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irinina [24]
3 years ago
14

Si en un laboratorio hay 300 moles de plomo y 300 moles de plata en cual de los dos casos la masa en mayor?

Chemistry
1 answer:
docker41 [41]3 years ago
4 0

Answer:

Esto significa que la masa en las moles de plomo es mayor que en las de plata.

Explanation:

Las moles de los átomos representan la cantidad de átomos presentes. Para saber la masa de los átomos debemos convertir las moles a gramos usando el peso molecular de cada uno de los átomos (207.2g/mol para el plomo y 107.82g/mol para la plata).

La masa de plomo en 300 moles es:

300 moles * (207.2g / mol) = 62160g de plomo

Y la masa de plata es:

300 moles * (107.82g / mol) = 32346g de plata

<h3>Esto significa que la masa en las moles de plomo es mayor que en las de plata</h3>
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Calculate the concentration in % (m/m) of a solution containing 30.0g of mgcl2 dissolved in 270.0g of h20
qaws [65]
As the question tells you, you need to use the formula

% mass= mass of solute/ mass of solution x 100

mass solute= 30.0 g
mass of solution= 30.0 + 270.0= 300.0 g

% mass= 30.0/ 300.0 x 100= 10%

answer is B
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3 years ago
Calculate the molarity of a solution that contains 3.00 grams Na2SO4 in 25 mL of solution.
Naddik [55]

Answer:

M Na2SO4 sln = 0.8448 M

Explanation:

  • molarity (M) [=] mol/L

∴ mass Na2SO4 = 3.00 g

∴ volume soln = 25 mL = 0.025 L

∴ molar mass Na2SO4 = 142.04 g/mol

⇒ mol Na2SO4 = (3.00 g)*(mol/142.04 g) = 0.02112 mol

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What statement best describes the compresibility of a gas?​
Misha Larkins [42]

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The majority of the elements on the periodic table are categorized as
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What volume is occupied by 3.760 g of oxygen gas at a pressure of 88.4
Lorico [155]

Answer:

3.2 L

Explanation:

Given data:

Mass of oxygen = 3.760 g

Pressure of gas = 88.4 Kpa (88.4×1000 = 88400 Nm⁻²)

Temperature = 19°C (19+273.15 = 292.15 K)

R = 8.314 Nm K⁻¹ mol⁻¹

Volume occupied = ?

Solution:

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 3.760 g/ 32 g/mol

Number of moles = 0.12 mol

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant

T = temperature in kelvin

V = nRT/P

V = 0.12 mol ×  8.314 Nm K⁻¹ mol⁻¹ × 292.15 K /88400 Nm⁻²

V = 291.472 Nm /88400 Nm⁻²

V = 0.0032 m³

m³ to L:

V = 0.0032×1000 = 3.2 L

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3 years ago
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