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Vikki [24]
3 years ago
11

SOMEONE HELP MEEEE

Chemistry
2 answers:
slavikrds [6]3 years ago
8 0
These are not questions but the directions or guide to completing your assignment.

I will provide you will with a choice of concept maps you can use:

- tree map

- circle maps

-comparison map ( might be best)

Please vote my answer branliest! Thanks.
Arlecino [84]3 years ago
6 0

The answer is correct

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Which of the following frequencies corresponds to light with the longest wavelength? A) 3.00 times 10^13 s^-1 B) 4.12 times 10^5
Vsevolod [243]

Answer:

1) B. 4.12 times 10^5 s^-1

2) B. frequency-v

3) C. 1.18 times 10^15 s^-1

4) B. 4.39 times 10^-19 J

5) B. Energy is absorbed

6) C. 3

7) 1s^2  2s^2  2p^6  3s^2  3p^6  4s^2  3d^10  4p^6  5s^2  4d^10  5p^6     6s^2  

Explanation:

1)

The wavelength is inversely proportional to the frequency. Thus, the smallest frequency shall correspond to the longest wavelength. Thus, the correct answer is <u>B. 4.12 times 10^5 s^-1</u>

2)

<u>B. frequency-v</u> are wrongly paired because frequency is represented by f. Thus, correct pair will be frequency-f.

3)

The relationship between wavelength (λ) and frequency (f) is:

c = fλ

f = c/λ

where, c = speed of light = 3 x 10^8 m/s

f = (3x10^8 m/s) / (254 x 10^-9 m)

f = 1.18 x 10^15 s^-1

Thus, the correct option is <u>C. 1.18 times 10^15 s^-1.</u>

4)

The energy of photon is given as:

E = hc/λ

where, c = speed of light = 3 x 10^8 m/s

            h = Plank's Constant = 6.625 x 10^-34 J.s

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s) / (453 x 10^-9 m)

E = 4.39 x 10^-19 J

Thus, the correct option is <u>B. 4.39 times 10^-19 J</u>.

5)

Since, the higher energy levels away from nucleus have higher energies. So, in order to move an electron from lower to higher or distant energy level, it must absorb energy from external source. So, the correct option is <u>B. Energy is absorbed</u>.

6)

The P-Sub-level has three orbitals, each having two electrons. Thus, P-Sub-Level accommodates total of 6 electrons. The correct option is <u>C. 3</u>

7)

The electronic configuration of barium atom is:

<u>1s^2  2s^2  2p^6  3s^2  3p^6  4s^2  3d^10  4p^6  5s^2  4d^10  5p^6  6s^2  </u>

8 0
3 years ago
Which is not a reason why mass of an atom is expressed in atomic mass units rather than in grams?
Tasya [4]
What are your answer choices? Scientists express the weight in AMU’s so that they can easily think about the weight of the atom. If it is expressed in grams, they’re constantly dealing with a weight that’s around 10^-24 grams, which is hard to deal with.
8 0
3 years ago
What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha
sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

<u>Given:</u>

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

<u>To determine:</u>

The mean free path, λ

<u>Calculation:</u>

The mean free path is related to the collision cross section by the following equation:

\lambda =\frac{1}{n\sigma }------(1)

where n = number density

n = \frac{P}{kT}-----(2)

Substituting for P, k and T in equation (2) gives:

n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\  m^{3}

Next, substituting for n and σ in equation (1) gives:

\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m

6 0
3 years ago
You are a researcher with an interest in understanding the mechanism of serine protease enzymes. To test the effect of the role
yawa3891 [41]
Ngl didn’t read it so I’m not gonna answer it. Don’t report please just want to ask a question but have to answer two
5 0
3 years ago
Which numerical setup can be used to calculate the<br> atomic mass of the element bromine?
Gekata [30.6K]

Answer:

From the numerical steps highlighted under explanation, the average atomic mass of bromine is 79.91 u

Explanation:

The steps to be taken will involve;

1) Find the number of isotopes of bromine.

2) Identify the atomic mass and relative abundance of each of the isotopes.

3) Multiply the atomic mass of each of the isotopes by their corresponding values relative abundance value.

4) Add the value in step 3 above to get the average atomic mass of bromine.

Now;

Bromine has 2 isotopes namely;

Isotope 1: Atomic mass = 78.92amu and a relative abundance of 50.69%.

Isotope 2: Atomic mass = 80.92amu and a relative abundance of 49.31%.

Using step 3 above, we have;

(78.92 × 50.69%)

And (80.92 × 49.31%)

Using step 4 above, we have;

(78.92 × 50.69%) + (80.92 × 49.31%) ≈ 79.91 u

6 0
2 years ago
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