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kodGreya [7K]
3 years ago
11

What is the density of lead (in g/cm^3 3 ) if a rectangular bar measuring 0.500 cm in height, 1.55 cm in width, and 25.00 cm in

length has a mass of 220.9 g? Report the answer with the units of g/cm^3 3 and the proper number of significant figures. Numbers in bold count towards significant figures. Type your answer here with correct significant figures and units:
Chemistry
1 answer:
Viktor [21]3 years ago
3 0

Answer:

Density = 11.4 g/cm³

Explanation:

Given data:

Density of lead = ?

Height of lead bar = 0.500 cm

Width of lead bar = 1.55 cm

Length of lead bar = 25.00 cm

Mass of lead bar = 220.9 g

Solution:

Density = mass/ volume

Volume of bar = length × width × height

Volume of bar = 25.00 cm × 1.55 cm × 0.500 cm

Volume of bar = 19.4 cm³

Density of bar:

Density = 220.9 g/ 19.4 cm³

Density = 11.4 g/cm³

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Bingel [31]

Answer:

8.73 L

Explanation:

First, you need to convert grams to moles using the molar mass.

Molar Mass (N₂): 2(14.009 g/mol)

Molar Mass (N₂): 28.018 g/mol

12.2 grams N₂               1 mole
----------------------  x  ------------------------  =  0.435 moles N₂
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To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (torr)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 torr*L/mol*K)

-----> T = temperature (K)

After converting the temperature from Celsius to Kelvin, you can plug the given values into the equation.

P = 1132 torr                                   R = 62.36 torr*L/mol*K

V = ? L                                            T = 91 °C + 273.15 = 364.15 K

n = 0.435 moles

PV = nRT

(1132 torr)V = (0.435 moles)(62.36 torr*L/mol*K)(364.15 K)

(1132 torr)V = 9888.015

V = 8.73 L

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2 years ago
The complete combustion (reaction with oxygen) of liquid octane (c8h18) a component typical of the hydrocarbons in gasoline, pro
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Answer:

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