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3 years ago

Why do scientists need to study seismic waves to analyze the interior of the earth and the layers of the earth?

1 answer:

3 0

Geologists use these records to establish the structure of Earth's interior. The two principal types of seismic waves are P-waves (pressure; goes through liquid and solid) and S-waves (shear or secondary; goes only through solid - not through liquid).

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The scientific method helps scientists avoid injecting which of the following into their experiments? a. information b. conclusi

yanalaym [24]

The answer is D. Bias

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3 years ago

Read 2 more answers

At equilibrium, ________. At equilibrium, ________. all chemical reactions have ceased the rate constants of the forward and rev

balandron [24]

Answer:

<em>At equilibrium, the rate of the forward, and the reverse reactions are equal.</em>

Explanation:

In an equilibrium chemical reaction, the rate of forward reaction, is equal to the rate of reverse reaction. Note that the reactions does not cease at equilibrium, but rather, the reactants are converted to product, at the same rate at which the product is also being converted into the reactants in the reaction. When chemical equilibrium is reached, a careful calculation of the value of equilibrium constant is approximately equal to 1.

NB: If the value of equilibrium constant is far far greater than 1, then the reaction will favors more of the forward reaction, and if far far less than 1, the reaction will favor more of the reverse reaction.

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3 years ago

Please answer these for me

Dennis_Churaev [7]

What do you want an answer to?

8 0

4 years ago

When an excited electron in a hydrogen atom falls from ????=5 to ????=2, a photon of blue light is emitted. If an excited electr

poizon [28]

Answer:

n = 3

Explanation:

Given the formula for the transition energy of an atom with 1 electron:

$E=-13.6*Z^{2}*(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}} ) eV$

For the H transition n=5 to n=2:

$E=-13.6*(\frac{1}{4}-\frac{1}{25} ) eV=-2.856 eV$

Then we solve for nf with Z=2 (Helium)

$n_{f}=\sqrt{\frac{n_{i}^{2}*Z^{2}*13.6 eV }{2.856eV*n_{i}^{2}+13.6eV*Z^{2}} }$

$n_{f}=\sqrt{\frac{4^{2}*2^{2}*13.6 eV }{2.856eV*4^{2}+13.6eV*2^{2}} }=3$

Is near 3, actually the energy of the transitions are:

H (5⇒2) = -2.85 eV = 434 nm (Dark blue)

He (4⇒3) = -2.64 eV = 469 nm (Light blue)

I thought it was cool to see the actual colors. Included them.

4 0

3 years ago

A 52 gram sample of an unknown metal requires 714 Joules of energy to heat it from

ratelena [41]

Answer: Approximately $0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}$

===================================================

Work Shown:

We have the following variables

Q = 714 joules = heat required
m = 52 grams = mass
c = specific heat = unknown
$\Delta t$ = 82-30.5 = 51.5 = change in temperature

note: the symbol $\Delta$ is the uppercase Greek letter delta. It represents the difference or change in a value.

Apply those values into the formula below. Solve for c.

$Q = m*c*\Delta t\\\\714 = 52*c*51.5\\\\714 = 52*51.5*c\\\\714 = 2678*c\\\\2678*c = 714\\\\c = \frac{714}{2678}\\\\c \approx 0.26661687826737\\\\c \approx 0.267\\\\$

The specific heat of the unknown metal is roughly $0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}$

3 0

3 years ago