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11Alexandr11 [23.1K]
1 year ago
7

Select the correct set of quantum numbers (n, l, ml, ms) for the first electron removed in the formation of a cation for stronti

um, Sr. 5, 1, 1, ½ 5, 0, 0, –½ 5, 0, 1, ½ 5, 1, 0, ½ 5, 1, 0, –½
Chemistry
1 answer:
matrenka [14]1 year ago
7 0

Answer:

5,0,0,-1/2

Explanation:

The quantum numbers are a way to characterize the electrons, and so, identify the region that it's more probable to find it (orbital). They are:

- Principal quantum number (n): represents the shell or level, and varies from 1 to 7, and are represented by the letter K, L, M, N, O, P, and Q.

- Azimuthal quantum number (l): represents the subshell or sublevel, and is represented by 0,1,2,3.., and for the letters s, p, d, f,...

- Magnetic quantum number (ml): represents the orbital. It varies from -l to +l passing by 0. Each orbital can have 2 electrons.

- Spin quantum number (ms): represents the spin of the electron. It can be +1/2 or -1/2.

The strontium has an atomic number equal to 38, by the Linus Pauling's diagram, the electronic distribution is:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²

The valence electron is at the subshell 5s, which has only one magnetic quantum number: 0. Because it has 2 electrons, the first one has spin =1/2, and the other -1/2. So the first electron of the formation of cation has quantum numbers:

n = 5; l = 0; ml = 0; ms = -1/2

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Silver chloride is formed by mixing silver nitrate and barium chloride solutions. What volume of 1.50 M barium chloride solution
konstantin123 [22]

Answer:

1.22 mL

Explanation:

Let's consider the following balanced reaction.

2 AgNO₃ + BaCl₂ ⇄ Ba(NO₃)₂ + 2 AgCl

The molar mass of silver chloride is 143.32 g/mol. The moles corresponding to 0.525 g are:

0.525 g × (1 mol/143.32 g) = 3.66 × 10⁻³ mol

The molar ratio of AgCl to BaCl₂ is 2:1. The moles  of BaCl₂ are 1/2 × 3.66 × 10⁻³ mol = 1.83 × 10⁻³ mol.

The volume of 1.50 M barium chloride containing 1.83 × 10⁻³ moles is:

1.83 × 10⁻³ mol × (1 L/1.50 mol) = 1.22 × 10⁻³ L = 1.22 mL

8 0
1 year ago
When sample a of methane, ch4, decomposes, it produces 35.0 grams of c and 2.04 grams of h. when sample b of ch4 decomposes, it
Free_Kalibri [48]
Since both samples are pure CH4 (methane), the proportion of C to H that evolves from the decomposition should be equal. In equation form:
35.0 g C / 2.04 g H = 23.0 g C / x g H
Solving for x gives a value of x = 1.3406 g H
So 1.3406 grams of hydrogen will be produced from sample b.
7 0
2 years ago
How many grams are in 2.3 x10-4 moles of calcium phosphate ca3(po3)2
Airida [17]

Answer:

calcium phosphate has the formula Ca3(PO4)2, which has the mass of 310 grams/mol.

1 mol contains 310 grams

2.3*10^-4 moles contain  2.3*10^-4 * 310, which means 713*10^-4 grams, or 71.3 milligrams.

If you wrote the formula right and named the compound wrong, all you have to do is replace 310 with 278 and the answer will be 639.4*10^-4 grams, or 63.94 milligrams.

6 0
1 year ago
Calculate the mass of oxygen (in mg) dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air. Assume the
AleksAgata [21]

Answer:

50 mg

Explanation:

First, we have to calculate the partial pressure of O₂ (pO₂) using the following expression.

pO₂ = P × X(O₂) = 1.13 atm × 0.21 = 0.24 atm

where,

P: total pressure

X(O₂): mole fraction of oxygen

Then, we can calculate the concentration of O₂ in water (C) using Henry's law.

C = k × pO₂ = (1.3 × 10⁻³ M/atm) × 0.24 atm = 3.1 × 10⁻⁴ M

where,

k: Henry's constant for O₂

The mass of oxygen in a 5.00 L bucket with a concentration of 3.1 × 10⁻⁴ M is: (MM 32.0 g/mol)

5.00L.\frac{3.1 \times 10^{-4}mol}{L} .\frac{32.0 \times 10^{3}mg}{mol} =50mg

6 0
2 years ago
A gas is know to have a volume of 7.81 liters when the pressure is 754 torr what would be the volume when the pressure is change
Elis [28]

Answer:

The volume when the pressure is changed to 1.23 atm and temperature is constant will be <u><em>6.3075 L</em></u>.

Explanation:

Pressure and volume are related by Boyle's law that says:

"The volume occupied by a certain gas mass at a constant temperature is inversely proportional to the pressure"

Boyle's law is expressed mathematically as:

P * V = k

where:

  • P: Pressure
  • V: Volume
  • k: Constant

Assuming a certain volume of gas V1 is at a pressure P1 at the beginning of the experiment. By varying the volume of gas to a new V2 value, then the pressure will change to P2, and the following will be true:

P1 * V1 = P2 * V2

In this case you have:

  • P1= 754 torr= 0.9921 atm (1 atm=760 torr)
  • V1= 7.82 L
  • P2=1.23 atm
  • V2=?

Replacing:

0.9921 atm*7.82 L=1.23 atm*V2

Resolving:

V2=\frac{0.9921 atm*7.82 L}{1.23 atm}

V2≅6.3075 L

<u><em>The volume when the pressure is changed to 1.23 atm and temperature is constant will be 6.3075 L.</em></u>

6 0
1 year ago
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