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Kaylis [27]
2 years ago
9

n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form

of the acid?
Chemistry
1 answer:
Nady [450]2 years ago
5 0

Answer : The ratio of the protonated to the deprotonated form of the acid is, 100

Explanation : Given,

pK_a=8.0

pH = 6.0

To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}

Now put all the given values in this expression, we get:

6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}

\frac{[Deprotonated]}{[Protonated]}=0.01  

As per question, the ratio of the protonated to the deprotonated form of the acid will be:

\frac{[Protonated]}{[Deprotonated]}=100  

Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100

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If you add a neutron to an atom how does it change ?
Kobotan [32]

Answer:

Its atomic mass increases by 1

An isotope of that element is fotmed with mass differences by 1

Explanation:

7 0
2 years ago
What volume in mt, of 0.5a M1HCI solution is needed to neutralize 77 ml of 1.54 M NaOH solution?
Rainbow [258]

Answer:

237.2 mL.

Explanation:

  • We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

(XMV) acid = (XMV) base.

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

<em>(XMV) HCl = (XMV) NaOH.</em>

<em></em>

For HCl; X = 1, M = 0.5 M, V = ??? mL.

For NaOH, X = 1, M = 1.54 M, V = 77.0 mL.

<em>∴ V of HCl = (XMV) NaOH / (XV) HCl = (</em>1)(1.54 M)(77.0 mL) / (1)(0.5 M) = <em>237.2 mL.</em>

8 0
3 years ago
Calculate the density of each object
poizon [28]
Object one is 5.2 g/cm3
object two is 3.46g/ml
7 0
3 years ago
1. A 2-kg bowling ball sits on top of a building that is 40 meters tall. (5 pts) Circle one: KE / GPE / both ( can you show work
masha68 [24]
Potential energy is energy due to an object's height above the ground.
Potential energy = mass x gravity x height
Kinetic energy is energy due to the motion of the object.
Kinetic energy = 1/2 x mass x velocity²

1.
The ball is not moving and is at a height above the ground so it has only potential energy.
P.E = 2 x 9.81 x 40
P.E = 784.8 J

2.
The ball is moving and has a height above the Earth's surface so it has both kinetic and potential energy.
P.E = same as part 1 = 784.8 J
K.E = 1/2 x 2 x 5²
K.E = 25 J

3.
The ball has no height above the Earth's surface and is moving so it has only kinetic energy.
K.E = 1/2 x 2 x 10²
K.E = 100 J

4.
50000 = 1/2 x 1000 x v²
v = 10 m/s

5.
39200 = 200 x 9.81 x h
h = 20.0 m

6.
12.5 = 1/2 x 1 x v²
v = 5 m/s
98 = 1 x 9.81 x h
h = 10.0 m
3 0
3 years ago
Lowering an object decrease is potential​
Tresset [83]

Answer:

Lowering the object near the ground decreases its <u>potential energy.</u>

<u></u>

Explanation:

Potential Energy : Energy possessed by the object due to its shape ,Size and Position is called potential energy.

Example :

Change in shape and size :  When  you compress the spring , potential energy is introduced in it . So it expand quickly when you remove your hand.

Change in position : when you swing , after attaining maximum height (extreme ends) , the swing comes back on its on .This is because at maximum height ,the swing has<u> maximum Potential energy . </u>Hence it fall back on its on because it already has potential energy.

Potential energy(P) is given by the formula :

P = mgh

where ,

m= mass of the object

g = acceleration due to gravity

h = height of the object from the ground

If the height of the object increases from the ground , its potential energy also get increase.

<u><em>On lowering the object The height of the object from the ground reduces . So potential energy also reduces.</em></u>

8 0
3 years ago
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