A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr

Solution:

Washing Clothes & Dissolving Sugar

Clarification:

Consider each scenario:

1) For washing clothes, water is essential; without it, washing is ineffective.

2) Connecting brake pedals to brake pads requires solids, not liquids.

3) To deodorize a space, one would likely reach for an aerosol spray, which is a gas.

4) Sculpting involves solid tools and a solid medium.

5) Dissolving sugar necessitates a liquid to be effective!

6) While one might assert that paint is a liquid, it still might not fit the category; I would categorize this application as solid.

7) Gears employed in machinery are solid components!

N₀ signifies the quantity of C-14 atoms per kg of carbon in the original sample at time = 0 seconds, when the carbon composition matched that in today’s atmosphere. As time progresses to ts, the number of C-14 atoms per kg declines to N, due to radioactive decay. λ indicates the decay constant.
Hence, we have N = N₀e - λt, which is the equation for radioactive decay. Rearranging gives us N₀/N = e λt, or In(N₀/N) = - λt, which becomes equation 1.
The sample contains mc kg of carbon, leading to an activity measured as A/mc decay per kg. The variable r represents the initial mass of C-14 in the sample at t=0 relative to the total mass of carbon which is calculated as [(total number of C-14 atoms at t = 0) × ma] / total mass of carbon. Thus, N₀ equates to r/ma, which becomes equation 2.
The activity of the radioactive element is directly related to the atom count at the moment. The activity equation A = dN/dt = λ(N) indicates that: A = λ₁(N × mc). Rearranging provides N = A / (λmc), represented in equation 3.
By integrating equations 2 and 3, we can solve for t yielding
t = (1/λ) In(rλmc/m₀A).

Answer:

[Cl⁻] = 0.016M

Explanation:

To begin, we analyze the reaction:

Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓  +  Mg(NO₃)₂(aq)

This indicates a solubility equilibrium, resulting in the formation of lead(II) chloride precipitate. The salt can dissociate as follows:

           PbCl₂(s)  ⇄  Pb²⁺ (aq)  +  2Cl⁻ (aq)     Kps

Initial        x

React       s

Eq          x - s              s                  2s

Given that this is an equilibrium scenario, the Kps serves as the constant (Solubility product):

Kps = s. (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s =  ∛(1.7ₓ10⁻⁵. 1/4)

s = 0.016 M

Answer:

The molality of the solution is 1.08 m.

Explanation:

First, determine the mass of the solvent.

A 13% solution by mass indicates that 13 grams are found in every 100 grams of solution.

Thus, solution mass = solute mass + solvent mass

100 g = 13 g + solvent mass

Therefore, solvent mass = 100 g - 13 g → 87 g

Next, we calculate the moles of solute (mass / molar mass):

13 g / 138.2 g/mol = 0.094 moles

Finally, to find the molality, which is the moles of solute per 1 kg of solvent (mol/kg), we convert the solvent mass to kg:

87 g. 1 kg / 1000 g = 0.087 kg

Then, molality → 0.094 mol / 0.087 kg = 1.08 m

The chemical reaction is:

HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l).

The enthalpy change for this reaction is ΔH = -54 kJ.

This means that reacting one mole of HCl with one mole of NaOH releases 54 kJ of heat energy.

This is an exothermic reaction, and its energy profile resembles that shown in figure (1).

The question asks:

What is the enthalpy change when 10 mL of 1 M HCl reacts with 20 mL of 1 M NaOH?

Calculating moles of HCl: molarity × volume = 1 M × 10 mL = 10 millimoles.

Assuming complete reaction, the moles of NaOH reacted equals moles of HCl.

Therefore, total moles that reacted = 10 millimoles, producing the same amount of water.

Since one mole of acid-base reaction produces one mole of water, formation of 10 millimoles of water releases energy:

54 × 10 × 10⁻³ = 0.54 kJ

Reaction:

HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l).

Enthalpy change: ΔH = -54 kJ.

One mole of HCl reacts with one mole of NaOH and liberates 54 kJ heat.

The reaction is exothermic, with the graph similar to figure (1).

Question:

What is the enthalpy change when 10 mL of 1 M HCl reacts with 20 mL of 1 M NaOH?

Moles of HCl = molarity × volume = 1 M × 10 mL = 10 millimoles.

Moles of NaOH equate to moles reacted.

Total acid-base reaction moles = 10 mmol, equating to water produced.

Since one mole acid-base reaction produces 1 mole of water, the energy released for 10 mmol water is:

54 × 10 × 10⁻³ = 0.54 kJ