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max2010maxim [7]
3 years ago
5

Using the Rydberg equation, calculate the energy for the following electronic transitions in a hydrogen atom and label each as a

n absorption or an emission. When calculating, watch your signs!
n = 3 → n = 1

a. Energy calculation
b. absorption or emission?
Chemistry
1 answer:
Murljashka [212]3 years ago
7 0

Answer:

-1.94  * 10^-18 J

Since the electron moved from a higher to a lower energy level (n = 3 → n = 1) it is an emission.

Explanation:

From Rydberg equation;

E = -RH(1/n^2final - 1/n^2initial)

For a transition from  n = 3 → n = 1

RH = 2.18 * 10^-18 J

E = -(2.18 * 10^-18) (1/1^2 - (1/3^2)

E = -(2.18 * 10^-18) (1-1/9)

E= -(2.18 * 10^-18) (8/9)

E = -1.94  * 10^-18 J

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4 0
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Masteriza [31]

Answer:

\large \boxed{\text{-827.4 kJ}}

Explanation:

We have three equations:

1. 2H₂S(g)            + O₂(g)   ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic)  + O₂(g)   ⟶ SO₂(g);                                 ∆H = -296.8 kJ

3. PbO(s)             + H₂S(g) ⟶ PbS(s)               + SO₂(g);    ∆H =  -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s)            + 3O₂(g) ⟶2PbO(s)             + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.  

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ  

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.  

7. 2S(s, rhombic)  + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s)  + <u>2H₂O(g</u>)  ⟶ 2PbO(s) + <u>2H₂S(g</u>);  ∆H =  208.6 kJ

6. <u>2H₂S(g)</u> + O₂(g)        ⟶ <u>2S(s</u>)     + <u>2H₂O(g)</u> ; ∆H = -442.4 kJ

<u>7</u><u>. </u><u>2S(s)</u><u>      + 2O₂(g)      ⟶ 2SO₂(g);                   ∆H = -593.6 kJ </u>

4 . 2PbS(s) + 3O₂(g)      ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}

8 0
3 years ago
An enzyme is discovered that catalyzes the chemical reaction SAD ↔ HAPPY A team of motivated researchers sets out to study the e
Pepsi [2]

Answer:

The K_{m} of a substrate will be "10 μM".

Explanation:

The given values are:

E_{t} = 20 \ nM

[Substract] = 40 \ \mu M

K_{cat}=600 \ s^{-1}

Reaction velocity, Vo=9.6 \ \mu M s^{-1}

As we know,

⇒  Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}

On putting the estimated values, we get

⇒  9.6=\frac{600\times 20\times 10^{-3}\times 40}{K_{m}+40}

⇒  K_{m}+40=\frac{600\times 20\times 10^{-3}\times 40}{9.6}

⇒  K_{m}+40=50

On subtracting "40" from both sides, we get

⇒  K_{m}+40-40=50-40

⇒  K_{m}=10 \ \mu M

6 0
3 years ago
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