A) A cat dish has a mass of 1975 g and a volume of 7500ml. What is the overall density of the cat dish? B) find the maximum mass

The new pressure of the gas is calculated to be 40.7 kPa. Using the principle that P1 • V1 = P2 • V2, we can set 98.8 kPa (P1) multiplied by 21.7 mL (V1) equal to P2 (unknown pressure) multiplied by 52.7 mL (V2). To isolate P2, we rearrange the equation to P2 = (98.8 kPa • 21.7 mL) / 52.7 mL, resulting in P2 equal to 40.7 kPa.

The response to this inquiry involves energy release. The bonds holding molecule atoms act as energy reserves. One method of energy release occurs when these bonds are severed, allowing energy to disperse outward. This breaking leads to smaller molecules rather than the creation of a larger one.

<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>

The correct answer is: c. N2O, N2O4, N2O5.

According to the law of multiple proportions, also referred to as Dalton's Law, when two elements form compounds, the mass ratios of the second element that combine with a specific mass of the first yield small whole number ratios.

1) For NO, the mass ratio m(N): m(O) is 14: 16, simplified to 7: 8.

2) In N₂O, the ratio m(N): m(O) equates to 2·14: 16, which simplifies to 7: 4.

3) For NO₂, the masses yield m(N): m(O) = 14: 2·16, simplifying to 7: 16.

4) In N₂O₅, the ratio is (2·14): (5·16), which simplifies to 7: 20.

5) For NO₄, the mass ratio is m(N): m(O) = 14: (4·16), which simplifies to 7: 32.

6) N₂O₄ gives a ratio of m(N): m(O) as (2·14): (4·16), simplifying to 7: 16.

A) This means m(N): m(O) = 5.6 g: 3.2 g, simplifying results in 1.75: 1, which further translates to m(N): m(O) = 7: 4.

B) Here, m(N): m(O) is 3.5 g: 8.0 g. Dividing reveals a ratio of 1: 2.285, which alters to m(N): m(O) = 7: 16.

C) Lastly, for m(N): m(O) = 1.4 g: 4.0 g, then adjustments yield a ratio of m(N): m(O) = 7: 20.

1. The total moles of the solution is 0.3079193 mol. 2. The mole fraction for gold is 0.2473212, and for silver, it is 0.7526787. 3. The molar entropy of mixing for gold is 2.87285 j/K, while for silver, it is 1.77804 j/K. 4. The total entropy of mixing sums to 4.65089 j/K. 5. Molar free energy amounts to -2325.445 kJ. 6. Chemical potential for silver is -1750.31129 j/mol and for gold, it is -575.13185 j/mol. To elaborate: (1) The molar mass of silver stands at 107.8682 g/mol, and gold's at 196.96657 g/mol. Hence, calculating moles leads to mass/molar mass for silver: 25 g/107.8682 g/mol = 0.2317643 mol and for gold: 15 g/196.96657 g/mol = 0.076155 mol, resulting in a total of 0.30791193 mol. (2) For the mole fractions, silver's fraction is 0.2317643/0.3079193 = 0.7526787, and for gold, it's 0.076155/0.3079193 = 0.2473212. (3) To find molar entropy mixing ∆Sm, we use the formula ∆Sm = -R * Xi * ln(Xi) where R = 8.3144598. For silver, substituting gives us 1.77804 j/K, while for gold, we get 2.87285 j/K. (4) Overall entropy of mixing totals 4.65089 j/K thus calculated. (5) The Gibbs free energy at 500 °C can be derived through G = H - TS, accounting to H = 0 (as T is 500 + 273 = 773 K and S is 4.65089), resulting in G equating to -3595.138 kJ. (6) The chemical potentials calculated derive from multiplying the Gibbs free energy by their mole fractions.