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MA_775_DIABLO [31]
3 years ago
13

What is the correct name for (NH4)4P407?​

Chemistry
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer:

Diammonium hydrogen phosphate

Explanation:

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If the lab technician needs 30 liters of a 25% acid solution, how many liters of the 10% and the 30% acid solutions should she m
Mice21 [21]

Answer:

7.5 L of the 10% and 22.5 L of the 30% acid solution, she should mix.

Explanation:

Let the volume of 10% acid solution used to make the mixture = x L

So, the volume of 30% acid solution used to make the mixture = y L

Total volume of the mixture = <u>x + y = 30 L .................. (1) </u>

For 10% acid solution:

C₁ = 10% , V₁ = x L

For 30% acid solution :

C₂ = 30% , V₂ = y L

For the resultant solution of sulfuric acid:

C₃ = 25% , V₃ = 30 L

Using  

C₁V₁ + C₂V₂ = C₃V₃

10×x + 30×y = 25×30

So,  

<u>x + 3y = 75 .................. (2) </u>

Solving 1 and 2 we get,

<u>x = 7.5 L </u>

<u>y = 22.5 L</u>

6 0
3 years ago
So its not A so its B C or D plz help
Rainbow [258]

Answer:

the answer is B

hope this helped!

3 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
3 years ago
Propiedades químicas del óxido​
tangare [24]

Óxidos básicos: Son formados por metales. El metal presente en su fórmula puede presentar carga eléctrica +1 y +2, o sea, poseer carácter iónico. Ejemplos: Na2O (óxido de sodio), BaO (óxido de bario).

Óxidos neutros: Son compuestos por no metales. No reaccionan con agua, ácido o base, en razón del enlace covalente que une sus componentes; de ahí el por qué de ser llamados óxidos inertes. Ejemplos: monóxido de dinitrógeno (N2O) y monóxido de carbono (CO).

Óxidos ácidos: También conocidos como anhídridos de ácidos, son formados por no metales y presentan carácter covalente. En la presencia de agua, producen ácidos y en la presencia de bases, origina sal y agua. Ejemplo: CO2 (dióxido de carbono o gas carbono) y el SO2 (dióxido de azufre)

Óxidos dobles o mixtos: La combinación de dos óxidos de un mismo elemento, da origen a este tipo de óxidos. Ejemplo: magnetita (Fe2O4), unión de los óxidos de hierro (Fe) y oxígeno (O).

Óxidos anfóteros: Presentan ambigüedad, en la presencia de un ácido se comportan como óxidos básicos y en la presencia de una base, como óxidos ácidos. Ejemplos: óxido de aluminio (Al2O3 ) y el óxido de zinc (ZnO).

Peróxidos: Compuestos que poseen en su fórmula el grupo (O2)2- . Los peróxidos más comunes son formados por hidrógeno, metales alcalinos y metales alcalinos térreos. Ejemplos: agua oxigenada (H2O) y peróxido de sodio (Na2O2).

8 0
3 years ago
Please help
Snezhnost [94]
Relatively few hydrogen atoms
5 0
3 years ago
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