Where do electromagnetic waves move slowest?

determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth m

mamaluj [8]

This is an incomplete question, here is a complete question.

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.

(a) $1s^22s^22p^63s^23p^5$

(b) $1s^22s^22p^63s^23p^63d^74s^2$

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

(d) $1s^22s^22p^63s^23p^63d^4s^1$

Answer :

(a) $1s^22s^22p^63s^23p^5$ → Halogen

(b) $1s^22s^22p^63s^23p^63d^74s^2$ → Transition metal

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$ → Transition metal

(d) $1s^22s^22p^63s^23p^63d^4s^1$ → Transition metal

Explanation :

Inert gas : These are the gases which lie in group 18.

Their general electronic configuration is: $ns^2np^6$ where n is the outermost shell.

Halogen : These are the elements which lie in group 17.

Their general electronic configuration is: $ns^2np^5$ where n is the outermost shell.

An alkali metal : These are the elements which lie in group 1.

Their general electronic configuration is: $ns^1$ where n is the outermost shell.

An alkaline earth metal : These are the elements which lie in group 2.

Their general electronic configuration is: $ns^2$ where n is the outermost shell.

Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.

Their general electronic configuration is: $(n-1)d^{1-10}ns^{0-2}$ where n is the outermost shell.

(a) $1s^22s^22p^63s^23p^5$

The element having this electronic configuration belongs to the halogen family.

(b) $1s^22s^22p^63s^23p^63d^74s^2$

The element having this electronic configuration belongs to the transition family.

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

The element having this electronic configuration belongs to the transition family.

(d) $1s^22s^22p^63s^23p^63d^4s^1$

The element having this electronic configuration belongs to the transition family.

4 0

3 years ago

What would ne the acceleration in a body moving wit uniform velocity and why

meriva

Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.

Hope this helps :) :)

3 0

3 years ago

Read 2 more answers

Is a planar carbon "backbone" possible for propane? Explain.

tankabanditka [31]

Answer:

Cyclopropane has a planar carbon back bone while propane does not

Explanation:

We have to recognize that in straight chain saturated organic compounds, carbon atoms have a tetrahedral geometry. Each carbon atom is bonded to four other atoms.

However, carbon atoms in cyclic compounds are also sp3 hybridized with each carbon bonded to only four other atoms but the ring system is highly strained.

Cyclopropane is a necessarily planar molecule with a bond angle that is far less than the expected tetrahedral bond angle due to strain in the molecule. Hence, the carbon atoms may have have a "planar backbone".

4 0

2 years ago

A lithium flame has a characteristic red color due to emissions of wavelength 671 nm. What is the mass equivalence of 1 mol of p

e-lub [12.9K]

Answer:

1.98x10⁻¹² kg

Explanation:

The energy of a photon is given by:

E= hc/λ

h is Planck's constant, 6.626x10⁻³⁴ J·s

c is the speed of light, 3x10⁸ m/s

and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)

E = 6.626x10⁻³⁴ J·s * 3x10⁸ m/s ÷ 6.71x10⁻⁷m = 2.96x10⁻¹⁹ J

Now we multiply that value by Avogadro's number, to calculate the energy of 1 mol of such protons:

1 mol = 6.023x10²³ photons

2.96x10⁻¹⁹ J * 6.023x10²³ = 1.78x10⁵ J

Finally we calculate the mass equivalence using the equation:

E=m*c²

m=E/c²

m = 1.78x10⁵ J / (3x10⁸ m/s)² = 1.98x10⁻¹² kg

6 0

3 years ago

A student determines the molar mass of acetone, CH3COCH3, by the method used in this experiment. She found that the equilibrium

Alexandra [31]

Answer:

(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.

Explanation:

Without mincing words let's dive straight into the solution to the question.

(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;

The freezing point depression = [ 1 - (-3)]° C = 4°C.

(b). The molality can be Determine by using the formula below;

Molality = the number of moles found in the solute/ solvent's weight(kg).

Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.

(c). The mass of acetone that was in the decanted solution = 11.1 g.

(d). The mass of water that was in the decanted solution = 89.01 g.

(e). 2.4 = x/ 58 × (1000/1000).

x = 2.4 × 58 = 139.2 g.

(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.

7 0

3 years ago