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3 years ago

A term related to frequency that increases as frequency increases

Chemistry

2 answers:

HACTEHA [7]3 years ago

8 0

Answer: Energy.

Explanation: The energy of a wave is directly proportional to its frequency, but inversely proportional to its wavelength. In other words, the greater the energy, the larger the frequency and the shorter (smaller) the wavelength. Given the relationship between wavelength and frequency described above, it follows that short wavelengths are more energetic than long wavelengths.

zvonat [6]3 years ago

7 0

As wavelength increases so does frequency

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Three common gaseous compounds of nitrogen and oxygen of different elementary composition are known: (A) laughing gas containing

Ede4ka [16]

Answer:

Please find how these data prove the law of multiple proportions below

Explanation:

The law of multiple proportions was proposed by an English chemist called John Dalton. The law states that when two elements combine and to form more than one compound. The weights/masses of the second element in the two compounds, which combines with a fixed ratio of the first element, is in a simple whole number ratio.

In this question, Nitrogen is said to combine with oxygen to give three different compounds as follows:

A) laughing gas containing 63.65% nitrogen i.e. 0.6365g

This means that the mass of oxygen will be (1-0.6365) = 0.3635g

B) colorless gas containing 46.68% nitrogen i.e. 0.4668g

This means that the mass of oxygen will be 0.5332g

C) brown toxic gas containing 30.45% nitrogen i.e. 0.3045g

This means that the mass of oxygen will be 0.6955g

The ratios of oxygen in the three compounds is therefore:

0.3635: 0.5332: 0.6955

Divide this ratio by the smallest number (0.3635)

0.3635/0.3635 = 1

0.5332/0.3635 = 1.467

0.6955/0.3635 = 1.913

Multiply this ratio by 2, we have:

2: 2.9 : 3.8

Hence, the simple whole number ratio is 2:3:4.

This proves the law of multiple proportions that oxygen is in simple whole number ratio in the three different compounds.

4 0

3 years ago

What is the sequence of energy transformations that occur in a nuclear reactor?

fredd [130]

B. nuclear to thermal to mechanical to electrical

7 0

3 years ago

Read 2 more answers

Indicate which of the following statements is FALSE.a. Covalent bonds connect nucleotides in a strand; noncovalent interactions

LenaWriter [7]

Answer:

The false statement is d Avery,Macleod and McCarty showed that DNA is the genetic information of cells and RNA is the genetic information in the viruses .

Explanation:

Avery,Macleod and MacCarty showed that DNA is the genetic material of the cell.

On the other hand Fraenkel, Conrat and Sanger carried out their experiment on tobacco mosaic virus to prove that RNA act as genetic material in some viruses.

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3 years ago

The insulating and packing material Styrofoam is a dating and packing material Styrofoam is a polymer of styrene. Find the molec

Nataly [62]

Explanation:

In order to find the molecular formula, we have to find the empirical formula first of all.

It is known that is a styrofoam, elements present are carbon and hydrogen atoms.

Element % Atomic mass Molar ratio Simple ratio

C 92.25 12.01 $\frac{92.25}{12.01}$ = 7.68 $\frac{7.68}{7.68}$ = 1

H 7.75 1.008 $\frac{7.75}{1.008}$ = 7.68 $\frac{7.68}{7.68}$ = 1

As empirical formula of styrofoam is C_{x}H_{x}.

Hence, empirical mass = (12.01 + 1.008) g/mol = 13.018 g/mol = 13.0 g/mol (approx)

Molar mass given is 104 g/mol.

So, $\frac{\text{molar mass}}{\text{empirical formula mass}}$ = $\frac{104}{13.0}$ = 8

Thus, we can conclude that molecular formula of the given styrene is $C_{8}H_{8}$ .

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3 years ago

At what pressure does the mean free path of argon at 20 degrees celsius become comparable to the diameter of a 100 cm3vessel th

inn [45]

Answer:

The mean free path of argon molecules becomes comparable to the diameter of this container at a pressure of 0.195 Pa

Explanation:

<u>Step 1</u>: Calculate the volume of a spherical container V

V = (4π*r³)/3

r = (3V/4π)^1/3

2r = d = 2*(3V/4π)^1/3

with r= radius

with d= diameter

The diameter is:

d= 2*(3V/4π)^1/3

d= 2*(3*100cm³/4π)^1/3

d= 5.76 cm

<u>Step2 </u>: Define the free path lambda λ of argon

with λ =k*T/ σp

with p = kT/σλ

with T= temperature = 20°C = 293.15 Kelvin

with k = Boltzmann's constant = 1.381 * 10^-23 J/K

with p = the atmospheric pressure

with σ = 0.36 nm²

p = kT/σλ

p = (1.38 * 10^-23 J*K^-1 * 1Pa *m³/1J)*(293,15K) /(0.36 nm²*(10^-9/ 1nm)² *(5.76cm* 10^-2m/1cm)

p = 0.195 Pa

The mean free path of argon molecules becomes comparable to the diameter of this container at a pressure of 0.195 Pa

5 0

3 years ago