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3 years ago

How many protons are there in 20.02 mol of neon (ne)?

2 answers:

7 0

In order to determine the number of protons in 20.02 mol of Ne, we use Avogadro's number to convert the number of moles to number of atoms, 1 mol = 6.022 x 10^23 atoms. From there, we must know the number of protons in a Neon atom, which is 10. Thus, the formula will be:

(20.02 mol Ne)x(6.022 x 10^23 atoms/mol)x(10 protons/1 atom Ne) =
1.2056 x 10^26 protons

Tanya [424]3 years ago

5 0

Answer: There are 1.216×10^26 protons in 20.02mole of neon

Explanation:

Using Avogadro's number

1 mole of neon contains 6.023×10^23 neon atoms.

I atm contains 10 protons

Number of protons = (6.023×10^23)×20.02×10= 1.216×10^26

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Paul [167]

Answer:

1. Percent composition of Al = 13.423 %

Percent composition of Zn = 28.02 %

Percent composition of Cl = 30.6 %

Percent composition of O = 41.3 %

3. The empirical formula is C₅O₁₆

4. Molecular Formula= P₄O₆

Explanation:

Part first :

Data Given

Formula of the Molecule = Al₂ (CrO₄)₃

% of Al₂ = ?

> First of all find the atomic masses of each component in a molecule

For Al₂ (CrO₄)₃ atomic masses are given below

Al = 27 g/mol

Cr = 52 g/mol

O = 16 g/mol

> Then find the total masses of each component

2 atoms of Al = 27 g/mol x 2

= 54 g/mol

3 atoms of Cr = 52 g/mol x 3

= 156 g/mol

12 atoms of O = 16 g/mol x 12

= 192 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Al₂ (CrO₄)₃ = [27x2 + 52x3 + 16x12]

Molar Mass of Al₂ (CrO₄)₃ = 402

Now to find the mass percent of Al

Formula used to find the Mass percent of a component

Percent composition of Al = mass of Al in Molecula / molar mass of Al₂(CrO₄)₃ x 100%

Put the values

Percent composition of Al = 54 (g/mol) / 402 (g/mol) x 100%

Percent composition of Al = 13.423 %

_______________________________________

Part 2

Data Given

Formula of the Molecule = Zn(ClO₃)₂

% Zn = ?

% Cl = ?

% O = ?

> First of all find the atomic masses of each component in a molecule

For Zn(ClO₃)₂ atomic masses are given below

Zn = 65 g/mol

Cl = 35.5 g/mol

O = 16 g/mol

> Then find the total masses of each component

1 atoms of Zn= 65 g/mol x 1

= 65 g/mol

2 atoms of Cl = 35.5 g/mol x

= 71 g/mol

6 atoms of O = 16 g/mol x 6

= 96 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Zn(ClO₃)₂ = [65x1 + 35.5x2 + 16x6]

Molar Mass of Zn(ClO₃)₂ = 232g/mol

Now to find the mass percent of of each component one by one

1. Formula used to find the mass percent of Zn

Percent composition of Zn= mass of Zn in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Zn = 65(g/mol) / 232 (g/mol) x 100%

Percent composition of Zn = 28.02 %

-------------------

2. Formula used to find the mass percent of Cl

Percent composition of Cl = mass of Cl in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Cl = 71 (g/mol) / 232 (g/mol) x 100%

Percent composition of Cl = 30.6 %

---------------------

3. Formula used to find the mass percent of O

Percent composition of O = mass of O in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of O = 96 (g/mol) / 232 (g/mol) x 100%

Percent composition of O = 41.3 %

________________________________________

Part 3:

Data Given

Percentage of C = 27.3 %

Percentage of O = 72.7 %

Emperical Formula of the compound = ?

Solution:

So the compound has 27.3 % C and 72% O

First, find the mass of each of the elements in 100 g of the Compound.

C = 27.3 g

O = 72 g

Now find how many moles are there for each element in 100 g of compound

For this molar mass are required

That is

C = 12 g/mol

O = 16 g/mol

Formula Used

mole of C = mass of C / Molar mass of C

mole of C = 27.3 / 12 g/mol

mole of C = 2.275

Formula Used

mole of O = mass of O / Molar mass of O

mole of O = 72g / 16 g/mol

mole of O = 7.2

Divide each one by the smallest number of moles

C = 2.275 / 2.275

C = 1

O = 7.2 / 2.275

O = 3.2

Multiply the mole fraction to a number to get the whole number.

