Which atomic models in task 1 are not supported by Thomson’s experimental evidence? For each of these models, explain the experi

Response:

q < 0, w > 0, the sign of ΔE cannot be ascertained from the provided details

Explanation:

Assessing the sign of q

The water bath's temperature before the reaction is at 25 °C

Following the reaction, the water bath's temperature rises to 28 °C, suggesting that heat was released from the system during the reaction.

If heat is absorbed by the system, q is positive; if heat is released, q is negative, leading to the conclusion that q < 0 in this scenario.

Determining the sign of w

The downward movement of the piston indicates a reduction in the system's volume, meaning work was conducted on the system. When work is done on the system, w is considered positive; hence, w > 0 in this case.

Evaluating ΔE sign

According to the first law of thermodynamics, the relationship among ΔE, q, and w is represented as follows:

ΔE = q + w

In this instance, since q is negative and w is positive, the resulting sign of ΔE is determined by the relative magnitudes of q and w. Since we lack sufficient information to gauge these magnitudes, we cannot definitively determine the sign of ΔE based on the given data.

As a result, the correct option among the choices presented is c: q < 0, w > 0, the sign of ΔE cannot be discerned from the information given.

Solution:

The molecular formula is PbSO₄, indicating lead sulfate

Option c.

Explanation:

The percentage makeup shows that in 100 g of this compound, there are:

68.3 g of Pb, 10.6 g of S, and (100 - 68.3 - 10.6) = 21.1 g of O

To find the moles of each element, we divide by their molar masses:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

Next, we find the mole ratio by dividing each by the smallest number of moles:

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Thus, the molecular formula is PbSO₄, representing lead sulfate.

The result is 0.14303691.

Carbon-13 (¹³C) is a stable isotope of carbon with a mass number of 13, composed of six protons and seven neutrons.

Isotopes are elements that share the same atomic number but have different mass numbers, meaning they have a varying number of neutrons.

ω(¹³C) = 1.10% ÷ 100%.

ω(¹³C) = 0.0110; this indicates the natural abundance of carbon-13.

m(¹³C) = 13.003355; the atomic mass assigned to carbon-13.

ω(¹³C) · m(¹³C) = 0.0110 · 13.003355.

ω(¹³C) · m(¹³C) = 0.14303691.

Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

The half-reactions are:

Anode oxidation half reaction:  

Cathode reduction half reaction:  

Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature =

n = electrons exchanged in oxidation-reduction = 2

= standard electrode potential for the cell = +0.63 V

= cell potential for the reaction =?

= concentration of Zn²⁺ = 3.5 M

=

Now substituting all known values into the equation, we arrive at:

So, the resulting cell potential for this reaction is 0.50 V

The solution to your inquiry yields P = 17.73 atm. Explanation: The volume V is 250 ml, equivalent to 0.25 liters (L), with a mass of 3.4 g and a temperature of 45°C, which converts to 318°K. We utilize the ideal gas law PV = nRT for the calculations.