a 15.0 kg chunk of ice falls off the top of an iceberg if the chunk of ice Falls 8.00 to the surface of the water what is the ki

Answer:

The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.

Explanation:

Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Thus, the overall reaction becomes;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?

According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;

Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

Answer:

When HBCG and BCG^- are at the same concentration, the resulting color is green. This green shade initially becomes visible at a pH of 3.8.

Explanation:

HBCG serves as an indicator formed by dissolving solids in ethanol.

Since

Ka=[BCG−][H3O+][HBCG] When [BCG-] equals [HBCG], it follows that Ka = [H3O+].

Ka= 1.58 ×10^-4

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

10 mL of this solution is further diluted to 250 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

10 mL of this solution is further diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

5 mL of this solution is diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

10 mL of this solution is further diluted to 500 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

10 mL of this solution is further diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

Explanation:

Filtration serves as a method of separation where solid particles that are suspended in a liquid are isolated by passing the mixture through filter paper's pores. This process ensures that the solid particles accumulate on the filter paper and the liquid flows out through the filter paper's pores.

The ordered sequence of the steps provided is:

1. Measure and fold the filter paper.
2. Insert the filter paper into the funnel, then position the funnel above the Erlenmeyer flask.
3. Let the solid/liquid mixture pass through the filter.
4. Rinse the filter paper that holds the mixture with water.
5. Measure the weight of the dry filter paper along with the copper.

Result: The count of bonding electrons and non-bonding electrons amounts to (4, 18).

Explanation:

The Lewis-dot structure reveals the number of bonding and non-bonding electrons in .

Lewis-dot representation: It illustrates the valence electron count for atoms in a molecule and shows how they bond, as well as any lone pairs of electrons.

In this structure, 'Xe' is the central atom while 'F' is the terminal atom.

Xenon comprises 8 valence electrons, whereas fluorine contains 7.

The total number of valence electrons in is calculated as 8 + 2(7) = 22 electrons

From the Lewis-dot structure, we can determine

The count of electrons involved in bonding = 4

The count of electrons involved in non-bonding (lone-pairs) = 22 - 4 = 18

Thus, the bonding and non-bonding electron counts are (4, 18).

Below is the Lewis-dot structure for .