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zaharov [31]
3 years ago
11

Problem page an electron in an atom is known to be in a state with magnetic quantum number =ml−2 . what is the smallest possible

value of the principal quantum number n of the state?
Chemistry
2 answers:
vovikov84 [41]3 years ago
6 0

<em>n</em> = 3. The principal quantum number must be at least 3 to have a magnetic quantum number <em>ml</em> = -2

The rules are:

• <em>n</em> > 0

• <em>l </em>< <em>n</em>

• |<em>ml</em>| ≤ <em>l </em>

Thus, if <em>ml</em> = -2, <em>l</em> ≥ 2.

If <em>l </em>≥ 2 and n ><em> l</em>, the allowed values of <em>n</em> are 3, 4, 5, etc.

The smallest allowed value of <em>n</em> is <em>n</em>  = 3<em>. </em>

vfiekz [6]3 years ago
5 0

Answer:

n=3

Explanation:

Hello,

Magnetic number accounts for the energy levels of electrons within the subshells, in such a way, if ml=2 or -2, we realize that such atom has five subshells, -2,-1,0,1,2 (magnetic number equal to 2) which are contained into the third shell, this is n=3.

This is known due to the fact that the first shell has the subshell 0 and the second shell the subshells -1,0,1.

Best regards.

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Part A. Two containers, one at 305 K and the other at 295 K, are placed in contact with each other. 1. 1 J of heat flows from th
posledela

Answer:

0.00011 JK.

The process does NOT violate the second law of thermodynamics

Explanation:

The following parameters are given which are going to help in solving for the change in entropy of the system. The term "entropy'' simply means the degree of disorderliness of a system.

=> The temperature of container A = 305 K, the temperature of container B = 295 K and the amount of heat generated when the containers are placed in contact with each other = 1. 1 J.

The change in entropy of the hot container = -(1/305) = - 0.00328 J/K.

The change in entropy of the cold container = 1/295 = 0.00339 J/K.

Therefore, the change in the entropy of the system = - 0.00328 J/K + 0.00339 J/K = 0.00011 JK.

Note that the change in entropy of the system gives a positive value. Hence, this process does not violate the second law of thermodynamics.

The process does NOT violate the second law of thermodynamics.

7 0
3 years ago
What is the correct reaction for the dissociation (ionization) for a weak acid?
slava [35]

Answer: HA + H2O ⇌ H3O+ + H- (option #4)

Explanation: Since the acid is weak you have to use a ⇌ (equilibrium) sign. Equilibrium is denoted in a chemical equation by the ⇌ symbol. Also, when any acid dissolves into water, it produces hydronium (H3O+ or H+). Therefore, the fourth chemical reaction is your answer.

7 0
3 years ago
The number of grams of H2 in 1470 mL of H2 gas. ​
Lorico [155]

Answer:

0.1313 g.

Explanation:

  • It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
  • Suppose that hydrogen behaves ideally and at STP conditions.

<u><em>Using cross multiplication:</em></u>

1.0 mol of hydrogen occupies → 22.4 L.

??? mol of hydrogen occupies → 1.47 L.

∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.

  • Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:

<em>The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen</em> = (6.563 x 10⁻² mol)(2.0 g/mol) = <em>0.1313 g.</em>

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3 years ago
What are stars and galaxy made of?
goldfiish [28.3K]
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In a spiral galaxy like the Milky Way, the stars, gas, and dust are organized into a "bulge," a "disk" containing "spiral arms," and a "halo." Elliptical galaxies have a "bulge-shape" and a "halo," but do not have a "disk.


Hope it helped
4 0
3 years ago
Read 2 more answers
The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol
Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration PCl_3 and Cl_2.

\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}

\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M

\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}

\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M

The given equilibrium reaction is,

                            PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initially                 0.70        0.70              0

At equilibrium    (0.70-x)   (0.70-x)           x

The expression of K_c will be,

K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}

K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}

Now put all the given values in the above expression, we get:

49=\frac{(x)}{(0.70-x)\times (0.70-x)}

By solving the term x, we get

x=0.59\text{ and }0.83

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of PCl_3 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of Cl_2 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of PCl_5 at equilibrium = x = 0.59 M

Therefore, the concentration of PCl_3 at equilibrium is 0.11 M

3 0
4 years ago
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