Which is/are true about the atomic particles?[mark all correct answers]

Ultraviolet radiation and radiation of shorter wavelengths can damage biological molecules because they carry enough energy to b

DIA [1.3K]

<h3>Answer:</h3>

Longest wavelength = 343.7 nm

<h3>Solution and Explanation:</h3>

In this question we need to first use the concept of energy of a photon.

Energy of a photon, E, is given by the formula, E = hf, where h is the plank's constant, f is the frequency.

But since, f is given by dividing speed, c, by wavelength, λ, then;

E = hc/λ

We are given 348 kJ/mol required to break carbon-carbon bonds.

We know that; 1 mole of bonds = 6.022 × 10^23 bonds.

We are required to find the longest wavelength with enough energy to break the C-C bonds.

This can be worked out in simple steps:

Step 1: Energy required to break one bond (kJ/bond)

1 mole of bonds = 6.022 × 10^23 bonds.

Therefore;

348 kJ = 6.022 × 10^23 bonds.

Thus;

1 bond = 348 kJ ÷ 6.022 × 10^23 bonds.

= 5.778 x 10^-22 kJ

But; 1000 joules = 1 kJ

Hence; energy per bond = 5.778 x 10^-19 Joules

Step 2: Energy per photon

Breaking one bond requires energy equivalent to energy of a photon.

Therefore;

1 photon = 5.778 x 10^-19 Joules

= 5.778 x 10^-19 J/photon

Step 3: Calculating the wavelength

From the equation of energy of a photon;

E = hc/λ

h is the plank's constant = 6.626 × 10^-34 J/s

c is the speed of light in vacuum = 2.9998 × 10^8 m/s

E is the energy of a photon = 5.778 x 10^-19 Joules

Therefore, making λ (wavelength) the subject;

$wavelength = \frac{hc}{E}$

$= \frac{(6.626 . 10^{-34})(92.9998.10^8) }{(5.778 .10^{-19} )}$

$= 3.437. 10^{-7} m$

= 3.437 x 10^-7 m

But; 1 nm = 10^-9 m

Thus;

wavelength = 343.7 nm

Therefore, the longest wavelength of the radiation will be 343.7 nm

5 0

3 years ago

Help Please! Will give brainlyest for correct answers! 99 points!!

lapo4ka [179]

Answer:

1- 1.54 mol.

2- 271.9 kPa.

3- Yes, the tires will burst.

4- 235.67 kPa.

5- As, the temperature increased, the no. of molecules that has minimum kinetic energy increases as shown in image 1 that represents the Maxwell’s Distribution of Speeds of molecules. "Kindly, see the explanation and the attached images".

Explanation:

Q1- How many moles of nitrogen gas are in each tire?

To calculate the no. of moles of nitrogen gas in each tire, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the nitrogen gas (P = 247.0 kPa/101.325 = 2.44 atm),

V is the volume of the nitrogen gas (V = 15.2 L),

n is the no. of moles of the nitrogen gas (n = ??? mole),

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the nitrogen gas (T = 21°C + 273 = 294 K).

∴ n = PV/RT = (2.44 atm)(15.2 L)/(0.082 L/atm/mol.K)(294.0 K) = 1.54 mol.

Q2: What would the maximum tire pressure be at 50 degrees C?

Now, the temperature is raised to be 50°C (T = 50°C + 273 = 323 K).

The pressure can be calculated using the general gas law: PV = nRT.

∴ P = nRT/V = (1.54 atm)(0.082 L/atm/mol.K)(323.0 K)/(15.2 L) = 2.68 atm = 271.9 kPa.

Q3: Will the tires burst in Spokane? Explain.

Yes, the tires will burst because the internal pressure be 271.9 kPa that exceeds 270 kPa, the pressure above which the tires will burst.

Q4: If you must let nitrogen gas out of the tire before you go, to what pressure must you reduce the tires before you start your trip? (Assume no significant change in tire volume.)

To get the pressure that we must begin with:

Firstly, we should calculate the no. of moles at:

T = 55°C + 273 = 328 K,

Pressure = 270 kPa (the pressure above which the tires will burst). (P =270 kPa/101.325 = 2.66 atm).

V = 15.2 L, as there is no significant change in tire volume.

