It has been proposed that North America is moving west 2cm per year. How many kilometers would it move in 5,000 years?

Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol

prohojiy [21]

Hey there!:

From the given data ;

Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar = g/L * mol/g

[E]t = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹

Hope that helps!

4 0

3 years ago

8. Calculate the number of moles of eachsubstance.a. 5.45 x 1026 particles of methane, CH4

Klio2033 [76]

<em>ANSWER</em>

The number of moles of methane is 905.32 moles

STEP-BY-STEP EXPLANATION:

Given information

The number of particles of methane = 5.45 x 10^26 particles

Let x represents the number of moles of methane

To calculate the number of moles, we will be using the below formula

$\text{Number of particles = number of moles x Avogadro's constant}$

Recall that, the Avogadro's constant is given as

$6.02\cdot10^{23}$ $\begin{gathered} 5.45\cdot10^{26}\text{ = x }\cdot\text{ 6.02 }\cdot10^{23} \\ \text{Divide both sides by 6.02 }\cdot10^{23} \\ x\text{ = }\frac{5.45\cdot10^{26}}{6.02\cdot10^{23}} \\ x\text{ = }\frac{5.45}{6.02}\cdot10^{26\text{ - 23}} \\ x\text{ = 0.9053 }\cdot10^3 \\ x\text{ = 905.32 moles} \end{gathered}$

Therefore, the number of moles of methane is 905.32 moles

6 0

1 year ago

How many moles of calcium chloride are contained in a 333 gram sample?

Bogdan [553]

Answer & Explanation:

The molar mass of calcium chloride is 110.98 g/mol. We can use this information to solve this problem. We can set up our equation like this..

$\frac{333 g(CaCl2)}{} *\frac{1mol(CaCl2)}{110.98g(CaCl2)}$

Multiply straight across on the top and straight across on the bottom.

$\frac{333}{110.98}$

Now divide.

$\frac{333}{110.98}=3.00$

So, there are 3.00 moles of calcium chloride contained in a 33 gram sample which is answer choice D.

6 0

3 years ago

I need help with this for chemistry. I don’t understand now to do this.

alina1380 [7]

The ipR.O.B.O.T states

aA+bB⇌ cC+dD

the equilibrium constant is written as follows:

Kc=[C]c[D]d[A]a[B]b

The ICE Table

The easiest approach for calculating equilibrium concentrations is to use an ICE Table, which is an organized method to track which quantities are known and which need to be calculated. ICE stands for:

"I" is for the "initial" concentration or the initial amount

"C" is for the "change" in concentration or change in the amount from the initial state to equilibrium

"E" is for the "equilibrium" concentration or amount and represents the expression for the amounts at equilibrium.

For the gaseous hydrogenation reaction below, what is the concentration for each substance at equilibrium?

C2H4(g)+H2(g)⇌C2H6(g)(1)

with Kc=0.98 characterized from previous experiments and with the following initial concentrations:

[C2H4]0=0.33

[H2]0=0.53

SOLUTION

First the equilibrium expression is written for this reaction:

Kc=[C2H6][C2H4][H2]=0.98(2)

ICE Table

The concentrations for the reactants are added to the "Initial" row of the table. The initial amount of C2H6 is not mentioned, so it is given a value of 0. This amount will change over the course of the reaction.

ICE

C2H4

H2

C2H6

Initial

0.33

0.53

0

Change

Equilibrium

ICE

C2H4

H2

C2H6

Initial

0.33

0.53

0

Change

-x

+x

Equilibrium

Equilibrium is determined by adding "Initial" and "Change together.

ICE

C2H4

H2

C2H6

Initial

0.33

0.53

0

Change

-x

+x

Equilibrium

0.33-x

0.53-x

x

The expressions in the "Equilibrium" row are substituted into the equilibrium constant expression to find calculate the value of x. The equilibrium expression is simplified into a quadratic expression as shown:

0.98=x(0.33−x)(0.53−x)(3)

0.98=xx2−0.86x+0.1749(4)

0.98(x2−0.86x+0.1749)=x(5)

0.98x2−0.8428x+0.171402=x(6)

0.98x2−1.8428x+0.171402=0(7)

The quadratic formula can be used as follows to solve for x:

x=−b±b2−4ac−−−−−−−√2a(8)

x=−0.1572±(−0.1572)2−4(0.98)(0.171402)−−−−−−−−−−−−−−−−−−−−−−−−−√2(0.98)(9)

x=1.78 or0.098(10)

Because there are two possible solutions, each must be checked to determine which is the real solution. They are plugged into the expression in the "Equilibrium" row for [C2H4]Eq :

[C2H4]Eq=(0.33−1.78)=−1.45(11)

[C2H4]Eq=(0.33−0.098)=0.23(12)

If x=1.78 then [C2H4]Eq is negative, which is impossible, therefore, x must equal 0.098.

So:

[C2H4]Eq=0.23M(13)

[H2]Eq=(0.53−0.0981)=0.43M(14)

[C2H6]Eq=0.098M(15)

Problems

1. Find the concentration of iodine in the following reaction if the equilibrium constant is 3.76 X 103, and 2 mol of iodine are initially placed in a 2 L flask at 100 K.

