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bearhunter [10]

2 years ago

13

Barbituric acid, HC4H3N2O3 (which we shall abbreviate H-Bar), was discovered by the Nobel Prize-winning organic chemist Adolph v

on Bayer and named after his friend, Barbara. It is the parent compound of widely used sleeping drugs, the barbiturates. Its pKa is 4.01. What are the [H+] and pH of a 0.043 M solution of H-Bar?

1 answer:

Montano1993 [528]2 years ago

7 0

Answer:

The [H+] is 2.04*10-3 and pH is 2.68

Explanation:

Barbituric acid is a a weak acid. Using Bronsted-Lowry equation, Ka = [H+][A-]/[HA]. we can calculate [H+] , we need to take barbituric acid pKa and apply antilog [-pKa] (in calculator press button 10x or ^ with negative value pKa). In that way we obtain Ka=9.77*10-5. Considering [H+]=[A-], we have than:

[H+]=√Ka*Acid Concentration=√9.77*10-5*0.043=2.04*10-3

pH=-㏒[H+]=-㏒[2.04*10-3]=2.7

I hope my answer helps you

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Consider the following hypothetical reaction: 2 P + Q â†’ 2 R + S The following mechanism is proposed for this reaction: P + P Q

aleksandr82 [10.1K]

Answer: the answer is option (D). k[P]²[Q]

Explanation:

first of all, let us consider the reaction from the question;

2P + Q → 2R + S

and the reaction mechanism for the above reaction given thus,

P + P ⇄ T (fast)

Q + T → R + U (slow)

U → R + S (fast)

we would be applying the Rate law to determine the mechanism.

The mechanism above is a three step process where the slowest step seen is the rate determining step. From this, we can see that this slow step involves an intermediate T as reactant and is expressed in terms of a starting substance P.

It is important to understand that laws based on experiment do not allow for intermediate concentration.

The mechanism steps for the reactions in the question are given below when we add them by cancelling the intermediates on the opposite side of the equations then we get the overall reaction equation.

adding this steps gives a final overall reaction reaction.

2P + Q ------------˃ 2R + S

Thus the rate equation is given as

Rate (R) = K[P]²[Q]

cheers, i hope this helps

3 0

2 years ago

The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

2 years ago

Read 2 more answers

Calculate the mass of 25,000 molecules of nitrogen gas. (1 mole = 6.02 x 1023 molecules)

Ainat [17]

Hey there!

Molar mass N2 = 28.01 g/mol

Therefore:

28.01 g N2 -------------- 6.02*10²² molecules N2

( mass N2 ?? ) ----------- 25,000 molecules N2

mass N2 = ( 25,000 * 28.01 ) / ( 6.02*10²³ )

mass N2 = 700250 / 6.02*10²³

mass N2 = 1.163*10⁻¹⁸ g

Hope that helps!

7 0

2 years ago

Read 2 more answers

If one has a solution of 0.10 m silver nitrate and it is diluted by a factor of two, what is the new concentration

Angelina_Jolie [31]

Diluted by a factor of two means that we double the volume of the solution by adding an equal volume of the water.
if we diluted it by a factor of one so the new concentration = 0.1/2=0.05 M and diluted by a factor of two so, the new concentration will be 0.05/2 = 0.025 M

7 0

2 years ago

The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ΔH, ΔS, a

Nostrana [21]

The question is incomplete, the complete question is;

The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ∆H, ∆S, and ∆G for the boiling process at this temperature.

A. ∆H > 0, ∆S > 0, ∆G < 0

B. ∆H > 0, ∆S > 0, ∆G > 0

C. ∆H > 0, ∆S < 0, ∆G < 0

D. ∆H < 0, ∆S > 0, ∆G > 0

E. ∆H < 0, ∆S < 0, ∆G > 0

Answer:

A. ∆H > 0, ∆S > 0, ∆G < 0

Explanation:

During boiling, a liquid is converted to vapour. This is a phase change for which heat is absorbed because energy must be taken in to break the intermolecular bonds in the liquid before it can be converted to a gas. Hence ∆H>0

Secondly, a phase change from liquid to gas leads to an increase in entropy hence ∆S>0.

Thirdly, the process is spontaneous. For every spontaneous process ∆G<0

3 0

2 years ago