A precipitate of lead(II) chromate can be formed using the reaction below.

Scoring: Your score will be based on the number of correct matches minus the number of incorrect matches. There is no penalty fo

jarptica [38.1K]

Answer:

a. Too close to zero

b. ΔS > 0

c. Too close to zero

d. ΔS < 0

e. ΔS > 0

Explanation:

In order to answer this question we need to compare the change in number of mol in the gas state for the products vs the reactants.

An increase in number moles gas leads to a positive change in entropy. Conversely a decrease will mean ΔS is negative.

Now we can solve the parts in this question.

a. H₂ (g) + F₂ (g) ⇒ 2 HF (g)

We have no change in the number of moles gas products minus reactants. Therefore is too close to zero to decide since there is no change in mol gas.

b. 2 SO₃ ( g) ⇒ 2 SO₂ (g) + O₂ (g)

We have three moles of gas products starting with 2 mol gas SO₂, therefore ΔS is positive.

c. CH₄ (g) + 2 O₂ (g) ⇒ CO₂ (g) + 2 H₂O (g)

Again, we have the same number of moles gas in the products and the reactants (3), and it is too close to zero to decide

d. 2 H₂S (g) + 3 O₂(g) ⇒ 2 H2O (g) + 2 SO₂ (g)

Here the change in number of moles gas is negative ( -1 ), so will expect ΔS < 0

e. 2H₂O₂ (l) ⇒ 2H₂O(l) +- O2(g)

Here we have 1 mol gas and 2 mol liquid produced from 2 mol liquid reactants, thus the change in entropy is positive.

3 0

3 years ago

Why are atoms considered the building blocks life?

dmitriy555 [2]

Because atoms can join together to form molecules, which in turn form most of the objects around us.

7 0

2 years ago

What volume will 12.0 g of oxygen gas occupy at 25 c and a pressure of 52.7 kpa?

Stolb23 [73]

We can use the ideal gas law equation to find the volume occupied by oxygen gas
PV = nRT
where ;
P - pressure - 52.7 kPa
V - volume
n - number of oxygen moles - 12.0 g / 32 g/mol = 0.375 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values in the equation
52 700 Pa x V = 0.375 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
V = 17.6 L
volume of the gas is 17.6 L

8 0

3 years ago

Cuales son las fórmulas de la velocidad

AnnZ [28]

Answer:

s = d÷ t

Explanation:

Where s means speed, d means distance and t means time

7 0

3 years ago

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr

Alecsey [184]

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

$E_{n} = -\frac{m_{e}e^{4}}{2(4\pi\epsilon_{0})^2\hbar^{2}}\frac{1}{n^2} = E_{1}\frac{1}{n^{2}}$ , where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

E = Photon's energy

h = Planck's constant

c = speed of light in vacuum

[tex]\lambda[tex] = wavelength

Solving for [tex]\lambda[tex] and substituting the required values:

[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).

5 0

3 years ago