A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74

First scenario:

IV: soda, gatorade, orange juice, and water

DV: state of the liquids listed above

Control: freezer and ice tray

Second scenario:

IV: laundry detergent, water

DV: cleanliness of the squares post-wash

Control: chocolate, cloth type, cloth squares

Third scenario:

IV: type of water used, pea plant

DV: growth of the pea plant

Control: pots and daily water amount for the plant

Answer:

Explanation:

Given data:

Initial temperature T₁ = 25.2°C = 298.2K

Initial pressure P₁ = 0.6atm

Final temperature = 72.4°C = 345.4K

What we need to find:

Final pressure = ?

To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.

Now we can substitute the known variables:

P₂ = 0.7atm

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).

Result:

94.7 %

Explanation:

The balanced reaction is:

2 S + 3 O₂ → 2 SO₃

The stoichiometric mole ratio is:

S: 2 moles

O₂: 3 moles

Moles are calculated as mass divided by molar mass:

n = w / m

where n = moles, w = mass, m = molar mass.

Given:

For sulfur: w = 6.0 g, molar mass = 32 g/mol, so n = 6 / 32 = 0.1871 mol

For oxygen: w = 5.0 g, molar mass = 32 g/mol, thus n = 5 / 32 = 0.15625 mol

Comparing to stoichiometric ratios, sulfur is in excess, so oxygen is the limiting reagent, controlling product formation.

Using proportions:

3 mol O₂ produce 2 mol SO₃, so 1 mol O₂ yields 2/3 mol SO₃.

Therefore, 0.15625 mol O₂ yields (2/3) × 0.15625 = 0.1042 mol SO₃.

Mass of SO₃ produced = n × molar mass = 0.1042 mol × 80 g/mol = 8.340 g

The percentage yield is actual yield divided by theoretical yield times 100:

Percent yield = (7.9 g / 8.340 g) × 100 = 94.7 %

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