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suter [353]
3 years ago
11

In most atoms the charges of the protons and electrons cancel each other out and the atom has no what

Chemistry
1 answer:
11111nata11111 [884]3 years ago
7 0
The atom has no charge
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Determine the volume in liters occupied by 22.6 g of I2 gas at STP.
AleksandrR [38]

 The volume in liters  occupied  by 22.6 g  of I₂  gas  at STP  is  1.99 L (answer A)

 <u><em>calculation</em></u>

Step: find the  moles of I₂

moles= mass÷  molar mass

from  periodic table the  molar mass  of I₂  is  253.8 g/mol

moles = 22.6 g÷253.8 g/mol =0.089 moles

Step 2:find the volume  of I₂  at STP

At STP  1  moles =22.4 L

         0.089 moles= ? L

<em>by cross  multiplication</em>

={ (0.089 moles x 22.4 L) /1 mole} = 1.99 L


6 0
3 years ago
If the amount of radioactive sodium-24, used for studies of electrolytes within the body, in a sample decreases from 0.8 to 0.2
kirill [66]

Answer:

15 h

Explanation:

Okay, the first thing that we all have to know before we can answer this question is that this Topic that is, Chemistry of Radioactivity is related to kinetics in a way that Radioactive disintegration follows the first order of Reaction which is under kinetics. So, we will be using the first order kinetics rate law to answer this question. Using the equation (1) below;

k =[ 2.303/ t ]×log ([N°}/ [Nr]) --------(1).

We are given from the question that N° = initial sample = 0.8 mg and Nr= sample remaining = 0.2 and the time taken = t= 30.0 h.

k= (2.303/ 30.0 h ) × log (0.8/0.2).

k=0.076768 h^-1 × log (4).

k= 0.076768 h^-1 × 0.6021.

k= 0.0462 h^-1.

Therefore, using the formula for Calculating half life below for first order kinetics we will be able to find out answer.

k = ln 2/ t(1/2). Where t(1/2) is the half life.

t(1/2) = ln 2/ k.

t(1/2) = ln 2 / 0.0462 h^-1.

t(1/2)= 0.6931/0.0462 h^-1.

t(1/2)=15 h

6 0
3 years ago
To show the electron configuration for an atom, what is the advantage of using an orbital notation compared to a dot structure?
stich3 [128]
The advantage of using an orbital notation is that it shows the electron distribution in shells. 

<span>Dot structures only show the valence electrons of an atom which are the electrons found at the outermost shell. The orbital notation gives a more detailed depiction of the electrons in each shell. This is most advantageous for atoms that have special cases. </span>

<span>Some examples of atoms that have special electronic configurations are copper and chromium. For example, copper is more stable when the 3d subshell is completely filled. This leaves the 4s subshell with only 1 electron. Chromium is also more stable when its s and d subshells are only half full. The orbital notation depicts these special cases, whereas the dot structure does not.</span>
6 0
3 years ago
Read 2 more answers
How is the Scientific Method used in science?
xeze [42]

Answer:

to collect measurable, empirical evidence in an experiment related to a hypothesis the results aiming to support or contradict a theory.

Explanation:

7 0
3 years ago
(a) Calculate the wavelength of light in vacuum that has a frequency of 5.49 ✕ 1018 Hz. 0.0546 Correct: Your answer is correct.
vfiekz [6]

Answer:

a) Wavelength of the light in vacuum = (5.46 × 10⁻¹¹) m = 0.0546 nm

b) Wavelength of the light in diamond = (2.26 × 10⁻¹¹) m = 0.0226 nm

c) Energy of one photon in vacuum = (3.638 × 10⁻¹⁵) J = (2.271 × 10⁴) eV

d) No, the energy of the photon doesn't change when it is travelling inside diamond.

Explanation:

Wavelength (λ), frequency (f) and velocity of light (v) are related as thus

v = fλ

a) v = fλ

v = velocity of light in vacuum = (3.0 × 10⁸) m/s

f = frequency of the light = (5.49 × 10¹⁸) Hz

λ = wavelength of the light = ?

λ = (v/f) = (3.0 × 10⁸) ÷ (5.49 × 10¹⁸)

= (5.46 × 10⁻¹¹) m = 0.0546 nm

b) To find the wavelength of the light in diamond, we need the refractive index of diamond. This is because light, just like all other waves, change their velocities and subsequently their wavelengths in different materials according to the refractive index of the materials.

Refractive index of diamond = 2.42 (from literature)

2.42 = (wavelength of light in vacuum) ÷ (wavelength of light in diamond)

2.42 = 0.0546 ÷ λ

λ = 0.0546 ÷ 2.42 = 0.0226 nm

c) Energy of a photon in vacuum is given as

E = hf

where E = energy in Joules = ?

h = Planck's constant = (6.626 × 10⁻³⁴) J.s

f = frequency of the light in vacuum = (5.49 × 10¹⁸) Hz

E = (6.626 × 10⁻³⁴) × (5.49 × 10¹⁸) = (3.638 × 10⁻¹⁵) J

1 eV = (1.602 × 10⁻¹⁹) J

The amount of the calculated energy in eV

= (3.638 × 10⁻¹⁵) ÷ (1.602 × 10⁻¹⁹) = (2.271 × 10⁴) eV

d) As light travels from material to material, it's velocity and wavelength changes from material to material, but the frequency of the light waves stay the same. Since the energy of the photon depends solely on this frequency, it shows that the energy of the photon stays consistent in whichever material.

Hope this Helps!!!

3 0
3 years ago
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