A reaction rate increases by a factor of 500. in the presence of a catalyst at 37oC. The activation energy of the original pathway is 106 kJ/mol. What is the activation energy of the new pathway, all other factor being equal

Answer:

[tex]\boxed {\boxed {\sf 0.078 \ L }}[/tex]

Explanation:

We are asked to find the volume of a solution given the moles of solute and molarity.

Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.

Substitute these values/variables into the formula.

[tex]1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}[/tex]

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

[tex]\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}[/tex]

[tex]1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl[/tex]

[tex]1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl[/tex]

Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.

[tex]\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}[/tex]

[tex]x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}[/tex]

The units of moles of potassium chloride cancel.

[tex]x= \frac{0.14 }{1.8 L}[/tex]

[tex]x=0.07777777778 \ L[/tex]

The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

[tex]x \approx 0.078 \ L[/tex]

There are approximately 0.078 liters of solution.

Answer:

[tex]m_{isoborneol }=0.39g\\\\m_{Camphor}=0.38g\\[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:

[tex]m_{isoborneol }=0.0025mol*\frac{154.25g}{1mol} =0.39g\\\\m_{Camphor}=0.0025mol*\frac{152.23 g}{1mol} =0.38g\\[/tex]

Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.

Regards!

Answer:

The right answer is "105.17 ppb".

Explanation:

According to the question,

The amount of [tex]Pb^{2+}[/tex] in ppb will be:

= [tex]0.5076\times 10^{-6}\times 207.2\times 106[/tex]

= [tex]105.17 \ ppb[/tex]

Thus, the amount of [tex]Pb^{2+}[/tex] is above action limit.

Answer:

The right answer is "6 gm".

Explanation:

Given:

Amount of Li,

= 0.3 gm

Battery rated,

= 5 Ah per cell

Total number of cells,

= 4

The total power limit will be:

= [tex]5\times 4[/tex]

= [tex]20 \ Ah[/tex]

hence,

The amount of Li in battery will be:

= [tex](0.3)\times 20[/tex]

= [tex]6 \ gm[/tex]

(Allowed for transportation).

Answer:

Attached below

Explanation:

The starting agents : attached below

There is no Solvent required to carry out this electrophilic alkene addition reaction

The products are :  attached below ( Cl )

The TLC values can only be determined by carrying out the experiment in the laboratory ( i.e. it is an experimental observation )

Attached below is the Mechanism showing the starting agents and products

Answer:

Element

Element : A pure substance composed of the same type of atom throughout. Compound : A substance made of two or more elements that are chemically combined in fixed amounts.

Explanation:

Answer:

False

Explanation:

Electrophilic addition reactions may be under kinetic or thermodynamic control. Whether the reaction is under kinetic or thermodynamic control is easily deducible from the reaction time.

Shorter reaction time often reflect kinetic control while longer reaction reaction times favour thermodynamic control.

Hence, electrophilic addition reaction of conjugated dienes that occur at high temperature and/or long reaction times (reversible conditions) are said to be under thermodynamic and not kinetic control.

Answer:

The mole ratio of AgNO3 to Ag2SO4 IS 2:1 .0.657 g Ag2SO4 x 1 mol / 312 g = 0.00211 mol Ag2SO4.

0.00211 mol Ag2SO4 x 2 mol AgNO3 / 1 mol Ag2SO4 = 0.00421 mol AgNO3

0.00421 mol AgNO3 x 1 L / 0.123 mol AgNO3 = 0.0342 L = 34.2 mL of AgNO3 solution.Therefore,34.2ml of 0.123M AgNO3 will be required.

Answer:

I don't know sorry yyyyyyy6yyyyyyyyyyyyyyyyyyyyyyyyyyy

Answer:

a mixture that doesn't really show the ingredients or things put into the material or food.

Answer:

Robert Millikan was a physicist who discovered the elementary charge of an electron using the oil-drop experiment

Answer:

the mass of an electron using the Oil-Drop experiment.

