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Vinil7 [7]
3 years ago
13

A 5.00-g sample of copper metal at 25.0 °C is heated by the addition of 133 J of energy. The final temperature of the copper is

________°C. The specific heat capacity of copper is 0.38 J/g°C.
Chemistry
1 answer:
vekshin13 years ago
3 0

<u>Answer:</u> The final temperature of the copper is 95°C.

<u>Explanation:</u>

To calculate the final temperature for the given amount of heat absorbed, we use the equation:

Q= m\times c\times \Delta T

Q = heat absorbed  = +133 J (heat is added to the system)

m = mass of copper = 5.00 g

c = specific heat capacity of copper = 0.38 J/g ° C      

\Delta T={\text{Change in temperature}}=T_2-T_1

T_1=25^oC

Putting values in above equation, we get:

+133J=5.00g\times 0.38J/g^oC\times (T_2-25)\\\\T_2=95^oC

Hence, the final temperature of the copper is 95°C.

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Calculate the root mean square velocity of F2,Cl2, and Br2 at 304 K .
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) Consider the starting materials and reagents. What do you expect to happen at the beginning of this reaction (Boxes 1-3 on the
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Check the explanation

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3 0
3 years ago
URGENT!! Chemistry gurus-- I summon thee!
Elina [12.6K]

1. Molarity : 0.25 M

2. mol CH₄ = 7.4 moles

mol O₂ = 14.8 moles

<h3>Further explanation</h3>

1.

Given

83.2 g CuCl2 in 2.5 liters of water

Required

the molarity

Solution

Molarity : mol solute per liter of solution(not per liter of solvent)

  • mol solute

mol solute = mol CuCl₂

mol CuCl₂ = mass : MW CuCl₂

mol CuCl₂ = 83.2 : 134.45

mol CuCl₂ = 0.619

  • Molarity

Molarity(M) = mol : V

Assume density CuCl₂ = 3.39 g/cm³

volume CuCl₂ = 8.32 g : 3.39 g/cm³ = 2.45 cm³=2.45 x 10⁻³ L

With this small volume value of CuCl₂, the volume of the solute is sometimes neglected in calculating molarity

volume of solution = 2.5 L + 2.45 x 10⁻³ L = 2.50245 L

Molarity(M) = mol : V

M = 0.619 : 2.50245 L = 0.247≈0.25

2.

Given

Reaction

The correct balanced reaction:

CH4 + 2O2 → CO2 + 2H2O

7.4 moles CO2

Required

moles of methane (CH4) and oxygen gas (O2)

Solution

From the equation, mol ratio of CO₂ : CH₄ : O₂ = 1 : 1 : 2

mol CH₄ = mol CO₂ = 7.4 moles

mol O₂ = 2 x mol CO₂ = 2 x 7.4 moles = 14.8 moles

3 0
3 years ago
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