All categories

14 days ago

Think Critically. Describe two ways you could influence the following equilibrium to produce more ethanal (CH3CH). Use Le Chatel

ier's principle to explain why each of your methods would produce the desired result.

Chemistry

You might be interested in

Write a balanced equation depicting the formation of one mole of NaBr(s) from its elements in their standard states.

alisha [2963]

Answer:

Refer to the explanation.

Explanation:

Formation reactions involve the creation of one mole of a compound from its elements in their standard states.

NaBr (s)

The equation for the standard formation is

Na (s) + (1/2)Br₂ (g) → NaBr (s)

As per appendix C, the standard heat of formation for NaBr(s) is

ΔH∘f = -359.8 kJ/mol.

SO₃ (g)

The equation for the standard formation is

S (s) + (3/2) O₂ (g) → SO₃ (g)

ΔH∘f = -395.2 kJ/mol.

Pb(NO₃)₂ (s)

The equation for the standard formation is

Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)

According to appendix C, the standard heat of formation for Pb(NO₃)₂(s) is

ΔH∘f = -451.9 kJ/mol.

I hope this is helpful!

</paccording>

6 0

3 months ago

Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1

alisha [2963]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

To begin, this isn't really a chemistry forum, but anyway.

This represents a limiting reagent scenario.

Set it up as a Dimensional Analysis issue.

Begin with your desired outcome.

Your goal is to find the mass of acrylonitrile (C3H3N)

so you should initiate with that (I'll abbreviate Acrylonitrile as ACL for convenience)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

If you calculate that, you will find that 15 grams of C3H6 yields 18.9 grams of acrylonitrile produced.

Utilize the same approach for the remaining two reactants.

So, I figured it out, and for

oxygen, I calculated 11.04 grams

and for ammonia, I calculated 15.29 grams

This means that the maximum possible production is 11.04 grams, since to create any additional amount, more O2 would be necessary, but with only 10 grams available, that's the upper limit for this reaction.

The other two reactants are in excess.

Please rate as brainliest!

3 0

3 months ago

Sea water's density can be calculated as a function of the compressibility, B, where p = po exp[(p - Patm)/B]. Calculate the pre

lions [2927]

Answer:

A pressure of 137.14 MPa exists 10,000 m beneath the ocean surface.

At this same depth, the density measures 2039 kg/m3.

Explanation:

P0 and ρ0 symbolize the pressure and density at sea level (indicative of atmospheric conditions). With an increase in ocean depth, both pressure and density likewise rise.

The relationship between pressure and density can be expressed as:

$\frac{dP}{dy}=\rho*g=\rho_0*g*e^{(P-P_0)/\beta\\\\$

By rearranging

$\frac{dP}{e^{(P-P_0)/\beta}}= \rho_0*g*dy\\\\\int\limits^{P}_{P_0} {e^{-(P-P_0)/\beta}}dP =\int\limits^y_0 {\rho_0*g*dy}\\\\(-\beta*e^{-(P-P_0)/\beta})-(\beta*e^0)=\rho_0*g*(y-0)\\\\-\beta*(e^{-(P-P_0)/\beta}-1)=\rho_0*g*y\\\\e^{-(P-P_0)/\beta}=1-\frac{\rho_0*g*y}{\beta}\\\\-\frac{P-P_0}{\beta} =ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\$

This equation allows for computation of P at 10,000 m beneath the ocean's surface:

$P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-200MPa*ln(1-\frac{1027kg/m^3*9.81m/s^2*10,000m}{200MPa})\\\\P-P_0=-200MPa*ln(1-\frac{1027*9.81*10,000Pa}{200*10^6Pa})\\\\P-P_0=-200MPa*ln(1-0.5037)\\\\P-P_0=-200MPa*(-0.6857)=137.14MPa$

The density found at a depth of 10,000 m in the ocean is

$\rho=\rho_0*e^{(P-P_0)/\beta}\\\rho=1027kg/m^3*e^{(137.14/200)}=1027*e^{0.686}kg/m^3\\\rho=1027*1.985 kg/m^3\\\rho=2039\,kg/m^3$

4 0

3 months ago

1. For which of these elements would the first ionization energy of the atom be higher than that of the diatomic molecule?

alisha [2963]

Answer: The correct selection is (b).

Explanation:

The energy required to detach an electron from an atom or ion in its gaseous state is termed ionization energy.

This indicates that a smaller atom necessitates a greater amount of energy to remove its valence electron. The reason for this is that there exists a strong attraction between the nucleus and the electrons in smaller atoms or elements.

Therefore, a significant amount of energy is needed to dislodge the valence electrons.

The electronic configuration for helium is $1s^{2}$ . Hence, due to its fully occupied valence shell, it exhibits greater stability.

Consequently, a large amount of energy is needed to remove an electron from a helium atom.

In conclusion, from the choices provided, the ionization energy of helium will be greater than that of the diatomic molecule.

7 0

3 months ago