A different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity

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.

Answer:

Explanation:

1. Molar concentration

Designate chloroform as C and acetone as A.

The molar concentration for C is derived from Moles of C per Litres of solution.

(a) Moles of C

We are assuming there are 0.187 moles of C.

This resolves that step.

(b) Litres of solution

Next, identify 0.813 moles of A.

(i) Mass of each component

(ii) Volume of each component

(iii) Volume of solution

Assuming mixing doesn't alter the total volume.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

2. Molal concentration of C

Molal concentration is calculated as moles of solute per kilograms of solvent.

Total moles of C = 0.187 mol.

Mass of A = 47.22 g = 0.047 22 kg.

Answer:

a. H₂O (conjugate acid); b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid); c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base); d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base); f. No conjugate acid or base exists; g. H₂S (conjugate acid), S⁻² (conjugate base);

h. H₄N₂ (conjugate base)

Explanation:

a. OH⁻ + H⁺ ⇄ H₂O

The hydroxide functions as a Bronsted-Lowry base, allowing it to capture a proton, thus water serves as the conjugate acid.

b. H₂O is amphoteric, capable of acting as either an acid or a base. As a base, its conjugate acid is H₃O⁺, whereas as an acid, its conjugate base is OH⁻.

c. HCO₃⁻ + H⁺ ⇄ H₂CO₃

HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺

Bicarbonate is also amphoteric. When it captures a proton, it forms carbonic acid as the conjugate acid when acting as a base. When HCO₃⁻ acts as an acid and releases a proton, carbonate becomes the conjugate base.

d. Ammonia functions as a weak base, with ammonium being the conjugate strong acid.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

e. Another amphoteric compound. Acid sulfate can function as both an acid and a base.

(similar to bicarbonate). Acting as a base yields sulfuric acid as the conjugate acid, while acting as an acid leads to sulfate as the conjugate base.

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

HSO₄⁻ + H⁺ ⇄ H₂SO₄

f. H₂O₂ does not accept H⁺ or OH⁻ nor does it expel H⁺. It’s neutral and does not function as an acid or base.

g. HS⁻ is amphoteric.

HS⁻ + H⁺ ⇄ H₂S

HS⁻ + H₂O ⇄ S⁻² + H₃O⁺

This is similar to the case of bicarbonate or acid sulfate.

h. H₅N₂⁺ + H₂O ⇄ H₄N₂ + H₃O⁺

Hydrazinium acts as an acid, making hydrazine its conjugate base.

Answer:

The correct answer is "Speed".

Explanation:

• An intensive or individualized physical property is identified when "speed" is observed as the excretion of an individual in a confined area, capable of reaching someone one meter away after sneezing or coughing.
• This measure is represented in the unit of "meter per second", indicating its intensive nature.

<span>128 g/mol Applying Graham's law of effusion, we can utilize the formula: r1/r2 = sqrt(m2/m1) where r1 = effusion rate of gas 1 r2 = effusion rate of gas 2 m1 = molar mass of gas 1 m2 = molar mass of gas 2 Given that the atomic weight of oxygen is 15.999, the molar mass of O2 = 2 * 15.999 = 31.998. We can now insert the known values into Graham's equation to find m2. r1/r2 = sqrt(m2/m1) 2/1 = sqrt(m2/31.998) 4/1 = m2/31.998 Thus, we find m2 to be 127.992. Rounding to three significant figures yields 128 g/mol</span>