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3 years ago

Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =3 to =1.

Chemistry

1 answer:

never [62]3 years ago

6 0

Answer:

$1.936\times 10^{-18}\ \text{J}$

Explanation:

$R_h$ = Rydberg constant = $2.178\times 10^{-18}\ \text{J}$

$n_i$ = Initial shell = 3

$n_f$ = Final shell = 1

We have the relation

$\Delta E=R_h(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})\\\Rightarrow \Delta E=2.178\times 10^{-18}(\dfrac{1}{1^2}-\dfrac{1}{3^2})\\\Rightarrow \Delta E=1.936\times 10^{-18}\ \text{J}$

The energy of the photon emitted here is $1.936\times 10^{-18}\ \text{J}$ .

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If 3.5 grams of NaN3 decomposed, how many grams of N2 would be produced?

Wewaii [24]

Answer:

5.25 moles.

Explanation:

The decomposition reaction of NaN₃ is as follows :

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

We need to find how many grams of N₂ produced in the process.

From the above balanced chemical reaction, we conclude that the ratio of moles of sodium azide and nitrogen gas are 2 : 3.

2 moles of sodium azide decomposes to give 3 moles of nitrogen gas. So,

3.5 moles of sodium azide decomposes to give $\dfrac{3}{2}\times 3.5=5.25$ moles of nitrogen gas.

Hence, the number of moles produced is 5.25 moles.

6 0

3 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago

For the reaction: 3 H2(g) + N2(g) <--> 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [

masya89 [10]

The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.