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3 years ago

14

How many joules of heat is lost when a 64 g piece of copper cools from 375 oC to 26 oC? The specific heat of copper is 0.38 J/go

C.

1 answer:

Svetlanka [38]3 years ago

7 0

Answer: 8500 J lost or -8500 J

Explanation:

q=cmt

t=375-26

t=349

q=0.38(64)(349)

q=8487.68 J

Answer must have 2 sig figs, which means it rounds to 8500 J.

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4 0

1 year ago

1. A light bulb is filled with 3.54 g of argon at 25 C and 725 mmHg. What is

Ugo [173]

Answer: The volume of given gas is 2.27 L.

Explanation:

Given: Mass = 3.54 g

Temperature = $25^{o}C$ = (25 + 273) K = 298 K

Pressure = 725 mm Hg (1 mm Hg = 0.00131579) = 0.95 atm

As moles is the mass of substance divided by its molar mass.

So, moles of argon (molar mass = 40 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{3.54 g}{40 g/mol}\\= 0.0885 mol$

Formula used to calculate the volume of given gas is as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\0.95 atm \times V = 0.0885 mol \times 0.0821 L atm/mol K \times 298 K\\V = 2.27 L$

Thus, we can conclude that the volume of given gas is 2.27 L.

3 0

3 years ago

At a particular temperature, Kp 0.25 for the reaction a. A ask containing only N2O4 at an initial pressure of 4.5 atm is allowed

Alex

Answer:

a. pNO₂ = 1 atm pN₂O₄ = 4 atm

b. pNO₂ = 1 atm pN₂O₄ = 4 atm

c. It does not matter.

Explanation:

From the information given in this question we know the equilibrium involved is

N₂O₄ (g) ⇄ 2 NO₂ (g)

with Kp given by

Kp = p NO₂²/ p N₂O₄ = 0.25

We know that if we place 4.5 atm of N₂O₄ is placed in a flask, a quantity x is going to be consumed producing 2x atm of NO₂ and we can setup the following equation:

0.25 = p NO₂²/ p N₂O₄ = (2x)² / (4.5 - x)

0.25 x (4.5 - x) = 4x²

4x² + 0.25 x - 1.125 = 0

after solving this quadratic equation, we get two roots

x₁ = 0.5

x₂ = -0.56

the second root is physically impossible, and the partial pressures for x₁ = 0.5 will be

pNO₂ = 2 x 0.5 atm = 1.0 atm

pN₂O₄ = (4.5 - 0.5) atm = 4.0 atm

Similarly for part b, we get the equilibrium equation

0.25 = (9- 2x)² / x

0.25x = 81 - 36x + 4x²

the roots of this equation are:

x₁ = 5.0625

x₂ = 4

the first root is physically impossible since it will give us a negative partial pressure of N₂O₄ :

p N₂O₄ = 9 - 2(5.0625) = -1.13

the second root give us the following partial pressures:

p N₂O₄ = (9 - 2x4) atm = 1 atm

p NO₂ = 4 atm

The partial pressures are the same, it does not matter from which direction an equilibrium position is reached since what is essential is that the partial pressures of the gasses N₂O₄ and NO₂ obey the equilibrium equation.

8 0

2 years ago