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olga55 [171]
2 years ago
7

A cylinder with a movable piston originally has a volume of 2805 mL and is filled with nitrogen to a pressure of 4.00

Chemistry
1 answer:
Sergio039 [100]2 years ago
6 0

This problem is providing the initial volume and pressure of nitrogen in a piston-cylinder system and asks for the final pressure it will have when the volume increases. At the end, the answer turns out to be 2.90 atm.

<h3>Boyle's law</h3>

In chemistry, gas laws are used so as to understand the volume-pressure-temperature-moles behavior in ideal gases and relate different pairs of variables.

In this case, we focus on the Boyle's law as an inversely proportional relationship between both pressure and volume at constant both temperature and moles:

P_1V_1=P_2V_2

Thus, we solve for the final pressure by dividing both sides by V2:

P_2=\frac{P_1V_1}{V_2}

Hence, we plug in both the initial pressure and volume and final volume in order to calculate the final pressure:

P_2=\frac{2805mL*4.00atm}{3864mL}\\ \\P_2=2.90atm

Learn more about ideal gases: brainly.com/question/8711877

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Hope this helps!

8 0
3 years ago
Read 2 more answers
The average atomic masses of some elements may vary, depending upon the sources of their ores. Naturally occurring boron consist
frez [133]
Sorry I don’t know but good luck!
6 0
3 years ago
Read 2 more answers
1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en
Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂    

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂              

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.  

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles      

n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles

m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

\% = \frac{R_{r}}{R_{T}}*100

<u>Donde</u>:

R_{r}: es el rendimiento real

R_{T}: es el rendimiento teórico

\% = \frac{3,5}{5,043}*100 = 69,4

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!        

5 0
3 years ago
\"Moving down group 2A (Alkaline Earth Metals), which element has the largest first ionization energy?\" Is the answer to this:
lesya692 [45]

Ionization energy is the energy required to remove the outermost electron from one mole of gaseous atom to produce 1 mole of gaseous in to produce a charge of 1. The greater the ionization energy, the greater is the chance f the electron to be removed from the nucleus. In this casse, Radium has the largest ionization energy.

6 0
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g Nitrogen in the atmosphere consists of two nitrogen atoms covalently bonded together (N2). Knowing that nitrogen is atomic num
vichka [17]

Explanation:

Since, the atomic number of nitrogen is 7 and its electronic distribution is 2, 5. So, in order to attain stability it needs to gain 3 electrons.

Hence, when it chemically combines another nitrogen atom then as both the atoms are non-metals. So, sharing of electrons will take place.

Also, there is no difference in electronegativity of two nitrogen atoms. Hence, compound formed N_{2} is non-polar covalent in nature.

5 0
3 years ago
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