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Semenov [28]
2 years ago
11

(100 POINTS) Write the chemical formula of the ionic compound magnesium chlorate. The magnesium (Mg) ion has a +2 charge. Chlora

te (ClO3) has a −1 charge.
A) Mg (ClO3)2
B) Mg2 (ClO3)
C) Mg2 (ClO3)
D) Mg (ClO3)
Chemistry
1 answer:
antoniya [11.8K]2 years ago
6 0

Answer:

A

Explanation:

since Mg has a charge of +2 and ClO3 has a charge of 1-, you need 2 ClO3 to cancel out the +2 since 2 ClO3 ions would have a 2- charge

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How many moles of cl- ions are needed to completely combine with 0.25 moles of mg+2 ions? 0.175 0.75 0.50 0.25 0.125?
Solnce55 [7]
When magnesium ions and chloride ions form a compound the chemical formula would be MgCl2 which means that for every 1 atom of Mg+ there would be 2 atoms of Cl- bonded. This would be proportional to units of moles.

Moles Cl- = 0.25 mol Mg2+ ( 2 mol Cl - / 1 mol Mg2+) = 0.50 mol Cl- would be needed
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3 years ago
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murzikaleks [220]

Answer:

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Explanation:

8 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!        

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The substance is followed by H2O
8 0
3 years ago
How do we get energy from the food we eat?
expeople1 [14]

Answer:

A

Explanation:

I used my resources and they was explaining and I came to a resolution that it was A

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