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Veseljchak [2.6K]
2 years ago
9

34 grams of carbon react with an unlimited amount of H2O. The reaction is: C + H2O → CO + H2 What is the ending substance?

Chemistry
2 answers:
katovenus [111]2 years ago
7 0
H2c+o
hope this helps :0

Tamiku [17]2 years ago
6 0

The answer is hydrogen.

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Which one is a true statement about endothermic reactions? A. I need you from an outside source is continuously being added. B.
tekilochka [14]
Correct answer: "A. Energy from an outside source is continuously being added."
An endothermic reaction is a reaction that is characterised by the system absorbing energy from its surroundings. That energy is usually in heat form. For example, when mixing water<span> with potassium chloride, this reaction will absorb heat and the container will feel cold - endothermic reaction.</span>
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3 years ago
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How many moles in 4 g of Ca3N2?<br><br> .027 mol<br><br> .613 mol<br><br> 2.05 mol<br><br> .760 mol
Arada [10]

Answer:

0.027mole

Explanation:

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1 year ago
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2 years ago
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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
2 years ago
_________ is the movement of materials across a cell membrane using cellular energy.
Airida [17]

Answer:

active transport

Explanation:

passive transport does not involve energy

4 0
2 years ago
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