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tatuchka [14]
3 years ago
7

Fill in the coefficients that will balance the following reaction: a0NH4Cl + a1Ag3PO4 → a2AgCl + a3(NH4)3PO4

Chemistry
2 answers:
kodGreya [7K]3 years ago
6 0

Answer:

a₀ - 3

a₁ - 1

a₂- 3

a₃ -

Explanation:

sergiy2304 [10]3 years ago
5 0
<h3>Answer:</h3>

a₀ - 3

a₁ - 1

a₂- 3

a₃ - 1

<h3>Explanation:</h3>
  • The balanced chemical equation for the reaction will be;

3NH₄Cl + Ag₃PO₄ → 3AgCl + (NH4)₃PO₄

  • Balancing of chemical equations is a try and error method that involves putting coefficients on reactants and products.
  • It ensures that the number of atoms of each element is equal on both sides of the equation.
  • Therefore, a balanced equation will obey the law of conservation of mass in chemical equations, since the mass of the reactants and that of products will be equal.
  • The above equation is balanced as the number of atoms of each element are equal on both sides of the equation.
  • That is, 3 nitrogen atoms, 12 hydrogen atoms, 3 chlorine atoms, 3 silver atoms, 1 phosphorus atom and 4 oxygen atoms on both side of the equation.
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Cuántos mililitros de agua hay en 3 litros de una botella de naranjas y las solucion tiene una concentracion de 20% v/v
sertanlavr [38]

Respuesta:

2400 mL

Explicación:

Paso 1: Información dada

  • Volumen de solución: 3 L (3000 mL)
  • Concentración de naranja: 20 % v/v

Paso 2: Calcular el volumen de naranja

La concentración de naranja es de 20 % v/v, es decir, cada 100 mL de solución hay 20 mL de naranja.

3000 mL Sol × 20 mL Naranja/100 mL Solución = 600 mL Naranja

Paso 3: Calcular el volumn de agua

El volumen de soluciónes igual a la suma de los volúmenes de naranja y agua.

VSolución = VNaranja + VAgua

VAgua = VSolución - VNaranja

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A chemist makes 940. mL of sodium carbonate (Na_2CO_3) working solution by adding distilled water to 200. mL of a 0.171 m stock
Alexxandr [17]

Answer:

The answer is 0.0698 M

Explanation:

The concentration was prepared by a serial dilution method.

The formula for the preparation I M1V1 = M2V2

M1= the concentration of the stock solution = 0.171 M

V1= volume of the stock solution taken = 200 mL

M2 = the concentration produced

V2 = the volume of the solution produced = 940 mL

Substitute these values in the formula

0.171 × 200 = 490 × M2

34.2 = 490 × M2

Make M2 the subject of the formula

M2 = 34.2/490

M2 = 0.069795

M2 = 0.0698 M ( 3 s.f)

The concentration of the Chemist's working solution to 3 significant figures is 0.0698M

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