Calculate the frequency of blue light of wavelength 4.5x10^2 nm

Question 2 (1 point) At the center of the atom is the electron. True False

Stells [14]

Answer:

False the electrons are on the outside of the atoms

Explanation:

5 0

3 years ago

The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make

natta225 [31]

Answer:

the energy of the third excited rotational state $\mathbf{E_3 = 16.041 \ meV}$

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = $\dfrac{m_1 \times m_2}{m_1 + m_2}$

μ = $\dfrac{1 \times 35}{1 + 35}$

μ = $\dfrac{35}{36}$

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ = $\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg$

μ = 1.6139 × 10⁻²⁷ kg

$r_o = 127 \ pm = 127*10^{-12} \ m$

The rotational level Energy can be expressed by the equation:

$E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)$

where ;

J = 3 ( i.e third excited state) &

$I = \mu r^2_o$

$E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)$

$E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)$

$E_3= 2.5665 \times 10^{-21} \ J$

We know that :

1 J = $\dfrac{1}{1.6 \times 10^{-19}}eV$

$E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV$

$E_3 = 16.041 \times 10 ^{-3} \ eV$

$\mathbf{E_3 = 16.041 \ meV}$

8 0

2 years ago

What is the relationship between chemical equilibrium and the rates of forward and reverse reaction?

Nataly_w [17]

Answer:

c

Explanation:

the rate of a forward process must be exactly balanced by the rate of the reverse process.

3 0

3 years ago

Read 2 more answers

CS2 (s) + 3 O2 (g) → CO2 (g) + 2 SO2 (g)

Natasha_Volkova [10]

Answer:

2.067 L ≅ 2.07 L.

Explanation:

The balanced equation for the mentioned reaction is:

CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),

It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.

At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

using cross multiplication:

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 3.1 L.

∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.

To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:

Using cross multiplication:

3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.

0.1384 mol of O₂ produce → ??? mol of SO₂.

∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.

Again, using cross multiplication:

1.0 mol of SO₂ represents → 22.4 L, at STP.

0.09227 mol of SO₂ represents → ??? L.

∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.

8 0

3 years ago

6 How many moles are in a 321 g sample of magnesium?

Tatiana [17]

Answer:

13.4mol of Mg

Explanation:

Given parameters:

Mass of magnesium = 321g

Unknown:

Number of moles = ?

Solution:

The number of moles of a substance is given as;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Mg = 24g/mol

Insert the parameters and solve;

Number of moles = $\frac{321}{24}$ = 13.4mol of Mg

3 0

2 years ago