Explanation:

Answer:

Molar mass = 180 g/mol

Explanation:

Relative Atomic Mass of C = 12

of H = 1

of O =16

Let Molar mass be mm

mm of C6H12O6 = 6(12) + 12(1) + 6(16)

= 72 + 12 + 96 = 180 g/mol

Explanation:

[tex]57816 \: moles[/tex]

are present in 4.20x10^24 atoms of Pb

Answer:

7 moles

Explanation:

(4.2*10^24)/(6*10^23)=7

Answer:

"They established the laws of planetary motion"

Explanation:

Mr. Kepler was the astronomer who came up with the "Laws of Planetary Motion."

Answer:

90.99 or 91.0

Explanation:

Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.

Answer: The volume of oxygen gas is 91.4 L.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of ethanol = 38.5 g

Molar mass of ethanol = 46 g/mol

Plugging values in equation 1:

[tex]\text{Moles of ethanol}=\frac{38.5g}{46g/mol}=0.840 mol[/tex]

The given chemical equation follows:

[tex]CH_3CH_2OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)[/tex]

By stoichiometry of the reaction:

If 1 mole of ethanol produces 3 moles of water

So, 0.840 moles of ethanol will produce = [tex]\frac{3}{1}\times 0.840=2.52mol[/tex] of water

The ideal gas equation is given as:

[tex]PV=nRT[/tex] .......(2)

where

P = pressure = 1.75 atm

V = volume of oxygen gas = ?

n = number of moles= 2.52 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = temperature of the tank = [tex]500^oC=[500+273]K=773K[/tex]

Putting values in equation 2, we get:

[tex]1.75 atm\times V=2.52mol\times 0.0821L.atm/mol.K\times 773K\\\\V=\frac{2.52\times 0.0821\times 773}{1.75}=91.4L[/tex]

Hence, the volume of oxygen gas is 91.4 L.

Answer:

M of HNO₃ is 0.119M

Explanation:

A basic concept of titration is that in equivalence point:

mmoles of acid = mmoles of base

We have data from base and we only have data from volume of acid.

In a case our titration is a strong acid against a strong base.

We apply formula:

M of acid . Vol of acid = M of base . Vol of base

M of acid . 21.8 mL = 0.137M . 19 mL

M of acid = (0.137M . 19 mL) / 21.8 mL

M of acid = 0.119 M

When we neutralize all the titrant we reach the equivalence point.

At this point, pH = 7

2HNO₃  +  Ca(OH)₂ → Ca(NO₃)₂ +  2H₂O

Answer:

the % yield is 82%

Explanation:

Given the data in the question,

we know that;

Molar mass of benzil is 210.23 g·mol−1

Molar mass of dibenzyl ketone is 210.27 g·mol−1

Molar mass of tetraphenylcyclopentadienone is 384.5 g·mol−1

Now,

2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole

2.2 g dibenzyl ketone = 2.2 / 210.27 = 0.0105 mole

3.0 g of tetraphenylcyclopentadienone = 3 / 384.5  = 0.0078 mole

Now, the limiting reagent is benzil. 0.0095 mole can reacts wiyh 0.0095 mole of dibenzyl ketone

percentage yield = ( 0.0078 mole / 0.0095 mole ) × 100%

= 0.82 × 100%

= 82%

Therefore, the % yield is 82%

Where are the questions?

A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change?

The pKa of acetic acid is 4.76.

Chemistry Buffer Calculations

1 Answer

Stefan V.

May 8, 2016

Δ

pH

=

0.29

Explanation:

!! LONG ANSWER !!

The idea here is that you need to use the Henderson-Hasselbalch equation to determine the ratio that exists between the concentration of the weak acid and of its conjugate base in the buffer solution.

Once you know that, you can use the total molarity of the acid and of the conjugate base to find the number of moles of these two chemical species present in the buffer.

