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Kryger [21]
3 years ago
6

Calculate the number of gold atoms in a sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and r

ound it to significant digits.
Chemistry
1 answer:
Fudgin [204]3 years ago
7 0

Answer: N = 2.78 × 10^23 atoms

There are N = 2.78 × 10^23 atoms in 70g of Au2cl6

Completed Question:

Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits

Explanation:

Given:

Molar mass of Au2cl6 = 303.33g/mol

Mass of Au2cl6 = 70g

Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol

According to the chemical formula of Au2cl6,

1 mole of Au2cl6 contains 2 moles of Au

Number of moles of Au = 2 × 0.231mol = 0.462mole

There are 6.022 × 10^23 atoms in 1 mole of an element.

Number of Atom of gold in 0.462 mole of gold is:

N = 0.462 mol × 6.022 × 10^23 atoms/mol

N = 2.78 × 10^23 atoms

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C because first comes the source or producer than the first eater then the second and then it decomposes
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How many atoms of Kr (Krypton) are in a balloon that contains 2.00 mol of Kr? (4)
Svetllana [295]

Answer:

atoms= 1.204x10^{24}atoms

Explanation:

Hello!

In this case, according to the Avogadro's number, it is possible to compute the atoms of Kr in 2.00 moles as shown below:

atoms=2.00mol*\frac{6.022x10^{23}atoms}{1mol} \\\\atoms= 1.204x10^{24}atoms

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3 years ago
A) An iron nail is attracted to a magnet.
11111nata11111 [884]

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3 years ago
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How far will an object move in 12 seconds if it is traveling at a velocity of 8 m/s?
xxMikexx [17]

Answer:

96 m/s.

Explanation:

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5 0
3 years ago
Determina el grado de pureza de un marmol (CaCO3), si al descomponerse 125 g del mismo se desprenden 20 litros de dióxido de car
jarptica [38.1K]

Answer:

67.8%

Explanation:

La reacción de descomposición del CaCO₃ es:

CaCO₃ → CO₂ + CaO

<em>Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.</em>

Usando la ley general de los gases, las moles de dioxido de carbono son:

PV = nRT.

<em>Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). </em>Reemplazando los valores en la ecuación:

PV / RT = n

1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles

Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.

La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:

0.846moles ₓ (100.087g / mol) = <em>84.7g de CaCO₃</em>

Así, la pureza del marmol es:

(84.7g de CaCO₃ / 125g) ₓ 100<em> = </em>

<h3>67.8%</h3>
7 0
3 years ago
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