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Diano4ka-milaya [45]
4 years ago
13

How does the distance between the planet Earth and the Sun compare with the distances between the planet Earth and other stars?

A. The Sun and all the other stars are the same distance from Earth. B. About half of the other stars are closer to Earth than the Sun. C. The Sun is many thousands of times closer to Earth than any other star. D. The other stars are many thousands of times closer to Earth than the Sun.
Chemistry
2 answers:
Paha777 [63]4 years ago
7 0
I think C, but I could be wrong
Virty [35]4 years ago
4 0

the correct answer is C


The average distance from the Earth to the Sun is approximately 150 million kilometers. Proxima Centauri is the next closest star to Earth, and it is located approximately 40 trillion kilometers away.

It takes only a few minutes for light from the Sun to reach the Earth, but it takes a few years for light from Proxima Centauri to reach the Earth. Also, it would take thousands of years for the fastest rocket to travel from the Earth to Proxima Centauri.  (study Island)


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How many of protien consumed by humans around the world is fish?
Alex17521 [72]

The answer would be 20% (B)

3 0
4 years ago
An unknown gas diffuses 0.25 times as fast as helium. What is it’s molar mass? What steps are to be done?
dmitriy555 [2]

Answer:

64.0 g/mol.

Explanation:

  • Thomas Graham found that, at a constant  temperature and pressure the rates of effusion  of various gases are inversely proportional to  the square root of their masses.

<em>∨ ∝ 1/√M.</em>

where, ∨ is the rate of diffusion of the gas.

M is the molar mass of the gas.

<em>∨₁/∨₂ = √(M₂/M₁)</em>

∨₁ is the rate of effusion of the unknown gas.

∨₂ is the rate of effusion of He gas.

M₁ is the molar mass of the unknown gas.

M₂ is the molar mass of He gas (M₂ = 4.0 g/mol).

<em>∨₁/∨₂ = 0.25.</em>

∵ ∨₁/∨₂ = √(M₂/M₁)

∴ (0.25) =√(4.0 g/mol)/(M₁)

<u><em>By squaring the both sides:</em></u>

∴ (0.25)² = (4.0 g/mol)/(M₁)

∴ M₁ = (4.0 g/mol)/(0.25)² = 64.0 g/mol.

4 0
3 years ago
Phosphorus pentachloride decomposes according to the chemical equation PCl 5 ( g ) − ⇀ ↽ − PCl 3 ( g ) + Cl 2 ( g ) K c = 1.80 a
mezya [45]

Answer:

See explanation for answer

Explanation:

First, let's write the equation again:

PCl5 <---------> PCl3 + Cl2      Kc = 1.8

The expression for equilibrium is the following:

Kc = [PCl3] [Cl2] / [PCl5]

We only know the data for PCl5 at the beggining, and to solve this, we need to write what happens before and after the reaction.

At first, we only have the 0.342 moles of PCl5 and nothing of the products. This is logic, because there is no reaction yet.

Concentration of PCl5 at this point (With a volume of 3.65 L) is:

[PCl5] = 0.342 / 3.65 = 0.094 M

Keep in mind, that this is not the concentration in equlibrium. This is only the concentration at the beggining. To know it's concentration in equilibrium, we'll do the following:

       PCl5 <---------> PCl3 + Cl2      Kc = 1.8

i)      0.094                  0        0

eq)   0.094 - x             x        x

Now, we will replace this values in the equilibrium equation:

1.8 = x * x / 0.094 - x

Solving for x we have:

1.8(0.094 - x) = x²

0.1692 - 1.8x = x²

x² + 1.8x - 0.1692 = 0

Using the general equation for x we have:

x = -1.8 ± √(1.8)² - 4 * 1 * (-0.1692) / 2

x = -1.8 ± √3.9168 / 2

x = -1.8 ± 1.98 / 2

x1 = -1.8 + 1.98 / 2 = 0.09

x2 = -1.8 - 1.98 / 2 = -1.89

This means that the value of x is 0.09 therefore, the concentrations in equilibrium for each species is:

[PCl5] = 0.094 - 0.09 = 0.04 M

[PCl3] = [Cl2] = 0.09 M

7 0
4 years ago
A photon with 2.3 eV of energy can eject an electron from potassium. What is the corresponding wavelength of this type of light?
Ira Lisetskai [31]

Answer:

\lambda=540.16\ nm

Explanation:

Given that:

The energy of the photon = 2.3 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.60 × 10⁻¹⁹ J

So, Energy = 2.3\times 1.60\times 10^{-19}\ J=3.68\times 10^{-19}\ J

Considering

Energy=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength of the light being bombarded

Thus,  

3.68\times 10^{-19}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\frac{3.68}{10^{19}}=\frac{19.878}{10^{26}\lambda}

3.68\times \:10^{26}\lambda=1.9878\times 10^{20}

\lambda=5.40163\times 10^{-7}\ m=540.16\times 10^{-9}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=540.16\ nm

3 0
4 years ago
You have been asked to organize the following aqueous solutions: HCl, NaOH, H2SO4,Ca(OH)2, LiOH, and HCIO3.How would you group t
frutty [35]

Answer & Explanation:

You could group them as acids and bases:

<u>Acids</u>

HCl

H₂SO₄

HClO₃

<u>Bases</u>

NaOH

Ca(OH)₂

LiOH

7 0
3 years ago
Read 2 more answers
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