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Misha Larkins [42]
4 years ago
14

An electron in an atom is known to be in a state with magnetic quantum number ml=0. What is the smallest possible value of the p

rincipal quantum number of the state?
Chemistry
1 answer:
rusak2 [61]4 years ago
3 0

Answer:

1

Explanation:

The principal quantum number, n, showsthe principal electron shell. which can inturn be describe as the most probable distance of the electrons from the nucleus, the larger the number n is, the farther the electron is from the nucleus, the larger the size of the orbital, and the larger the atom is. n can be any positive integer starting at 1, as n=1 designates the first principal shell (the innermost shell). When an electron is in an excited state or it gains energy, it may jump to the second principle shell, where n=2

As the energy of the electron increases, so does the principal quantum number, e.g., n = 3 indicates the third principal shell, n = 4 indicates the fourth principal shell, and so on. n=1,2,3,4…

On the other hand, the magnetic quantum number ml determines the number of orbitals and their orientation within a subshell. Consequently, its value depends on the orbital angular momentum quantum number l. Given a certain l, ml is an interval ranging from –l to +l, so it can be zero, a negative integer, or a positive integer. ml=−l,(−l+1),(−l+2),…,−2,−1,0,1,2,…(l–1),(l–2),+l

note also that: The orbital angular momentum quantum number l determines the shape of an orbital, and therefore the angular distribution. The number of angular nodes is equal to the value of the angular momentum quantum number l. Each value of l indicates a specific s, p, d, f subshell (each unique in shape.) The value of l is dependent on the principal quantum number n. Unlike n, the value of l can be zero. It can also be a positive integer, but it cannot be larger than one less than the principal quantum number (n-1): l=0,1,2,3,4…,(n−1).

So the answer is 1

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21 pts. Sodium reacts with oxygen to produce sodium oxide as described by the balanced equation below. If 54.1g of sodium reacts
Aleksandr [31]
First, calculate the amount fo Na2O that should be produced with the given amount of sodium using proper dimensional analysis and  the balanced chemical reaction,

   theoretical amount of Na2O = (54.1 g of Na)(1 mol Na/23 g Na)(2 mos Na2O/4 mols Na)(62 g Na2O/ 1mol Na2O)
       
      theoretical amount of Na2O = 72.9 g of Na2O


Calculation of percent yield,

                % yield = (61.8 g Na2O) / (72.9 g of Na2O) x 100%
                 % yield = 84.75%

Answer: 84.75%
8 0
3 years ago
For the reaction ? NO + ? O2 → ? NO2 , what is the maximum amount of NO2 which could be formed from 16.42 mol of NO and 14.47 mo
stira [4]

Answer:

1a. The balanced equation is given below:

2NO + O2 → 2NO2

The coefficients are 2, 1, 2

1b. 755.32g of NO2

2a. The balanced equation is given below:

2C6H6 + 15O2 → 12CO2 + 6H2O

The coefficients are 2, 15, 12, 6

2b. 126.25g of CO2

Explanation:

1a. Step 1:

Equation for the reaction. This is given below:

NO + O2 → NO2

1a. Step 2:

Balancing the equation. This is illustrated below:

NO + O2 → NO2

There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by putting 2 in front of NO and 2 in front of NO2 as shown below:

2NO + O2 → 2NO2

The equation is balanced.

The coefficients are 2, 1, 2

1b. Step 1:

Determination of the limiting reactant. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO required 1 mole of O2.

Therefore, 16.42 moles of NO will require = 16.42/2 = 8.21 moles of O2.

From the calculations made above, there are leftover for O2 as 8.21 moles out of 14.47 moles reacted. Therefore, NO is the limiting reactant and O2 is the excess reactant.

1b. Step 2:

Determination of the maximum amount of NO2 produced. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO produced 2 moles of NO2.

Therefore, 16.42 moles of NO will also produce 16.42 moles of NO2.

1b. Step 3:

Conversion of 16.42 moles of NO2 to grams. This is illustrated below:

Molar Mass of NO2 = 14 + (2x16) = 14 + 32 = 46g/mol

Mole of NO2 = 16.42 moles

Mass of NO2 =?

Mass = number of mole x molar Mass

Mass of NO2 = 16.42 x 46

Mass of NO2 = 755.32g

Therefore, the maximum amount of NO2 produced is 755.32g

2a. Step 1:

The equation for the reaction.

C6H6 + O2 → CO2 + H2O

2a. Step 2:

Balancing the equation:

C6H6 + O2 → CO2 + H2O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by 6 in front of CO2 as shown below:

C6H6 + O2 → 6CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:

C6H6 + O2 → 6CO2 + 3H2O

There are a total of 15 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 15/2 in front of O2 as shown below:

C6H6 + 15/2O2 → 6CO2 + 3H2O

Multiply through by 2 to clear the fraction.

2C6H6 + 15O2 → 12CO2 + 6H2O

Now, the equation is balanced.

The coefficients are 2, 15, 12, 6

2b. Step 1:

Determination of the mass of C6H6 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of C6H6 = (12x6) + (6x1) = 72 + 6 = 78g/mol

Mass of C6H6 from the balanced equation = 2 x 78 = 156g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 15 x 32 = 480g

2b. Step 2:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

156g of C6H6 required 480g of O2.

Therefore, 37.3g of C6H6 will require = (37.3x480)/156 = 114.77g of O2.

From the calculations made above, there are leftover for O2 as 114.77g out of 126.1g reacted. Therefore, O2 is the excess reactant and C6H6 is the limiting reactant.

2b. Step 3:

Determination of mass of CO2 produced from the balanced equation. This is illustrated belowb

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 12 x 44 = 528g

2b. Step 4:

Determination of the mass of CO2 produced by reacting 37.3g of C6H6 and 126.1g O2. This is illustrated below:

From the balanced equation above,

156g of C6H6 produced 528g of CO2.

Therefore, 37.3g of C6H6 will produce = (37.3x528)/156 = 126.25g of CO2

5 0
4 years ago
112,300 joules heat is transferred when 240 g of a
Flura [38]

Answer:

0.54 J/goC

Explanation:

Recall that we define the heat transferred as ;

H= mcθ

Where;

H= heat transferred = 112300J

m= mass of the metal= 240g or 0.24 Kg

c= specific heat capacity of the metal = the unknown

θ= change in temperature = (880°C-13°C) = 867°C

Substituting values and making the specific heat capacity the subject of the formula;

c= H/mθ

c= 112300/ 240×867

c= 0.54 J/goC

Hence, the specific heat capacity of the metal is 0.54 J/goC.

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