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3 years ago

Characteristics of bipolar disorder typically include

Chemistry

2 answers:

crimeas [40]3 years ago

7 0

The correct answer is option b, that is, difficulty concentrating.

Manic depression also called bipolar disorder refers to a mental disorder, which leads to periods of unusually elevated mood and the periods of depression. The elevated mood is substantial and is called hypomania or mania, on the basis of its severity, or whether the signs of psychosis are there.

One of the usual sign of bipolar disorder is the difficulty in concentrating or diminished the tendency to think and indecisiveness. The individual suffering with the condition seems to make wrong decisions due to this sign, and as they do not think completely, the patients suffering from bipolar disorder possess poor decision making.

I am Lyosha [343]3 years ago

3 0

C. Difficulty in concentrating.
One of the common symptom for Bipolar disorder is the difficulty in concentration or decreased ability to think and indecisiveness. They tend to make wrong decisions because of this symptom. And because they do not think thoroughly, bipolar disorder patients have poor decision making.

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5. Which contains more nitrogen: 60g of urea, (NH2)2CO, or 100g of ammonium sulphate, (NH4)2SO4

topjm [15]

Answer:

Urea contains more nitrogen

Explanation:

1 mole of Urea contains 2 moles of Nitrogen and 1 mole of ammonium sulfate contains, also, 2 moles of nitrogen.

60g of urea (Molar mass: 60g/mol) contains:

60g × (1mol / 60g) = 1 mole. As 1 mole of urea contains 2 moles of nitrogen, moles of nitrogen are 2.

100g of ammonium sulfate (Molar mass: 132g/mol) contains:

100g × (1mol / 132g) = 0.758 moles.

As 1 mole of urea contains 2 moles of nitrogen, moles of nitrogen are 0.758×2 = 1.516 moles.

That means, urea contains more nitrogen.

5 0

3 years ago

Where are most of the nonmetals found in the periodic table?. . A.. at the top. . B.. on the right side. . C.. in the bottom lef

mote1985 [20]

The correct answer among the choices listed is option B. Most of the nonmetals found are found in the right side of the periodic table. There are seventeen nonmetal elements that can found on that side. Nonmetal elements are poor conductors of heat and electricity.

6 0

3 years ago

an electron in the 3rd shell of an Aluminium atom moves to the first shell in a bombardment process. Calculate the frequency of

Liula [17]

Following the quantic theory, the energy of a photon equals the radiation frequency multiplied by the universal constant. ν = 2.923x10¹⁵ Hz. E = 3.09x10¹⁵Hz.

<h3>What is quantum mechanic?</h3>

It is the branch of physics that studies objects and forces at a very low scale, at atoms, subatoms, and particles levels.

Quantum mechanics states that the elemental particles that constitute matter -electrons, neutrons, protons- have the properties of a wave and a particle.

It emerges from the quantic theory exposed by Max Planck (1922), in which he affirmed that light propagates in energy packages or photons.

He discovered the Universal Planck constant, h, used to calculate the energy of a photon.

He stated that the energy of a photon (E) equals the radiation frequency (ν) multiplied by the universal constant (h).

E = νh

In the exposed example, we need to calculate the energy required to change from the 3rd shell to the first shell.

To do it, we should know that the energy in a level (Eₙ) equals the energy associated to an electron in the most inferior energy level (E₁) divided by the square of the shell number (n²).

Eₙ = E₁ / n²

E₁ is a constant. We can express it in Joules or electroVolts

E₁ = -2.18x10⁻¹⁸ J
E₁ = -13.6 eV

So, let us calculate the energy at level 1 and 3

Eₙ = E₁ / n²

E₁ = -2.18x10⁻¹⁸ J / 1² = -2.18x10⁻¹⁸ J

E₁ = -13.6 eV / 1² = -13.6 eV

E₃ = -2.18x10⁻¹⁸ J / 3² = -2.18x10⁻¹⁸ J / 9 = - 2.42x10⁻¹⁹ J

E₃ = -13.6 eV / 3² = -13.6 eV / 9 = - 1.51 eV

The change of energy can be calculated in two ways,

Option 1

ΔE = E₁ - E₃ = 2.18x10⁻¹⁸ - 2.42x10⁻¹⁹ = 1.93x10⁻¹⁸J

ΔE = E₁ - E₃ = 13.6 - 1.51 = 12.09 eV

Option 2

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE =-13.6 eV (1/nf² - 1/ni²)

Where nf is the final level and ni is the initial level. When the electron passes from its initial level to its final level it is called electronic transition.

ni = 3
nf = 1

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE = -2.18x10⁻¹⁸ J (1/1² - 1/3²)

ΔE = -2.18x10⁻¹⁸ J (1 - 0.111)

ΔE = -2.18x10⁻¹⁸ J (0.888)

ΔE = - 1.937x10⁻¹⁸ J

ΔE = -13.6 eV (1/nf² - 1/ni²)

ΔE = -13.6 eV (1/1² - 1/3²)

ΔE = -13.6 eV (1 - 0.111)

ΔE = -13.6 eV (0.888)

ΔE = -12.08 eV

This is the energy required for the electron to go from n= 3 to n = 1. The negative sign (-) means energy (as light or photons) released or emitted.

If we want to express the result in Hz, we just need to make a conversion.

1Hz ⇔ 6.626x10⁻³⁴J ⇔ 4.136x10¹⁵ eV.

The energy required for the electron to go from n= 3 to n = 1 is 3.09x10¹⁵ Hz.

Now, we need to calculate the frequency, ν. This is, how many times the wave oscillates back and foward per second.

To do it, we will use the universal Planck constant, h, and the absolute value of the energy, E.

ν = E/h = 1.937x10⁻¹⁸ J / 6.626x10⁻³⁴ Js = 2.923x10¹⁵ 1/s = 2.923x10¹⁵ Hz.

Answer: