What phase would this substance be in at a pressure of 1.25atm and a temperature of 700C

15A.4(a) What is the temperature of a two-level system of energy separation equivalent to 400 cm−1 when the population of the up

maksim [4K]

Answer:

T = 525K

Explanation:

The temperature of the two-level system can be calculated using the equation of Boltzmann distribution:

$\frac{N_{i}}{N} = e^{-\Delta E/kT}$ (1)

where Ni: is the number of particles in the state i, N: is the total number of particles, ΔE: is the energy separation between the two levels, k: is the Boltzmann constant, and T: is the temperature of the system

The energy between the two levels (ΔE) is:

$\Delta E = hck$

where h: is the Planck constant, c: is the speed of light and k: is the wavenumber

$\Delta E = 6.63\cdot 10^{-34} J.s \cdot 3\cdot 10^{8}m/s \cdot 4 \cdot 10^{4}m^{-1} = 7.96 \cdot 10^{-21}J$

Solving the equation (1) for T:

$T = \frac{-\Delta E}{k \cdot Ln(N_{i}/N)}$

With Ni = N/3 and k = 1.38x10⁻²³ J/K, the temperature of the two-level system is:

$T = \frac{-7.96 \cdot 10^{-21}J}{1.38 \cdot 10^{-23} J/K \cdot Ln(N/3N)} = 525K$

I hope it helps you!

3 0

3 years ago

You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of

Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$

Amount of BaSO₄ from K₂SO₄ is;

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$ = $X\times\frac{233.38 }{142.04}$ = 1.643·X

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$ = $Y\times\frac{233.38 }{174.259 }$ = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0

2 years ago

In which reaction are electrons transferred from one reactant to another reactant?

aliina [53]

The answer is (1). The electrons transfer form one reactant to another means that there is the change of valence. For other choices, the valence of reactant does not change.

5 0

2 years ago

Read 2 more answers

4NH3 + 3O2 mc004-1.jpg 2N2 + 6H2O

ioda

Assuming the complete question is;
Consider the balanced equation below;
4NH3 + 3O2 = 2N2 +6H2O
What is the mole ratio of NH3 to N2;
From the balanced equation; the mole ratio of NH3 to N2 is 4:2.
The mole ratio is the ratio of moles of one substance to the number of moles of another substance in a balanced equation. In the above equation; 4 moles of ammonia requires 3 moles of oxygen to produce two moles of nitrogen and six moles of water.

4 0

2 years ago

How many mL of 1.76 M sodium hydroxide solution (strong base) must be added to 0.75

erma4kov [3.2K]

Answer: Special care is required to prepare a solution of sodium hydroxide or NaOH in water because considerable heat is liberated by the exothermic reaction. The solution may splatter or boil. Here is how to make a sodium hydroxide solution safely, along with recipes for several common concentrations of NaOH solution.

Explanation:

Hope this helps!

5 0

2 years ago