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ANEK [815]
3 years ago
15

My teacher laid this much out for us but I don’t know how to get the products in each one.

Chemistry
1 answer:
irakobra [83]3 years ago
3 0

Answer:dang

Explanation:

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What is the empirical formula for a compound that is 29.44\% calcium, 23.55% sulfur, and 47.01% oxygen? This compound is a commo
PSYCHO15rus [73]

Answer:

Empirical formula is CaSO₄.

Explanation:

Given data:

Percentage of calcium =29.44%

Percentage of sulfur = 23.55%

Percentage of oxygen = 47.01%

Empirical formula = ?

Solution:

Number of gram atoms of Ca = 29.44 / 40 = 0.74

Number of gram atoms of S = 23.55 / 32 = 0.74

Number of gram atoms of O = 47.01 / 16 = 3

Atomic ratio:

            Ca                      :        S                :         O

           0.74/0.74           :     0.74/0.74      :       3/0.74

               1                     :          1              :          4

Ca : S : O = 1 : 1 : 4

Empirical formula is CaSO₄.

3 0
3 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
Determine the Density of a mass = 2000g and volume = 80ml
sergey [27]

Answer:

Explanation:

D = m/v

D = 2000/80 = 25

5 0
2 years ago
Explain the arrangement of particles in solids, liquids and gases.​
Alex787 [66]
Solids are tightly compacted
Liquids are medium
And gases are very spaced and floating around
6 0
3 years ago
Which of the following statements is true for radioactive reactions:
sasho [114]

Answer:

d) Nucleus of the element does not change during the react

hope it helps (^^)

# Cary on learning

8 0
2 years ago
Read 2 more answers
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