So, the Henderson-Hasselbalch equation looks like this

∣

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

pH

=

p

K

a

+

log

(

[

conjugate base

]

[

weak acid

]

)

a

a

∣

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

In your case, you have acetic acid,

CH

3

COOH

, as the weak acid and the acetate anion,

CH

3

COO

−

, as its conjugate base. The

p

K

a

of the acid is said to be equal to

4.76

, which means that you have

pH

=

4.76

+

log

(

[

CH

3

COO

−

]

[

CH

3

COOH

]

)

The pH is equal to

5

, and so

5.00

=

4.76

+

log

(

[

CH

3

COO

−

]

[

CH

3

COOH

]

)

log

(

[

CH

3

COO

−

]

[

CH

3

COOH

]

)

=

0.24

This will be equivalent to

10

log

(

[

CH

3

COO

−

]

[

CH

3

COOH

]

)

=

10

0.24

which will give you

[

CH

3

COO

−

]

[

CH

3

COOH

]

=

1.74

This means that your buffer contains

1.74

times more conjugate base than weak acid

[

CH

3

COO

−

]

=

1.74

×

[

CH

3

COOH

]

Now, because both chemical species share the same volume,

120 mL

, this can be rewritten as

n

C

H

3

C

O

O

−

120

⋅

10

−

3

L

=

1.74

×

n

C

H

3

C

O

O

H

120

⋅

10

−

3

L

which is

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

n

C

H

3

C

O

O

−

=

1.74

×

n

C

H

3

C

O

O

H

a

a

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

(

1

)

So, the buffer contains

1.74

times more moles of acetate anions that of acetic acid.

Now, the total molarity of the buffer is said to be equal to

0.1 M

. You thus have

[

CH

3

COOH

]

+

[

CH

3

COO

−

]

=

0.10 M

Once again, use the volume of the buffer to write

n

C

H

3

C

O

O

H

120

⋅

10

−

3

L

+

n

C

H

3

C

O

O

−

120

⋅

10

−

3

L

=

0.1

moles

L

This will be equivalent to

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

n

C

H

3

C

O

O

−

+

n

C

H

3

C

O

O

H

=

0.012

a

a

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

(

2

)

Use equations

(

1

)

and

(

2

)

to find how many moles of acetate ions you have in the buffer

1.74

⋅

n

C

H

3

C

O

O

H

+

n

C

H

3

C

O

O

H

=

0.012

n

C

H

3

C

O

O

H

=

0.012

1.74

+

1

=

0.004380 moles CH

3

COOH

This means that you have

n

C

H

3

C

O

O

−

=

1.74

⋅

0.004380 moles

n

C

H

3

C

O

O

−

=

0.007621 moles CH

3

COO

−

Now, hydrochloric acid,

HCl

, will react with the acetate anions to form acetic acid and chloride anions,

Cl

−

H

Cl

(

a

q

)

+

CH

3

COO

−

(

a

q

)

→

CH

3

COO

H

(

a

q

)

+

Cl

−

(

a

q

)

Notice that the reaction consumes hydrochloric acid and acetate ions in a

1

:

1

mole ratio, and produces acetic acid in a

1

:

1

mole ratio.

Use the molarity and volume of the hydrochloric acid solution to determine how many moles of strong acid you have

∣

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

c

=

n

solute

V

solution

⇒

n

solute

=

c

⋅

V

solution

a

a

∣

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

In your case, this gets you

n

H

C

l

=

0.300 mol

L

−

1

⋅

volume in liters

6.60

⋅

10

−

3

L

n

H

C

l

=

0.001980 moles HCl

The hydrochloric acid will be completely consumed by the reaction, and the resulting solution will contain

n

H

C

l

=

0 moles

→

completely consumed

n

C

H

3

C

O

O

−

=

0.007621 moles

−

0.001980 moles

=

0.005641 moles CH

3

COO

−

n

C

H

3

C

O

O

H

=

0.004380 moles

+

0.001980 moles

=

0.006360 moles CH

3

COOH

The total volume of the solution will now be

V

total

=

120 mL

+

6.60 mL

=

126.6 mL

The concentrations of acetic acid and acetate ions will be

[

CH

3

COOH

]