I2(g)⇌2I−(aq)(16)

2. What is the concentration of silver ions in 1.00 L of solution with 0.020 mol of AgCl and 0.020 mol of Cl- in the following reaction? The equilibrium constant is 1.8 x 10-10.

AgCl(s)⇌Ag+(aq)+Cl−(aq)(17)

3. What are the equilibrium concentrations of the products and reactants for the following equilibrium reaction?

Initial concentrations: [HSO−4]0=0.4 [H3O+]0=0.01 [SO2−4]0=0.07 K=.012

HSO−4(aq)+H2O(l)⇌H3O+(aq)+SO2−4(aq)(18)

4. The initial concentration of HCO3 is 0.16 M in the following reaction. What is the H+ concentration at equilibrium? Kc=0.20.

H2CO3⇌H+(aq)+CO2−3(aq)(19)

5.The initial concentration of PCl5 is 0.200 moles per liter and there are no products in the system when the reaction starts. If the equilibrium constant is 0.030, calculate all the concentrations at equilibrium.

Solutions

1.

I2

I−

Initial

2mol/2L = 1 M

0

Change

−x

+2x

Equilibrium

1−x

2x

At equilibrium

Kc=[I−]2[I2]

3.76×103=(2x)21−x=4x21−x

cross multiply

4x2+3.76.103x−3.76×103=0

apply the quadratic formula:

−b±b2−4ac−−−−−−−√2a

with: a=4 , b=3.76×103 c=−3.76×103 .

The formula gives solutions of of x=0.999 and -940. The latter solution is unphysical (a negative concentration). Therefore, x=0.999 at equilibrium.

[I−]=2x=1.99M(20)

[I2]=1−x=1−.999=0.001M(21)

2.

Ag+

Cl−

Initial

0

0.02mol/1.00 L = 0.02 M

Change

+x

Equilibrium

0.02+x

Kc=[Ag−][Cl−](22)

1.8×10−10=(x)(0.02+x)(23)

x2+0.02x−1.8×1010=0(24)

x=9×10−9(25)

[Ag−]=x=9×10−9(26)

[Cl−]=0.02+x=0.020(27)

3.

H2CO3

SO2−4

H3O+

Initial

0.4

0.01

0.07

Change

−x

Equilibrium

0.4−x

0.01+x

0.07+x

Kc=[SO2−4][H3O+]H2CO3(28)

0.012=(0.01+x)(0.07+x)0.4−x(29)

cross multiply and get:

x2+0.2x−0.0041=0(30)

apply the quadratic formula

x = 0.0328

[H2CO3]=0.4-x=0.4-0.0328=0.3672

[S042-]=0.01+x=0.01+0.0328=0.0428

[H30]=0.07+x=0.07+0.0328=0.1028

4.

H2CO3

H+

CO2−3

Initial

.16

0

Change

-x

Equilibrium

.16-x

apply the quadratic equation

x=0.1049

[H+]=x=0.1049

5. First write out the balanced equation:

PCl5(g)⇌PCl3(g)+Cl2(g)

PCl5

PCl3

Cl2

Initial

0.2

0

Change

-x

Equilibrium

0.2-x

Kc=[PC3][Cl2][PCl5](31)

0.30=x20.2−x(32)

Cross multiply:

x2+0.03x−0.006=0(33)

Apply the quadratic formula:

x=0.064

[PCl5]=0.2-x=0.136

[PCl3]=0.064

[Cl2]=0.064

Information is verified by Brainly Incorporations.

Do not copy this information without the consent of Brainly Inc.

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4 0

3 years ago

Which step in the free-radical chlorination of methane do you expect to be the most exothermic? attempt this problem without loo

stiks02 [169]

The termination step of the free-radical chlorination of methane is the most stable one among all three steps.

The free-radical substitution reaction between chlorine and methane features three major steps:

Initiation, during which chlorine molecules undergo homolytic fission to produce chlorine free radicals. Ultraviolet radiations are typically applied to supply the energy required for breaking the chlorine-chlorine single bonds. The initiation step is thus <em>endothermic</em>.

Propagation, a process in which chlorine free radicals react with methane molecules and remove a hydrogen atom from the alkane to produce hydrogen chloride and an alkyl radical e.g., $\cdot \text{CH}_3$ . The carbon-containing free radical would react with chlorine molecules to produce chloromethane and yet another chlorine free radical. This process can well repeat itself to chlorinate a significant number of methane molecules.

Termination. Free radicals combine to produce molecules. For example, two chlorine free radicals would combine to produce a chlorine molecule, whereas two alkyl free radicals would combine to produce an alkane with two-carbon atoms in its backbone.

Chemical processes that increase the stability of a substance reduces its chemical potential energy. Energy conserves, thus such processes would also release energy equal to the potential energy lost in quantity. Free radicals are unstable and- as seen in the propagation step- compete readily with neutral molecules for their electrons. The propagation step keeps the number of free radicals constant and is therefore more exothermic than the initiation step. The termination step reduces the number of free radicals, increase the stability of the system by the greatest extent, and is therefore the most exothermic step among the three.

3 0

3 years ago