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GaryK [48]
3 years ago
8

What is the final temperature of a 93.9 g block of copper (whose specific heat is .385 J/g0C) that starts at a temperature of 45

.0 degrees Celsius and absorbs 1,450.3 J of energy?
Chemistry
1 answer:
Digiron [165]3 years ago
8 0

Answer:

7.01 hope this is right !

Explanation:

This problem can be summarized thusly:

qlost by copper = qgained by water + qgained by calorimeter

2) Therefore:

(610. g) (95.3 °C − x) (0.387 J g¯1 K¯1) = (45.0 g) (x − 36.5 °C) (4.184 J g¯1 °C¯1) + [(10.0 J/K) (x − 36.5 °C)]

Comment: The K and the °C cancel because the °C in this problem is a temperature difference (not one single specific value) and the "size" of one K = one °C.

22497.471 − 236.07x = 198.28x − 7237.22

424.35x = 29734.691

x = 70.1 °C

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Which of he following compounds is always part of an aqueous solution
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Suppose that Daniel has a 3.00 3.00 L bottle that contains a mixture of O 2 O2 , N 2 N2 , and CO 2 CO2 under a total pressure of
Alenkinab [10]

Answer:

Partial pressure O₂ → 2.74 atm

Explanation:

Let's analyse the data given:

Volume → 3L

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Total pressure is 4.80 atm

Let's apply the Ideal Gases Law to determine the total moles of the mixture

P . V = n .  R. T

4.80 atm . 3L = n . 0.082 . 273K

n = 4.80 atm . 3L / 0.082 . 273K → 0.643 moles

We apply the concept of mole fraction:

Mole fraction of a gas X = moles of gas X / Total moles

Mole fraction of a gas X = Partial pressure X / Total pressure

In a mixture, sum of mole fraction of each gas = 1

We determine mole fraction of N₂ → 0.230 / 0.643 = 0.357

We determine mole fraction of CO₂ → 0.350 atm / 4.80 atm = 0.0729

1 - mole fraction N₂ - mole fraction CO₂ = mole fraction O₂

1 - 0.357 - 0.0729 = 0.5701 → mole fraction O₂

We replace in the formula: Mole fraction O₂ = Partial pressure O₂ / 4.80 atm

0.5701 . 4.80 atm = Partial pressure O₂ → 2.74 atm

5 0
3 years ago
Read 2 more answers
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