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3 years ago

Neutrons have neither charge nor mass. TRUE FALSE

Chemistry

1 answer:

iren2701 [21]3 years ago

7 0

Answer:

FALSE

nuetrons do indeed have no charge, however they are nearly 2000 more times massive than electrons.

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What constitutes only 16% of earth's volume but 32% of earth's mass

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The earth's core makes up 16% of the earths volume, but 32% of the earths mass.

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4 years ago

Write the formula for each compound (the composition is given after each name):

erastovalidia [21]

Aluminum bromide should be written as AlBr₃ since we are told that there are 3 atoms of Br for every atom of Al.

I hope this helps. Let me know if anything is unclear.

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4 years ago

When metallic sodium is dissolved in liquid sodium chloride, electrons are released into the liquid. These dissolved electrons a

qaws [65]

Answer:

The edge of the length is $\mathbf{L = 8.54 \times 10^{-10} \ m}$

Explanation:

From the given information:

The associated energy for a particle in three - dimensional box can be expressed as:

$E_n = \dfrac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2)$

here;

h = planck's constant = $6.626 \times 10^{-34} \ Js$

$n_i$ = the quantum no in a specified direction

m = mass (of particle)

L = length of the box

At the ground state $n_x = n_y = n_z=1$

The energy at the ground state can be calculated by using the formula:

$E_1 =\dfrac{3h^2}{8mL^2}$

At first excited energy level, one of the quantum values will be 2 and the others will be 1.

Thus, the first excited energy will be: 2,1,1

∴

$E_2 =\dfrac{(2^2+1^2+1^2)h^2}{8mL^2}$

$E_2 =\dfrac{(4+1+1)h^2}{8mL^2}$

$E_2 =\dfrac{(6)h^2}{8mL^2}$

The transition energy needed to move from the ground to the excited state is:

$\Delta E= E_2 - E_1$

$\Delta E= \dfrac{6h^2}{8mL^2}- \dfrac{3h^2}{8mL^2}$

$\Delta E= \dfrac{3h^2}{8mL^2}}$ ----- (1)

Recall that:

the wavelength identified with the electronic transition is: 800 nm

800 nm = 8.0 × 10⁻⁷ m

However, the energy-related with the electronic transition is:

$\Delta E =\dfrac{hc}{\lambda}$

$\Delta E =\dfrac{6.626 \times 10^{-34} \times 2.99 \times 10^8}{8.0 \times 10^{-7} }$

$\Delta E =2.48 \times 10^{-19} \ J$

Replacing the value of $\Delta E$ in (1); then:

$2.48 \times 10^{-19}= \dfrac{3h^2}{8mL^2}}$

Making the edge length L the subject of the formula; we have:

$L = \sqrt{\dfrac{3h^2}{8m \times2.48 \times 10^{-19}} }$

$L = \sqrt{\dfrac{3\times (6.626 \times 10^{-34})^2}{8(9.1 \times 10^{-31} ) \times2.48 \times 10^{-19}} }$

$\mathbf{L = 8.54 \times 10^{-10} \ m}$

Thus, the edge of the length is $\mathbf{L = 8.54 \times 10^{-10} \ m}$

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3 years ago

A scientist is studying a shock wave from an earthquake. What kind of wave is being studying?

34kurt

The answer is an Electromagnetic Wave, because the Idiosyncratic Panacea in the earthquake, has a fluxuating Acrimony that allows the wave to Ostracize, which in turn creates an Electromagnetic charge that charges the wave.
The answer is: A<span>n electromagnetic wave
</span>Hope this Helps!

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3 years ago

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On a given day, the water vapor in the air is 2.5%. If the partial pressure of the vapor is 19.4 mm Hg, what is the atmospheric

erica [24]

Answer:775 mm Hg

Explanation:

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3 years ago