How many grams are in 4.5 moles of chlorine gas (Cl2)?

A chunk of zinc is added to a solution of gold (III) nitrate to extract the gold. The reaction forms,

alina1380 [7]

Answer:

it forms :

1. Gold ( Au )

2. Zinc nitrate ( Zn(NO3)2 )

Explanation:

When a chunk of zinc is added to a solution of gold (III) nitrate to extract the gold. The reaction forms Gold and Zinc nitrate .

it's a single displacement reaction,

here's the balanced equation for above reaction :

3 Zn + 2 Au(NO3)3 =》3 Zn(NO3)2 + 2 Au

7 0

3 years ago

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The molar mass of AI2CI6 is the mass of one mole of the compound. It is also which of the following?

Hunter-Best [27]

Answer:

Is their a picture with the question?

Explanation:

8 0

2 years ago

How water deposits soil sediment and rock

Oliga [24]

Water deposits soil ,sediments ,and rock by moving them to a different place .Say like a beach .The ocean is currently moving the bits of sand and (glass:sand)rock pieces and always ending up in new places.Same as a river.!!!!

6 0

3 years ago

What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0

3 years ago

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Electrons in conductors, like copper wire, are free to roam throughout the metal in the presence of the ions that have released

Scrat [10]

Answer:

<u><em>No, I would not consider a metal to be a plasma because plasma is just another state of matter, and the copper wire is in solid state.</em></u>

Explanation:

Metal is not a state of matter. Metals can be solid or liquid (molten) depending on their melting point and the temperature at which they are.

Plasma is a state of matter, similar to gas, but it is reached only at very high temperatures like in the Sun. The particles in plasma state are not neutral atoms or molecules but negatively charged ions and electrons.

The copper wire is yet a solid, thus it cannot be considered a plasma.

Metals can be in plasma state only if the temperature is too high, like the temperatures in the stars. In fact, the metals in the Sun and other hotter stars are in plasma state.

5 0

3 years ago