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Gre4nikov [31]
2 years ago
6

The atomic number is determined only by the number of ________ in the nucleus of an atom. But, in a neutral atom it also represe

nts the number of _________ in the electron cloud.
Chemistry
1 answer:
sveticcg [70]2 years ago
5 0

Hello!

The atomic number is determined only by the number of protons in the nucleus of an atom. But, in a neutral atom it also represents the number of electrons in the electron cloud.


Neutrons are only important in the nucleus for helping us find atomic weight, which varies as we move along the perodic table and does not always equal the same amount of it's atomic number. Which is why it would not be a suitable answer for the first blank space. Electrons do not work either as they do not exist inside the nucleus but rather outside the atom.

The second space, since it states is in the electron cloud, we can deduct that electrons would be an appropriate answer there.

If you need anymore help feel free to ask, but I hope this answers your question.

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Energy produced by nuclear fusion and fission
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Read 2 more answers
Calculate the values of ΔU, ΔH, and ΔS for the following process:
ladessa [460]

Answer:

ΔU = 45.814 KJ

ΔH = 46.4375 KJ

ΔS = 18.76 J/K

Explanation:

            H2O(l)        →          H2O(l)                →              H2O(steam)

   298.15K, 1atm   ΔHp     373.15K,1atm       ΔHv         373.15K,1 atm

∴ ΔHp = Qp = nCpΔT

∴ n H2O = 1 mol

∴ Cp,n = 75.3 J/mol.K

∴ ΔT = 373.25 - 298.15 = 75 K

⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J

⇒ ΔHp = 5647.5 J = 5.6475 KJ

⇒ ΔH = ΔHp + ΔHv

∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ  

⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ

ideal gas:

∴ ΔH = ΔU + PΔV

∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L

∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L

⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³

∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa

⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))

⇒ ΔU = 46.4375 KJ - 0.623 KJ

⇒ ΔU = 45.814 KJ

∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)

⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)

⇒ ΔS = 16.896 J/K + 1.863 J/K

⇒ ΔS = 18.76 J/K

3 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
What is the [OH-] if the [H3O+] is 1 x 10 -6?
Mariulka [41]

The answer is [OH⁻] = 1 x 10⁻⁸.

To find OH⁻, divide the ionic product of water by [H₃O⁺] as :

<u>OH⁻ + H₃O⁺ = H₂O</u>

<u />

  • [OH⁻] = 1 x 10⁻¹⁴ / 1 x 10⁻⁶
  • [OH⁻] = 1 x 10⁻⁸
5 0
2 years ago
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