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3 years ago

feel free to answer any of these. i need the answer to the appropriate amount of significant figures on this pleasee

Chemistry

1 answer:

jekas [21]3 years ago

4 0

Answer:

6. (2.7* $10^{5}$ )

8. (1.0* $10^{4}$ )

Explanation:

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If the mass of a radioactive substance is 8 grams and it has a half-life of 4 hours, how much mass remains after 8 hours?

Alborosie

The sample will lose half of its mass after 4 hours. The half life.
82=4
The sample will lose half of the remaining four after another half life.
42=2 Hope this helps! :)

3 0

3 years ago

Read 2 more answers

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wave

timofeeve [1]

Explanation:

It is given that,

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,

$n_i=8$

$\lambda=3745\ nm$

The amount of energy change during the transition is given by :

$\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]$

And

$\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]$

Plugging all the values we get :

$\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5$

So, the final level of the electron is 5.

4 0

2 years ago

PLEASE HELP. WILL MARK BRAINLIEST SERIOUSLY CONFUSED HELLPPPP PLEASE

Yuri [45]

Alright! Here are the answers:

1. C. Fluorine is more reactive than nitrogen because fluorine needs only one electron to fill its outermost shell.

2. Aluminum (Al)

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2 years ago

Which of the following was not a geological clue used to support the theory of continental drift

Andrei [34K]

The answer is:

D. Samples of ocean water.

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2 years ago

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O 1. 12 moles

mars1129 [50]

Answer:

$n_{H_2O}=5.94molH_2O$

Explanation:

Hello there!

In this case, according to the given question about stoichiometry, it is possible for us to calculate the required moles of water that will be produced by 12 grams of hydrogen, by using the molar mass of this reactant (2.02 g/mol as it is diatomic) and the 2:2 mole ratio in the chemical equation by solving the following setup:

$n_{H_2O}=12gH_2*\frac{1molH_2}{2.02gH_2} *\frac{2molH_2O}{2molH_2} \\\\n_{H_2O}=5.94molH_2O$

Regards!

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2 years ago