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nexus9112 [7]
3 years ago
5

Deduce the charge of Z ,in the compound of ZBr2

Chemistry
1 answer:
lyudmila [28]3 years ago
8 0

Answer:

2+

Explanation:

The charge on Z in the compound ZB₂ is a 2+  

  From the given compound:

   We know that Br is bromine with a charge of -1;

So;

   Using the combining power approach:

Atoms                              Z                                Br

Combining power           2                                1

Exchange of valency      1                                 2

So;

  The charge on Z is  2+

          2

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4. How many moles of atoms are present in 7 moles of sulfuric acid, H₂SO4?
Afina-wow [57]

Answer:

<u>49 moles of atoms</u>

Explanation:

There are 7 individual atoms in each molecule of H2SO4:  (2 hydrogens + 1 sulfur + 4 oxygens).

Therefore, if 7 moles are decomposed, there would be 7 times that amount of individual atoms:

7 x 7 = 49 moles of atoms

4 0
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
Treatment of butanedioic (succinic) anhydride with ammonia at elevated temperature leads to a compound of molecular formula C4H5
artcher [175]

Answer:

The product is Methyl cyanoacetate

Explanation: see structure attached

5 0
3 years ago
2. Calculate the mass of K in 90g of KOH​
ValentinkaMS [17]

KOH = 90g

KOH = 57 g/mol

K = 40g/mol

57g/mol contains 90 g

40g/mol will contain?

= 63.15 g

The mass of K is 63.15g

8 0
2 years ago
What type of energy results from the
jeka94

Answer:

C fghhtrsfeagutyrwraqedf

3 0
3 years ago
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