Which statement is TRUE regarding the macroscopic and

Answer:

La masa molar del compuesto es: 168.82 g/mol

La masa molar del gas es: 16.38 g/mol

Explanation:

(a)

Utilizando la ecuación de gases ideales:

donde,

P es la presión

V es el volumen

n es el número de moles

T es la temperatura

R es la constante de los gases, cuyo valor es = 0.0821 L.atm/K.mol

Además,

Moles = masa (m) / Masa molar (M)

La densidad (d) = Masa (m) / Volumen (V)

Así, la ecuación de gases ideales se puede expresar como:

Dado que:-

Presión = 20 kPa = 20000 Pa

La expresión para la conversión de presión en Pascal a presión en atm se muestra a continuación:

P (Pa) = P (atm)

20000 Pa = atm

Presión = 0.1974 atm

Temperatura = 330 K

d = 1.23 kg/m³ = 1.23 g/L

Masa molar =?

Aplica la fórmula:

0.1974 atm × M = 1.23 g/L × 0.0821 L.atm/K.mol × 330 K

⇒M = 168.82 g/mol

La masa molar del compuesto es: 168.82 g/mol

(b)

Dado que:

Presión = 152 Torr

Temperatura = 298 K

Volumen = 250 cm³ = 0.25 L

Utilizando la ecuación de gases ideales:

R =

Aplicando la fórmula:

152 Torr × 0.25 L = n × 62.3637 L.torr/K.mol × 298 K

⇒n = 0.002045 moles

Dado que:

Masa del gas = 33.5 mg = 0.0335 g

Masa molar =?

La fórmula para calcular los moles se muestra a continuación:

Así,

La masa molar del gas es: 16.38 g/mol

Answer:NH₃/NH₄Cl

The pH of a buffer can be determined using Henderson-Hasselbalch's equation.

When the concentration of acid equals that of the base, the pH aligns with the pKa of the buffer. The ideal pH range is pKa ± 1.

Below are the buffers and their corresponding pKa values:

• CH₃COONa/CH3COOH (pKa = 4.74)

• NH₃/NH₄Cl (pKa = 9.25)

• NaOCl/HOCl (pKa = 7.49)

• NaNO₂/HNO₂ (pKa = 3.35)

• NaCl/HCl Not a buffer

Thus, the ideal buffer is NH₃/NH₄Cl.

Answer:

Empirical formula: Na3S4O9

Molecular formula: Na6S8O18

Explanation:

1) Upang makuha ang empirical formula, na siyang pinaka-simpleng buong bilang na ratio ng bawat elemento sa hindi kilalang compound na ito, una nating hahatiin ang dami ng bawat elemento sa kanilang sariling atomic mass.

Na= 23g/mol, S= 32g/mol, O=16g/mol

Na = 20.2/23 = 0.878mol

S = 37.6/32 = 1.175mol

O = 42.2/16 = 2.638mol

Susundan ito sa pamamagitan ng paghahati ng bawat bilang ng moles sa pinakamababang halaga (0.878mol)

Na = 0.878/0.878 = 1

S = 1.175/0.878 = 1.34

O = 2.638/0.878 = 3.004

Susunod, paramihin ang bawat decimal na bahagi ng 3 upang maging buo ang mga numero:

Na = 3

S= 4.02 ~ 4

O= 9.012 ~ 9

Kaya, ang pinaka-simpleng ratio ng Na, S, at O ay 3:4:9 kaya, ang empirical formula ay Na3S4O9

2) Para makuha ang molecular formula, gagamitin natin ang molecular weight ng bawat elemento sa empirical ratio, at ang kabuuang molar mass ng compound i.e.

Molar mass ng compound= 682.8g/mol

molecular weight ng Na3S4O9 = (3 x 23g/mol) + (4 x 32 g/mol) + (9 x 16 g/mol)

= 69 + 128 + 144

= 341 g/mol

Kaya, (Na3S4O9)n = 682.8g/mol

= (341)n = 682.8

n = 682.8/341

n~ 2

Kaya, molecular formula = (Na3S4O9)2

= Na6S8O18

Answer:

7.3

Explanation:

Using the Henderson Hasselbalch equation, one can determine the pH or pOH of a solution via its pKa. Remember, , and pKa = -logKa, where Ka denotes the acid's equilibrium constant.

The Henderson Hasselbalch formula:

In this context, acid X possesses two ionic forms: the carboxyl group and an alternative form. Initially, we have 0.1 mol/L of acid in 100 mL, which gives:

n1 = (0.1 mol/L)×(0.1 L) = 0.01 mol

Upon dissociation, it yields 0.005 mol of the carboxyl form and 0.005 mol of the other form with stoichiometry assumed constant.

Introducing NaOH at a concentration of 0.1 mol/L and 75 mL, the moles of become:

n2 = (0.1 mol/L)×(0.075 L) = 0.0075 mol

Thus, 0.0075 mol of reacts with 0.005 mol of the carboxyl form, leading to 0.0025 mol of , which in turn reacts with 0.005 mol of the alternating group, leaving 0.0025 mol of the latter.

The new solution’s volume is 175 mL, but the concentrations of both forms remain unchanged in volume, so we can utilize the moles in the equation.

Thus, we arrive at:

6.72 = pKa - log 4

pKa - log 4 = 6.72

pKa = 6.72 + log 4

pKa = 6.72 + 0.6

pKa = 7.3

Answer:

Calcium phosphate has the formula Ca3(PO4)2, which has a molar mass of 310 grams per mole.

For 2.3*10^-4 moles, we calculate 2.3*10^-4 * 310, resulting in 713*10^-4 grams, or 71.3 milligrams.

If the formula is accurate but the compound's name is incorrect, simply replace 310 with 278, yielding 639.4*10^-4 grams, equivalently 63.94 milligrams.