91影视

Given the information, we need to find the moles of NaN obtained from the data. We know the total mass of the anhydrous sodium salt is 30.0 mg, and the volume of reducing agent required to react completely with NaN is 15.26 mL, with a molarity of 0.02313 M. Using the 1:1 mole ratio as stated in the problem: \( moles NaN = moles reducing agent = volume \times molarity \) \( moles NaN = 15.26 \times 0.02313 = 0.352954 \times 10^{-3} \)

Using the moles of NaN, we can find out the possible value of n. Considering there is a common oxyanion A with charge n-, we need to find the correct charge state of A in NaN. If n=1, it means the formula for the sodium salt is NaA. Otherwise, if n=2, the formula is Na鈧侫, and if n=3, Na鈧傾. The moles of Na equals moles of NaN. Let's calculate the mass of Na in NaN for each possible n. For n = 1 (NaA): \( mass Na = moles NaN \times M_{Na} \) \( mass Na = 0.352954 \times 10^{-3} \times 22.99 = 8.1066 \) For n = 2 (Na鈧侫): \( mass Na = 2 \times moles NaN \times M_{Na} \) \( mass Na = 2 \times 0.352954 \times 10^{-3} \times 22.99 = 16.2132 \) For n = 3 (Na鈧傾): \( mass Na = 3 \times moles NaN \times M_{Na} \) \( mass Na = 3 \times 0.352954 \times 10^{-3} \times 22.99 = 24.3198 \) We can notice n = 1, as the mass of Na is the smallest, which is consistent with the total mass of the anhydrous sodium salt (30.0 mg). Thus, the identity of the oxyanion \(\mathrm{A^{n-}}\) can be obtained by considering the mass difference between the total mass of the anhydrous sodium salt and the mass of Na in NaN: \( mass A = 30.0 - 8.1066 = 21.8935 \) Now, find the equivalent molar mass of A, considering the moles of NaN: \( A_{Molar\_mass} = \frac{mass A}{moles NaN} \) \( A_{Molar\_mass} = \frac{21.8935}{0.352954 \times 10^{-3}} = 62.03 \) A molar mass of 62.03 corresponds to sulfate ion (SO鈧劼测伝), with a molar mass of 32(S) + 4 脳 16(O) = 96. Thus, A = SO鈧劼测伝 and NaN = Na鈧係O鈧�.

The solid metal has a body-centered cubic unit cell and has an atomic radius of 198.4 pm. The density of the solid, pure metal is 5.243 g/cm鲁. Using these properties, let's calculate the edge length of the body-centered cubic unit cell: \( Atomic\_Radius = 198.4 * 10^{-12} m = 198.4 pm \) \( Edge\_length = \frac{4 * Atomic\_Radius}{\sqrt{3}} = \frac{4 * 198.4 * 10^{-12}}{\sqrt{3}} = 3.57795 * 10^{-10} m \) Now, find the volume of the unit cell: \( Volume\_cell = (Edge\_length)^{3} = (3.57795 * 10^{-10})^{3} = 4.57442 * 10^{-29} m^{3} \) We can calculate the mass of the unit cell using the density and volume: \( Mass\_cell = Density * Volume\_cell = 5.243 * 4.57442 * 10^{-29} = 2.39979 * 10^{-26} kg \) We know that there are two atoms in the BCC unit cell, calculate the molar mass of the metal: \( Metal\_Molar\_mass = \frac{Mass\_cell}{2 Atoms} * N_{A} = \frac{2.39979 * 10^{-26}}{2} * 6.022 * 10^{23} = 179.98 \) The metal M has a molar mass of 179.98 and is derived from a relatively expensive silvery-white metal. This corresponds to Yttrium (Y) with an oxidation state of +3.

To determine the formula of the strong electrolyte, first calculate the number of moles of the compound present by using the osmotic pressure equation: \( \pi = i\frac{n}{V}RT \) where \( \pi \) (osmotic pressure) is given as 558 torr, i is the van't Hoff factor, n is the number of moles of the compound, V is the solution volume (10.0 mL), R is the ideal gas constant (0.08206 L atm K鈦宦� mol鈦宦�), and T is the temperature (298.15 K). The strong electrolyte Mx(A)y 路 zH鈧侽 dissociates into ions and contributes to the osmotic pressure. Since M has an oxidation state of +3, the ionic formula can be written as M(A)y(SO鈧�)鈧冣伝 - a trivalent cation with a divalent oxyanion. Thus, the formula of the strong electrolyte is M(SO鈧�)鈧� (since we ignored zH鈧侽). Now we can find the van't Hoff factor (i). The electrolyte dissociates into two ions, M鲁鈦� and A鈦�, so i = 2. Now, let's solve for the number of moles of the compound: \( 558 \frac{torr}{760 \frac{atm}{torr}} = 2 \frac{n}{10.0 * 10^{-3}} * 0.08206 * 298.15 \) \( n = 0.001224 \;mol \) The molar mass of M(SO鈧�)鈧� can be calculated using the molar mass of Yttrium (Y) and sulfate (SO鈧劼测伝): \( Molar\_mass\ M(SO鈧�)鈧� = M_{Y} + 3 * M_{SO鈧剗 = 88.91 + 3 * 96 = 376.91 \) The mass of the compound present in the solution can be found using the number of moles and molar mass: \( Mass\;Compound = moles * Molar\_mass = 0.001224 * 376.91 = 0.461267 g \) This mass is very close to the provided compound mass (33.45 mg), which supports our assumption of the electrolyte formula. Hence, the strong electrolyte's general formula is \( \mathrm{M}_{x}(\mathrm{~A})_{y} \cdot z \mathrm{H}_{2} \mathrm{O} \), where M is Yttrium (Y), A is sulfate (SO鈧劼测伝), and x = 1, y = 3, and we ignored z. The strong electrolyte formula is: \( \mathrm{Y}_{1}(\mathrm{SO}_{4})_{3} \)