treatment with NaBH4 in water.
soldier1979 [14.2K]

Explanation:

Meso compounds are optically inactive stereoisomers. Despite of having chiral carbon, they do not show  optical activity because it has a plane of symmetry in its structure itself. It is superposable on its mirror image.

Aldo sugars on reaction which reducing agents such as , NaBH₄ reduces the carbonyl group in the sugar to alcohol and gives corresponding alditol.

<u>The D- aldohexose which on reduction gives a meso alditol are allose and galactose. </u>

<u>The structure is shown in the image below. Thus, in allitol and galactitol formed have internal plane of symmetry which makes the optically inactive.</u>

7 0
3 years ago
Is food coloring dissolving in water a chemical change
zubka84 [21]
Answer: There no chemical change, but there is a physical change.


Reason: When a substance dissolves in water, the substance retains its molecular identity and simply mixes throughout the water
6 0
3 years ago
Read 2 more answers
Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq)
Svetach [21]

Answer:

The solution 0.010 M has a higher percent ionization of the acid.

Explanation:

The percent ionization can be found using the following equation:

\% I = \frac{[H_{3}O^{+}]}{[CH_{3}COOH]} \times 100    

Since we know the acid concentration in the two cases, we need to find [H₃O⁺].          

By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:

1. Case 1 (0.1 M):

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq)   (1)

0.1 - x                                         x                     x

Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}  (2)

Where:

Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.

1.7 \cdot 10^{-5} = \frac{x^{2}}{0.1 - x}  

1.7 \cdot 10^{-5}*(0.1 - x) - x^{2} = 0

By solving the above equation for x we have:

x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]

Hence, the percent ionization is:      

\% I = \frac{1.29 \cdot 10^{-3} M}{0.1 M} \times 100 = 1.29 \%                      

   

2. Case 2 (0.01 M):

The dissociation constant from reaction (1) is:

Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}

With [CH₃COOH] = 0.01 M

1.7 \cdot 10^{-5} = \frac{x^{2}}{0.01 - x}  

1.7 \cdot 10^{-5}*(0.01 - x) - x^{2} = 0

By solving the above equation for x:

x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]    

Then, the percent ionization for this case is:

\% I = \frac{4.04 \cdot 10^{-4} M}{0.01 M} \times 100 = 4.04 \%

As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.

Therefore, the solution 0.010 M has a higher percent ionization of the acid.

I hope it helps you!    

4 0
3 years ago
You are performing a titration of a triprotic acid, when you spill water on your lab notebook. you can read that: pka 1 = 1.40,
eimsori [14]
According to the PH formula:
PH= Pka +㏒ [strong base/weak acid]
when we have PH at the first equivalence =3.35 and the Pka1 = 1.4
So, by substitution, we can get the value of ㏒[strong base / weak acid]
3.35 = 1.4 + ㏒[strong base/ weak acid]
∴㏒[strong base/weak acid] = 3.35-1.4 = 1.95 
to get the Pka2 we will substitute with the value of ㏒[strong base/ weak acid] and the value of PH of the second equivalence point
∴Pk2 = PH2 - ㏒[strong base/ weak acid]
          = 7.55 - 1.95 = 5.6 
5 0
3 years ago
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