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xxTIMURxx [149]
3 years ago
10

When atoms come together, there is _______ between the negative electrons clouds.

Chemistry
1 answer:
Verdich [7]3 years ago
7 0
I think the answer is ‘repulsion’
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Calculate the maximum amount of useful work that can be obtained and comment on the spontaneity for the reaction at 25C :
harkovskaia [24]

Answer:

1.41 *10^{3}  kJ/mol

Explanation:

First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)

Then we solve the ecuation for ΔH°reaction

ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)

ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol

ΔH°reaction = -1.41 *10^3  kJ/mol

3 0
3 years ago
"native," or elemental copper can be found in nature, but most copper is mined as oxide or sulfide minerals. chalcopyrite (cufes
max2010maxim [7]
I acually habe no idea but I nust want to point out that my toast's brand is Sarah Lee lol
7 0
3 years ago
Read 2 more answers
The concentration of urea in a solution prepared by dissolving 16 g of urea in 20 g of H2OH2O is ________% by mass. The molar ma
Arada [10]

Answer:

44

Explanation:

Given that :

Mass of solute = Mass of urea = 16g

Mass of water = 20g

Mass of solution = (mass of solute + mass of solvent) = (mass of urea + mass of water) = (16g + 20g) = 36g

Percentage Mass = (mass of solute / mass of solution) * 100%

Percentage Mass = (16 / 36) * 100%

Percentage Mass = 0.4444444 x 100%

Percentage Mass = 44.44%

Percentage Mass = 44%

4 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

+x

Equilibrium

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

ICE

C2H4  

H2  

C2H6  

Initial

0.33

0.53

0

Change

-x

-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  

+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.

ipR.O.B.O.T is an international Internet Protocol Recessive Observation Branch Organization Technologies

4 0
3 years ago
Chemists use letters of the alphabet as ____ for the elements.
d1i1m1o1n [39]

Answer:

The symbol is the right answer.

Explanation:

The “ Symbol” is the correct answer because chemist uses the letters of the alphabet to denote the element. For instance, the element oxygen is denoted by the letter of the alphabet “O”, the hydrogen is denoted by the letter of alphabet “H”, Boron is denoted by the letter of alphabet “B”, etc. Here these are the examples that use one letter but there are other elements that use more than 1 letter as the symbol. For example, the Chlorine is represented by the Cl.

8 0
3 years ago
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