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alex41 [277]
3 years ago
6

Hola, podrían reseolverme estos problemas? Principalmente el último. Gracias 1. Al mezclar una disolución acuosa de CaCl2 con ot

ra 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 %.
Chemistry
1 answer:
Aneli [31]3 years ago
5 0

Answer:

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.

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Not sure how in depth or what level of particles but I will go as deep as I know. The matter that makes up the world is comprised of 12 particles which are known as fermions. There are 12 fermions which are made up of 6 quarks (up, charm, top, Down, Strange, Bottom) 3 electrons (electron, muon, tau) and three neutrinos (e, muon, tau). Technically, only the up quark, down quark, electron, and electron neutrino are necessary to create all known matter since others would simply be very unstable and decay into those particles. The other type of particles are known as Bosons. These particles transmit forces and all sorts of different interactions. I have included a photo from online which describes the main characteristics of each elementary particle.

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3 years ago
A compound is found to contain 37.32 % phosphorus , 16.88 % nitrogen , and 45.79 % fluorine by
Alla [95]

Answer: 1. The empirical formula is PNF_2  

2. The molecular formula is PNF_2

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of P = 37.32 g

Mass of N = 16.88 g

Mass of F = 45.79 g

Step 1 : convert given masses into moles.

Moles of P =\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{37.32g}{31g/mole}=1.20moles

Moles of N =\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.88g}{14g/mole}=1.20moles

Moles of F =\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{45.79g}{19g/mole}=2.41moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For P = \frac{1.20}{1.20}=1

For N = \frac{1.20}{1.20}=1

For F =\frac{2.41}{1.20}=2

The ratio of P: N: F= 1: 1: 2  

Hence the empirical formula is PNF_2

The empirical weight of PNF_2= 1(31)+1(14)+2(19)= 82.98 g.

The molecular weight = 82.98 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{82.98}{82.98}=1

The molecular formula will be=1\times PNF_2=PNF_2

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