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AleksandrR [38]
3 years ago
7

2AlCl3 --> 2Al + 3Cl2

Chemistry
1 answer:
neonofarm [45]3 years ago
4 0
0.1125 moles

Total mass number of AlCl3 = 27 + (3x35.5) = 133.5

Number of moles equals mass divided by mass number

10:133.5=0.075 ( this is the number of moles of AlCl3)

To get the number of moles of Cl2, look at the coefficient of both AlCl3 and Cl2

0.075x3:2=0.1125 moles

I hope this helps
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Calculate the volume that 140 g of H2 gas will occupy at 305 K and 8.2 atm.
Alona [7]

Answejust put my mum said i didnt have to do this question

Explanation:

becuase im cool like that

6 0
3 years ago
How do the properties of an atom compare(similar or different) to the properties of :
RSB [31]
An isotope of an atom is a neighbour atom but which has more or less neutrons
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hope this helps!
6 0
4 years ago
Compare the de Broglie wavelength of an alpha particle moving at 3.40×107 miles per hour (1.52×107 m/s) to that of a baseball mo
denis-greek [22]

Answer:

1.74 x 10⁻²⁰(baseball) : 1.90 x 10⁴ (electron) : 1 (alpha particle)

baseball not detected

electron x ray

alpha particle beyond gama rays

Explanation:

The strategy for us here is to utilize deBroglie relation of wavelength and momentum:

λ = h/ mv

where  λ is wavelength,  h isPlanck´s constant, m is the mass of the particle, and v its velocity.

For determining the regions of spectrum the wavelengths fall into, we need to consult charts that describe the wavelengths vs region.

So  using the above  equation we will compute the wavelengths for the alpha particle, baseball,  and an electron.

Notice we are missing the masses of the alpha particle, and that of the electron, and baseball

The mass of the alpha particle is 4 times  amu  (4 x 1.66 x 10⁻²⁷ Kg ) since the alpha particle is esentially a helium nucleus which has atomic number 4.

The mass of the electron is 9.11 x 10⁻³¹ kg.

Mass of baseball per rules around 0.145 kg.

Notice we are working in the metric system, so use the velocities in m/s.

Now that we have all the data required, lets proceed to calculate the respective wavelengths.

λ ( alpha particle ) = 6.626 x 10⁻³⁴ J·s / ( 4 x 1.66 x 10⁻²⁷ kg x 1.52 x 10⁷ m/s )

= 6.57 x 10⁻¹⁵ m

λ ( baseball ) = 6.626 x 10⁻³⁴ J·s / ( 0.145 kg  x 40.2 m/s ) = 1.14 x 10⁻³⁴ m

λ ( electron ) = 6.626 x 10⁻³⁴ J·s / ( 9.11 x 10⁻³¹ kg x 5.81 x 10⁶ m/s )

= 1.25 x 10⁻¹⁰ m

Comparing the wavelengths from largest to smallest we have:

λ ( baseball) : λ ( electron )   : λ ( alpha particle )

1.14 x 10⁻³⁴  : 1.25 x 10⁻¹⁰ :  6.57 x 10⁻¹⁵

1.74 x 10⁻²⁰(baseball) : 1.90 x 10⁴ (electron) : 1 (alpha particle)

We can see the wavelength  of the baseball is very, very small compared to that of an electron and an alpha particle. For this small wavelength  we are not going to see effects such as diffraction or interference.This is the reason that for everyday macroscopic objects we do not talk about its associated wavelength. The wavelength can only be appreciated in  very small microscopic particles as exemplified for the cases of the electron and alpha particle in this question.

The wavelength of the baseball cannot be detected so a placement in the electromagnetic spectrum is undefined.

The wavelength of the electron in this question will fall into the x ray region of the spectrum (   region  around 10⁻⁹ to 10⁻¹² ) and the alpha particle will fall beyond the gamma rays ( that is wavelength shorter than 10⁻¹² m ).

3 0
3 years ago
Does the size of the sample of uranium-238 affect its half-life
AleksandrR [38]
Answer is: <span>the size of the sample of uranium-238 does not affect its half-life.
</span><span>
The half-life for the radioactive decay of U-238 is 4.5 billion years and is independent of initial concentration (size of the sample).
</span>Half-life <span>is the time required for a quantity (in this example number of radioactive uranium) to reduce to half its initial value.</span><span>

</span>
3 0
4 years ago
HELLLPPPP!!!!
AleksAgata [21]
A. Because you have to simplify
6 0
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