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Svetach [21]
3 years ago
8

(30 Points) You need to produce a buffer solution that has a pH of 5.02. You already have a solution that contains 10. mmol (mil

limoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74. Express your answer numerically in millimoles.
Chemistry
2 answers:
chubhunter [2.5K]3 years ago
8 0
<span>By definition:

pH = pKa + log [acetate]/ [acetic acid]

so

5.02 = 4.74 + log [acetate] / 10 mmole

10mmole = 10/1000 = 0.01 mole

5.02 = 4.74 + log [acetate] / 0.01

5.02 - 4.74 = 0.28 = log [acetate] /0.01


10^0.28 = </span><span>1.90546</span> = [acetate] / 0.01 <span>
[acetate] = 0.019 mole 

= 19 millimoles

</span>
ElenaW [278]3 years ago
3 0
<span>pH = pKa + log [acetate]/ [acetic acid] ( Handerso-Hasselbalch equation)

5.35 = 4.74 + log acetate / 0.010 moles

5.35 - 4.74 =0.61 = log acetate / 0.010

10^0.61 =4.1 = acetate / 0.010

moles acetate = 0.041

millimoles acetate required = 41</span>
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A 100.0 mLflask is filled with 0.065 moles of A and allowed to react to form B according to the reaction below. The following ex
kodGreya [7K]

The question is incomplete. The complete question is :

A 100.0 mL flask is filled with 0.065 moles of A and allowed to react to form B according to the reaction below. The following experimental data are obtained for the amount of A as the reaction proceeds. What is the average rate of appearance of B in units of M/s between t = 10 min. and t = 30 min.? Assume that the volume of the flask is constant. A(g) → B(g)

Time 0.0 10.0 20.0 30.0 40.0

Moles of A 0.065 0.051 0.042 0.036 0.031

Solution :

Consider the following reaction as follows :

$A \rightarrow B$

The experiment data is given as follows :

Time (min) :   0.0        10.0       20.0        30.0       40.0

Moles of A :  0.065    0.051    0.042      0.036     0.031

According to the rate of reaction concept, the rate can be expressed as a consumption of the reactant and formation of the product as follows :

Average rate : $= -\frac{d[A]}{dt} =  \frac{d[B]}{dt} $

Now we have to calculate the average rate between 10.0 to 30.0 min w.r.t. A as follows :

Rate  $=-\frac{(0.051-0.036) mol \times \frac{1}{0.1 \ L}}{(30.0-10.0) mol \times \frac{60 \ s}{1 \ min}}$

         $=\frac{0.15 \ M}{20 \ min \times \frac{60 \ s}{1 \ min}}$

         $= 1.25 \times 10^{-4 }\ M/s$

Therefore, the rate = $= 1.3 \times 10^{-4 }\ M/s$

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Do the ch3f molecules appear to attract each other much? briefly describe their interaction.
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Neither attract much nor allows it be free.

How the molecules behave in such way?

Because fluorine has a higher electronegativity than carbon and hydrogen, CH3F possesses a non-zero dipole moment. As a result, fluorine draws a partial negative charge whereas other atoms (such as carbon or hydrogen) draw a partial positive charge, leading to a non-zero dipole moment in CH3F. Lewis' structure, CH3F, is made up of one carbon atom, three hydrogen atoms, and one fluorine atom. Carbon is the key atom in the CH3F lewis structure because it is the least electronegative. The CH3F lewis dot structure consists of 3 lone pairs and a total of 4 bound pairs.

Properties of CH3F

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7 0
2 years ago
What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?
mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

C=\frac{n}{V} .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is K_{2}SO_{4} thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass

Putting the values,

n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol

Thus, number of moles of  K_{2}SO_{4} will be 0.1061\times 0.5=0.053 mol.

The volume of solution is 225 mL, converting this into L,

1 mL=10^{-3}L

Thus,

225 mL=0.225 L

Putting the values in equation (1),

C=\frac{(0.053 mol}{0.225 L}=0.236 M

Therefore, concentration of potassium sulfate solution is 0.236 M.


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3 years ago
Convert 5.3 g water to mL water
lord [1]

Answer:

5.3 mL

Explanation:

1 g = 1 mL

5.3 g

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