C = 1 x 5 = 5

O = 3.2 x 5 = 16

So, the empirical formula is C₅O₁₆

______________________________________

Part 4

Data Given

Percentage of P= 56.38 %

Percentage of O = 43.62%

Molar Mass = 219.9g

Molecular Formula of the compound = ?

Solution:

First, find the mass of each of the elements in 100 g of the Compound.

Mass of P= 56.38g

Mass of O = 43.62g

Now find how many moles are there for each element in 100 g of compound

find the moles in total compounds

Formula Used

mole of P = mass of / Molar mass of P

mole of P = 56.38 g / 31 g/mol

mole of P = 1.818

Formula Used

mole of O = mass of O / Molar mass of O

mole of O = 43. 62 / 16 g/mol

mole of O = 2.7262

Now

first find the Emperical formula

Divide each one by the smallest number of moles

P = 1.818 /1.818

P= 1

for oxygen

O = 2.7262 / 1.818

O = 1.5

Multiply the mole fraction to a number to get the whole number.

P = 1 x 2 = 2

O = 1.5 x 2 = 3

So, the empirical formula is P₂O₃

Now

Find molar mass of the empirical formula P₂O₃

2 (31) + 3 (16) = 62 + 48 = 110

Now find that how many empirical units are in a molecular unit.

(219.9 g/mol) / ( 110 g/mol) = empirical units per molecular unit

empirical units per molecular unit = 1.999 =2

A here we get two empirical units in a molecular unit,

So the molecular formula is:

2 (P₂O₃) = P₄O₆

7 0

3 years ago

Given K = 3.61 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.19 at 45°C for the reaction 2 A(g)

Firlakuza [10]

Answer:

K = 0.55

Kp = 0.55

mol fraction B = 0.27

Explanation:

We need to calculate the equilibrium constant for the reaction:

C(g) + D(g) ⇄ 2B(g) K₁= ? (1)

and we are given the following equilibria with their respective Ks

A(g) + B(g) ⇄ C(g) K₂= 3.61 (2)

2 A(g) + D(g) ⇄ C(g) K₃= 7.19 (3)

all at 45 ºC.

What we need to do to solve this question is to manipulate equations (2) and (3) algebraically to get our desired equilibrium (1).

We are allowed to reverse reactions, in that case we take the reciprocal of K as our new K' ; we can also add two equilibria together, and the new equilibrium constant will be the product of their respective Ks .

Finally if we multiply by a number then we raise the old constant to that factor to get the new equilibrium constant.

With all this in mind, lets try to solve our question.

Notice A is not in our goal equilibrium (3) and we want D as a reactant . That suggests we should reverse the first equilibria and multiply it by two since we have 2 moles of B as product in our equilibrium (1) . Finally we would add (2) and (3) to get (1) which is our final goal.

2C(g) ⇄ 2A(g) + 2B(g) K₂´= ( 1/ 3.61 )²

₊

2 A(g) + D(g) ⇄ C(g) K₃ = 7.19

C(g) + D(g) ⇄ 2B(g) K₁ = ( 1/ 3.61 )² x 7.19

K₁ = 0.55

Kp is the same as K = 0.55 since the equilibrium constant expression only involves gases.

To compute the last part lets setup the following mnemonic ICE table to determine the quantities at equilibrium:

pressure (atm) C D B

initial 1.64 1.64 0

change -x -x +2x

equilibrium 1.64-x 1.64- 2x

Thus since

Kp =0.55 = pB²/ (pC x pD) = (2x)²/ (1.64 -x)² where p= partial pressure

Taking square root to both sides of the equation we have

√0.55 = 2x/(1.64 - x)

solving for x we obtain a value of 0.44 atm.

Thus at equilibrium we have:

(1.64 - 0.44) atm = 1.20 atm = pC = p D

2(0.44) = 0.88 = pB

mole fraction of B = partial pressure of B divided into the total gas pressure:

X(B) = 0.88 / ( 1.20 + 1.20 + 0.88 ) = 0.27

8 0

3 years ago

A neutron collides with a nitrogen atom, resulting in a transmutation. Balance the equation. Superscript 1 Subscript 0 Baseline

o-na [289]

Answer:

A is 14

Z is 6

X is C

Explanation:

Have a great day!

7 0

2 years ago