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Marat540 [252]
3 years ago
5

3. You measure a cube and determine that its sides are 0.65m. You place the cube on a mass scale and determine that this cube ha

s a mass of 10,500 grams. What is the density of this cube in units of kg/m3 and in units of g/mL?
Chemistry
1 answer:
amid [387]3 years ago
7 0

The density of the cube in-

Kg/m³= 38.234 kg/m³

g/mL= 0.038234 g/mL

Explanation:

We know that volume of the cube equals m ³

Where “m” equals side of the cube

Given data-

Side of the cube=0.65m

mass of the cube= 10,500 gm

We know that 1000gm= 1 kg

Hence, 10,500 gm= 10.5 kg

Volume of the cube= (0.65)³

∴ Volume= 0.274625 m ³

We know that density = mass/volume

⇒Substituting the value of mass and volume, we get-

⇒Density= 10.5/0.274625= 38.234 kg/m³

We know that 1 kg/m³= 0.001 g/mL

Hence 38.234 kg/m³ would equal 0.038234 g/mL

Hence the density of the cube is 38.234 kg/m³ and 0.038234 g/mL

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Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
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Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

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Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
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Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

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