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zloy xaker [14]
3 years ago
9

5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wav

elengths of light that can be used to cause this phenomenon. Show your calculations with units of measure (dimensional analysis) and briefly explain your reasoning.
Chemistry
1 answer:
torisob [31]3 years ago
7 0

Answer: The range of wavelengths of light that can be used to cause given phenomenon is 8.953 \times 10^{21} m.

Explanation:

Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J

Formula used is as follows.

E = \frac{hc}{\lambda}

where,

E = energy

h = Planck's constant = 6.625 \times 10^{-25} Js

c = speed of light = 3 \times 10^{8} m/s

Substitute the values into above formula as follows.

E = \frac{hc}{\lambda}\\222000 J = \frac{6.625  \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m

Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is 8.953 \times 10^{21} m.

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grandymaker [24]

Answer:

1. 136 °C.

2. 0.21 atm.

Explanation:

1. Determination of the new temperature in °C.

Initial volume (V1) = 1.35L

Final volume (V2) = 1.95L

Initial temperature (T1) = 283 K

Final temperature (T2) =...?

Using the Charles' law equation, the new temperature of the gas can be obtained as follow:

V1 /T1 = V2 /T2

1.35/283 = 1.95/T2

Cross multiply

1.35 × T2 = 283 × 1.95

1.35 × T2 = 551.85

Divide both side by 1.35

T2 = 551.85/1.35

T2 = 408.8 ≈ 409 K

Finally, we shall convert 409 K to °C. This can be obtained as follow:

T (°C) = T(K) – 273

T(K) = 409 K

T (°C) = 409 – 273

T (°C) = 136 °C

Therefore, the new temperature of the gas is 136 °C.

2. Determination of the new pressure.

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1.67 L

Final pressure (P2) =.?

Next, we shall convert 1.67 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

1.67 L = 1.67 L × 1000 mL / 1 L

1.67 L = 1670 mL

Therefore, 1.67 L is equivalent to 1670 mL.

Finally, we shall determine the new pressure of the gas as follow:

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1670 mL

Final pressure (P2) =.?

P1V1 = P2V2

1.34 × 267 = P2 × 1670

357.78 = P2 × 1670

Divide both side by 1670.

P2 = 357.78 / 1670

P2 = 0.21 atm.

Therefore, the new pressure of the gas is 0.21 atm.

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2 years ago
How does coffee keep you awake?
ivanzaharov [21]

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2 years ago
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True or false: Nonmetals are good conductors of electricity?
Ostrovityanka [42]

Answer:

False

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2 years ago
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Cobalt-63 has a half-life of 5.3 years. If a pellet that has been in storage for 15.9 years contains 40.0g of Cobalt-63, how muc
Cerrena [4.2K]

Answer:

320 g  

Step-by-step explanation:

The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

  No. of               Fraction         Mass

half-lives   t/yr   Remaining   Remaining/g

      0        0              1

      1         5.3           ½

     2        10.6           ¼

     3        15.9           ⅛                 40.0

     4        21.2           ¹/₁₆

We see that 40.0 g remain after three half-lives.

This is one-eighth of the original mass.

The mass of the original sample was 8 × 40 g = 320 g

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