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Tom [10]
2 years ago
9

Hi! ❤️ , im looking for some help here. ill give brainliest if able to.

Chemistry
1 answer:
dexar [7]2 years ago
7 0

A student builds a model of a race car. The scale is 1:15. In the scale model, the car is 8 cm tall. How tall is the actual car?

<h2>Answers:</h2>

<h3>A. 120 cm</h3>

#CarryOnLearning

You might be interested in
NEED HELP WITH ONE QUESTION.
const2013 [10]

Answer:

all I have to say is yea 1 like he said

5 0
2 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 <em>Joules </em>or <em>electroVolts </em>

  • 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 =<u>  -2.18x10⁻¹⁸</u><u> J</u>

        E₁ =  -13.6 eV / 1² =<u>  -13.6 </u><u>eV</u>

  • E₃ =  -2.18x10⁻¹⁸ J / 3² =  -2.18x10⁻¹⁸ J / 9 =<u> - 2.42x10⁻¹⁹ </u><u>J</u>

        E₃ =  -13.6 eV / 3² =  -13.6 eV / 9 = <u>- 1.51 </u><u>eV</u>

The change of energy can be calculated in two ways,

<u>Option 1</u>

ΔE = E₁ - E₃ = 2.18x10⁻¹⁸ - 2.42x10⁻¹⁹ =<u> 1.93x10⁻¹⁸</u><u>J</u>

ΔE = E₁ - E₃ = 13.6 - 1.51 = <u>12.09 </u><u>eV</u>

<u>Option 2</u>

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)

<u>ΔE</u><u> = - 1.937x10⁻¹⁸ </u><u>J</u>

or

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)

<u>ΔE</u><u> = -12.08</u><u> eV</u>

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.

<u />

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 <u>3.09x10¹⁵ </u><u>Hz</u><u>.</u>

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 =  <u>2.923x10¹⁵ Hz</u>.

<u>Answer</u>:

  • Frequency, ν = E/h = <u>2.923x10¹⁵ </u><u>Hz</u>.
  • Energy, E = <u>3.09x10¹⁵ </u><u>Hz</u><u>.</u>

You can learn more about quantum mechanic at

brainly.com/question/11855107

brainly.com/question/23780112

brainly.com/question/11852353

5 0
1 year ago
I WILL RATE BRAINLIEST!!!! NEED QUICK HELP!!!!
Alik [6]
You will need the Gas Law:

pV = nRT

Since T and p are constant, R is constant too, then moles increases->volume will increase with the same ratio too!
4 0
3 years ago
In a lab, silver nitrate,AgNO₃, is dissolved in water until no solid is observed in the container. Then, a solution of sodium ch
leva [86]

Answer:

The answer to your question is: A, C, E

Explanation:

After reading the information given, we can conclude that both reactants react to form a new solid product and another product that do not precipitate.

The reaction is:

                         AgNO₃ + NaCl   ⇒      AgCl  +  NaNO₃

a. Formation of a solid indicates that a chemical reaction occurred  This option is correct, precipitation indicates that a changed occurred in the solution due to a chemical reaction.

b. The initial disappearance of silver nitrate in water indicates a chemical reaction occurred  No, what is happing is that silver nitrate is dissolving in water, this process is a physical change because matter do not change its nature.

c. The color change of the solution indicates a chemical reaction occurred  This option is true, when a color appears suggest the formation of a new product.

d. A chemical reaction occurred when silver nitrate was added to water

This option is not correct, in this process silver nitrate is dissolving in water but it is not a chemical reaction because matter does not change is nature.

e. A chemical reaction occurred when the solutions were combined  This option is true because silver nitrate and sodium chloride react to form a new product (solid formation and change in color).

f. The lack of temperature change indicates a chemical reaction didn't occur This is false, some reactions do not need a change of color to happen.

6 0
3 years ago
The following reaction produces ethanoic acid (CHACOOH) from methanol (CH3OH) and carbon
tigry1 [53]

Answer:

\boxed{\text{300 g}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:         32                          60

           CH₃OH + CO ⟶ CH₃COOH

m/g:        160

(a) Moles of CH₃OH

\text{Moles of CH$_{3}$OH} = \text{160 g CH$_{3}$OH }\times \dfrac{\text{1 mol CH$_{3}$OH }}{\text{32 g CH$_{3}$OH}}= \text{5.00 mol CH$_{3}$OH}

(b) Moles of CH₃COOH

\text{Moles of CH$_{3}$COOH} = \text{5.00 mol CH$_{3}$OH } \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{1 mol CH$_{3}$OH }} = \text{5.00 mol CH$_{3}$COOH}

(c) Mass of CH₃COOH

\text{Mass of CH$_{3}$COOH} =\text{5.00 mol CH$_{3}$COOH} \times \dfrac{\text{60 g CH$_{3}$COOH}}{\text{1 mol CH$_{3}$COOH}} = \textbf{300 g CH$_{3}$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}

3 0
3 years ago
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