=

0.006360 moles

126.6

⋅

10

−

3

L

=

0.05024 M

[

CH

3

COO

−

]

=

0.005641 moles

126.6

⋅

10

−

3

L

=

0.04456 M

Use the Henderson-Hasselbalch equation to find the new pH of the solution

pH

=

4.76

+

log

(

0.04456

M

0.05024

M

)

pH

=

4.71

Therefore, the pH of the solution decreased by

Δ

pH

=

|

4.71

−

5.00

|

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

0.29 units

a

a

∣

∣

−−−−−−−−−−−−−

Answer link

Related topic

Buffer Calculations

Questions

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Answer:

A₅B₄

Explanation:

Since we have one atom of element A at the center of each face of the unit cell, since the unit cell is a cubic cell, we have 6 faces. Since the atom on the face of the unit cell is shared with another cell, we have half of it in the unit cell is shared So, the number of atoms per face is 1/2 atom/face × 6 faces = 4 atoms on the faces of the unit cell.

Also, we have 1 atom at each corner of the cubic unit cell. Since there are 8 corner in the cubic unit cell. Also, each atom at the corner is shared with 8 unit cells, so we have 1/8 atom per corner. So, the number of atoms per unit cell is 1/8 atom/corner × 8 corners = 1 atoms at the corners of the unit cell.

So, in total we have 4 + 1 = 5 atoms of element A in the unit cell.

Also, there are 4 atoms of element B in the unit cell.

So, the ratio of atoms of element A to element B is 5 : 4.

A:B = 5:4

So, the empirical formula of the compound containing elements A and B is A₅B₄

Answer:

5.75L is the volume of the sample after the reaction

Explanation:

Based on the reaction, 1 mole of H2 reacts with 1 mole of O2 to produce 1 mole of H2O2.

As in the reaction, 0.1200 moles of H2 and 0.1200 moles of O2 are added, 0.1200 moles of H2O2 are produced.

Before the reaction, the moles of gas are 0.2400 moles and after the reaction the moles are 0.1200 moles of gas.

Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:

V1/n1 = V2/n2

Where V is volume and n are moles of 1, initial state and 2, final state.

Replacing:

V1 = 11.5L

n1 = 0.2400 moles

V2 = ?

n2 = 0.1200 moles

11.5L*0.1200 moles / 0.2400 moles = V2

V2 = 5.75L is the volume of the sample after the reaction

[tex]\\ \large\sf\longmapsto H_2SO_4[/tex]

[tex]\\ \large\sf\longmapsto 2(1u)+32u+4(16u)[/tex]

[tex]\\ \large\sf\longmapsto 2u+32u+64u[/tex]

[tex]\\ \large\sf\longmapsto 98u[/tex]

[tex]\\ \large\sf\longmapsto 98g/mol[/tex]

[tex]\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:Mass}}[/tex]

[tex]\\ \large\sf\longmapsto No\:of\:moles=\dfrac{245}{98}[/tex]

[tex]\\ \large\sf\longmapsto No\:of\:moles=2.5mol[/tex]

Now

[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Volume\:of\:solution\;in\;L}}[/tex]

[tex]\\ \large\sf\longmapsto Molarity=\dfrac{2.5}{1}[/tex]

[tex]\\ \large\sf\longmapsto Molarity=2.5M[/tex]

Answer:

Explanation:  Độ thẩm thấu của NaCl 0.9% và glucose 5% lần lượt là 308 và 278 ... Dung dịch natri clorid sử dụng trong pha thuốc tiêm truyền thường dùng

The name of the compound shown below is 1- pentene. The correct answer is option D.

A compound is a substance made up of two or more different elements chemically bonded together in a fixed ratio.

1-pentene is an unsaturated hydrocarbon with the chemical formula [tex]\rm C_5H_{10}[/tex]. It is an alkene, which means it contains a carbon-carbon double bond.

Therefore, option D. 1-pentene is the name of the compound shown.

Learn more about compound here:

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