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abruzzese [7]
3 years ago
9

Fission and fusion reactions both release energy. However, only fusion reactions

Chemistry
1 answer:
miv72 [106K]3 years ago
5 0
Answer:
<span>D) combine light nuclei
</span>
Explanation:
Fission and fusion both deal with the alteration of atoms in order to produce energy. However, they are actually opposite processes, this is because:
1- Fission occurs when an atomic nucleus is split/divided into two forming two atoms
2- Fusion occurs when two light atomic nuclei combine together to form a single atom.

Hope this helps :) 
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A 32.0 mL sample of hydrogen is collected over water at 20.0°C and 750.0 mm Hg. What is the volume of the dry gas at STP?
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bro thanks for the points tho

Explanation:

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If a transverse wave travels 10 meters in 5 seconds what is its speed
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2 meters per second. You do 10 divided by 5 to find your answer, which is 2.
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A small piece of hot metal is placed in cooler water. The metal is left in the water
scoundrel [369]

Answer: The amount of energy lost by the metal is equal to the amount of energy gained by the water

Explanation:

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2 years ago
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
The National Environmental Policy Act was established in 1965. True or false
Mariana [72]
The answer to your question is false

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3 years ago
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