∴ n = PV/RT = (2.66 atm)(15.2 L)/(0.082 L.atm/mol.K)(328 K) = 1.5 mol.

1.5562 moles of N₂ in the tires will give a pressure of 270 kPa at 55°C, so this is the minimum moles of N₂ that will make the tires burst.

Now, we can enter this number of moles into the original starting conditions to tell us what pressure the tires will be at if we start with this number of moles of N₂.

P = ???

V = 15.6 L.

n = 1.5 mol

T = 21°C + 273 = 294.0 K

R = 0.0821 L.atm/mol.K.

∴ P = nRT/V = (1.5 mol x 0.082 x 294.0 K) / (15.6 L) = 2.2325 atm = 235.67 kPa.

So, the starting pressure needs to be 235.67 kPa or just under in order for the tires not to burst.

Q5: Create a drawing of the tire and show a molecular view of the air molecules in the tire at 247 kpa vs the molecular view of the air molecules after the tires have been heated. Be mindful of the number of molecules that you use in your drawing in the before and after scenarios. Use a caption to describe the average kinetic energy of the molecules in both scenarios.

As, the temperature increased, the no. of molecules that has minimum kinetic energy increases as shown in “image 1” that represents the Maxwell’s Distribution of Speeds of molecules.

The no. of molecules that possess a critical K.E. of molecules increases due to increasing the temperature activate the motion of molecules with high velocity as

(K.E. = 3RT/2), K.E. directly proportional to the temperature of the molecules (see image 2).

Also, the average speed of molecules increases as the K.E of the molecules increases (see image 3).

5 0

3 years ago

A student increases the temperature of a 556 cm3 balloon from 278 K to 308 K. Assuming constant pressure, what should the new vo

Mandarinka [93]

The answer is: [D]: " 417 cm³ " .
_____________________________________________________
Explanation: Use the formula:

V₁ /T₁= V₂ /T₂ ;

in which: V₁ = initial volume = 556 cm³ ;
                T₁ = initial temperature = 278 K ;
                V₂ = final ("new") temperature = 308 K
                T₂ = final ("new:) volume = ?

Solve for "V₂" ;

Since: V₁ /T₁= V₂ /T₂ ;

We can rearrange this "equation/formula" to isolate "V₂" on one side of the equation; and then we can plug in our know values to solve for "V₂" ;
_______________________________________________________
       V₁ /T₁= V₂ /T₂ ; Multiply EACH side of the equation by "T₂ " :

          → T₂ (V₁ /T₁) = T₂  (V₂ /T₂) ;
______________________________
to get:

↔ T₂  (V₂ /T₂) = T₂ (V₁ /T₁) ;

     → V₂ = T₂ (V₁ /T₁) ;
______________________________
Now, plug in our known values, to solve for "V₂" ;
______________________________
    → V₂ = T₂ (V₁ /T₁) ;
______________________________
    → V₂ = 308 K ( 556 cm³ /278 K) ;
             → The units of "K" cancel to "1" ; and we have:
________________________________________________________
    → V₂ = 308*( 556 cm³ / 278 ) = [(208 * 556) / 278 ] cm³ ;
Note: We will keep the units of volume as: "cm³ "; since all the answer choices given are in units of: "cm³ " ; {that is, "cubic centimeters"}.

   → [(208 * 556) / 278 ] cm³ = [ (115,648) / (278) ] cm³ ;

              → For the "(115,648)" ; round to "3 (three significant figures)" ;
                        → "(115,648)" → rounds to: "116,000" ;
____________________________________________________
              →      (116,000) / (278) = 417.2661870503597122 ;
                                 → round to 3 significant figures; → "417 cm³ " ;
                                               → which corresponds with "choice [D]".
______________________________________________________
The answer is: [D]: "417 cm³ " .
______________________________________________________

3 0

3 years ago

The outermost electrons are the_____electrons

Lynna [10]

Answer:

Valence electrons. they can be used to determine the group of an element.

Explanation:

6 0

3 years ago

If a gas is cooled from 325K to 275K and the volume is constant, what would the final pressure be if it was originally 750mmHg?

salantis [7]

Answer:

635

p/T is a constant

(750/325)×275

3 0

4 years ago