Hola, podrían reseolverme estos problemas? Principalmente el último. Gracias 1. Al mezclar una disolución acuosa de CaCl2 con otra de AgNO3 se forma un precipitado de cloruro de plata: Se mezcla un volumen de 15,0 ml de una disolución 0,30 M de CaCl2 con 30 ml de una disolución 0,05 M de AgNO3. Calcula los gramos de AgCl que precipitarán. 2. Determina el grado de pureza de un mármol (CaCO3) si al descomponerse 125 g del mismo se desprenden 20 litros de dióxido de carbono medidos a 15 ºC y 1 atm. 3. A partir de 120 g de KClO3 se obtuvieron a 18 ºC y a la presión de 740 mmHg, 20 L de oxígeno. ¿Cuál fue el rendimiento de la reacción? 4. Para la obtención de bromobenceno se hacen reaccionar 300 cm3 de benceno (d = 0,89 g/cm3) en exceso de bromo. Determina la masa de bromobenceno obtenido si el rendimiento de la reacción es del 65 %.

0.215 g AgCl produced.

Explanation:

First, the balanced chemical reaction must be written ,

2 AgNO₃ + CaCl₂ ⇒ 2 AgCl + Ca (NO₃) ₂.

- The next step is to calculate the amount of moles of the reagents , for this the molarity is multiplied by the volume in liters:

CaCl₂: 0.15 mol / L * 0.03 L = 0.0045 mol CaCl₂.

AgNO₃: 0.1 mol / L * 0.015 L = 0.0015 mol AgNO₃ .

- The next step is to identify the limiting reagent

For this, the moles of the reagents are divided by their stoichiometric coefficient of the balanced chemical reaction and the lowest value corresponds to the

limiting reagent: CaCl₂: 0.0045 / 1 = 0.0045

AgNO₃: 0.0015 / 2 = 0.00075  ⇒ Limiting reagent. 

- The next step is to determine the amount of moles that form from the precipitate, for this the stoichiometric coefficients are used , in this case we have that 2 moles of AgNO₃ produce 2 moles of AgCl, therefore:

0.0015 mol AgNO₃ * (2 mol AgCl / 2 mol AgNO₃) = 0.0015 mol AgCl.

- Finally, the grams are calculated using the molar mass of the AgCl:

0.0015 mol AgCl * (143.32 g / 1 mol)

= 0.215 g AgCl produced.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.