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Tanzania [10]
2 years ago
12

Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomi

c gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are diff erent. Determine the ratio KEKrypton KEArgon of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom.

Chemistry
1 answer:
azamat2 years ago
7 0

Answer:

2.1

Explanation:

The solution is in the attached file below

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Electron A falls from energy level X to energy level Y and releases blue light. Electron B falls from energy level Y to energy l
MrMuchimi

Answer: Transition from X to Y will have greater energy difference.

Explanation: For studying the energy difference, we require Planck's equation.

                                E=\frac{hc}{\lambda}

where, h = Planck's Constant

c = Speed of light

E = Energy

\lambda = Wavelength of particle

From the equation, it is visible that the energy and wavelength follow inverse relation which means that with low wavelength value, energy will be the highest and vice-versa.

As electron A falls from X-energy level to Y-energy level, it releases blue light which has low wavelength value (around 470 nm) which means that it has high energy.

Similarly, Electron B releases red light when it falls from Y-energy level to Z-energy level, which has high wavelength value (around 700 nm), giving it a low energy value.

Energy Difference between X-energy level and Y-energy level will be more.


5 0
3 years ago
How many grams are in 1.76 x 10^23 atoms of iodine
Mariana [72]

Answer:

\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

6.022*10^{23}

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Multiply the given number of atoms by Avogadro's number.

1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Flip the fraction so the atoms of iodine will cancel out.

1.76*10^{23} \ atoms \ I*\frac{  1 \ mol \ I}{6.022*10^{23} \ atoms \ I}

1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }

Multiply so the problem condenses into 1 fraction.

\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }

0.2922617071 \ mol \ I

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

  • Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply this fraction by the number of moles found above.

0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply. The moles of iodine will cancel.

0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }

The 1 as a denominator is insignificant.

0.2922617071 *{ 126.9045 \ g\ I }

37.08932581 \ g \ I

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

37.08932581 \ g \ I

The 8 in the hundredth place tells us to round the 0 up to a 1.

\approx 37.1\ g \ I

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

5 0
3 years ago
What is chemisorption?
Blababa [14]
When the substance being absorbed is held together by chemical bonds.
6 0
3 years ago
In a chemical reaction, energy can be A. released. B. absorbed. C. released or absorbed. D. neither released nor absorbed.
Inessa05 [86]
C. released or absorbed.

When the energy is released the reaction is called exotermic reaction
When the energy is absorbed, the reaction is called endothermic reaction

An example of exotermic reaction is between water and H2SO4
4 0
3 years ago
Read 2 more answers
Explain how both nucleic acids and proteins are both polymers?
jenyasd209 [6]
Because there are parts of them just repeats them selfs over and over, classifying them as polymers.
7 0
2 years ago
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