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daser333 [38]
3 years ago
9

Calculate the pH of a buffer prepared by mixing 20.0 mL of 0.10 M acetic acid and 55.0 mL of 0.10 M sodium acetate

Chemistry
1 answer:
MatroZZZ [7]3 years ago
3 0

Answer:

Calculate the pH of a buffer prepared by mixing 30.0 mL of 0.10 M acetic acid and 40.0 mL of 0.10 M sodium acetate.

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At room temperature iodine is a solid and bromine is a liquid.
vichka [17]

Answer:

<u>Kinetic particle theory</u>

Arrangement and motion of solid particles

-> Solid particles are packed closely with each other in an orderly manner. They vibrate vigorously in their fixed positions.

Arrangement and motion of liquid particles

-> Liquid particles are packed less closely with each other as compared to solid particles in a disorderly manner. They move around in a random motion; sliding past each other.

3 0
3 years ago
X2o3 express you answer as a whole number
kobusy [5.1K]
 <span>2.40 - 1.68 =0.72 g of oxigen 
moles = 0.72/16 g/mol=0.045 

moles x = 1.68/ 55.9=0.03 

0.03/0.03 = 1 = x 
0.045 / 0.03 = 1.5 = O 

to get whole numbers multiply by 2 

x2O3 

X2O3 +3 CO = 2 X + 3 CO2</span>
4 0
3 years ago
The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0
3 years ago
Which of the following is NOT true?
ivanzaharov [21]
B, hope this helps :)
4 0
3 years ago
Read 2 more answers
Help please fast will mark as Brainliest
wlad13 [49]

Answer:

The answer is 2i on right hand side.

Explanation:

We should star by checking the equation from right.

First we check how many Zn r there in left hand side. Which is 1. Let us check how many Znr there in right hand side, there is 1.So Zn is balanced, and don't worry about Znplus2 on right hand side it is just the ions not how many zinc r there.

Now let us check how many I are there left hand side. Which is 2. Now how many I are there in right hand side? Only 1.

So we put 2 behind I.

Now there r 2 I on both sides.

Its simple actually.

3 0
3 years ago
Read 2 more answers
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