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Lapatulllka [165]
3 years ago
8

Given that RH= 2.18 x 10⁻¹⁸J, 1 nm = 1 x 10⁻⁹m, h = 6.63 x 10⁻³⁴J·s, and c = 3.00 x 10⁸m/s:

Chemistry
1 answer:
Bas_tet [7]3 years ago
3 0

Answer:

The wavelength the light emitted by a hydrogen atom during a transition is 1006 nm.

Explanation:

By using Rydberg's Equation we cab determine the wavelength of the light:

\Delta E=R_H\times Z^2\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\Delta E = Energy difference

R_H = Rydberg's Constant

n_f = Final energy level

n_i= Initial energy level

We have : n_i=7,n_f=3 , Z = 1

R_H=2.18\times 10^{-18} J

\Delta E=2.18\times 10^{-18} J\times 1^2\left(\frac{1}{7^2}-\frac{1}{3^2} \right )

\Delta E=1.9773\times 10^{-19} J

Now by using Plank's equation we can determine the wavelength of the light emitted.

E=\frac{hc}{\lambda }

E = Energy of the emitted light

h = Planck's constant = 6.63\times 10^{-34} Js

c = speed of light = 3.00\times 0^8 m/s

For the given transition the energy of the light = E

E =1.9773\times 10^{-19} J

\lambda=\frac{hc}{E}=\frac{6.63\times 10^{-34} Js\times 3.00\times 0^8 m/s}{1.9773\times 10^{-19} J}

\lambda =1.006\times 10^{-6} m =1.006\times 10^{-6}\times 10^9=1006 nm

The wavelength the light emitted by a hydrogen atom during a transition is 1006 nm.

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Gold cylinder has a mass of 75 g and a specific heat of 0.129J/G degrees Celsius it is heated to 65°C and then put in 500 g of w
nadezda [96]
<h3>Answer:</h3>

89.88° C

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of gold cylinder as 75 g
  • specific heat of gold is 0.129 J/g°C
  • Initial temperature of gold cylinder is 65°C
  • Mass of water is 500 g
  • Initial temperature of water is 90 °C

We are required to calculate the final temperature;

  • We know that Quantity of heat is given by the product of mass, specific heat capacity and change in temperature.
  • That is, Q = m × c × ΔT
<h3>Step 1: Calculate the quantity of heat absorbed by the Gold cylinder</h3>

Assuming the final temperature is X° C

Then; ΔT = (X-65)°C

Therefore;

Q = 75 g × 0.129 J/g°C × (X-65)°C

   = 9.675X - 628.875 Joules

<h3>Step 2: Calculate the quantity of heat released by water</h3>

Taking the final temperature as X° C

Change in temperature, ΔT = (90 - X)° C

Specific heat capacity of water is 4.184 J/g°C

Therefore;

Q = 500 g × 4.184 J/g°C × (90 - X)° C

  = 188,280 -2092X joules

<h3>Step 3: Calculate the final temperature, X°C</h3>

we know that the heat gained by gold cylinder is equal to the heat released by water.

9.675X - 628.875 Joules = 188,280 -2092X joules

2101.675 X = 188908.875

              X = 89.88° C

Thus, the final temperature is 89.88° C

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