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8_murik_8 [283]
3 years ago
15

How much energy in joules is released when 18.5 grams of copper cools from 285 °C down to 45 °C

Chemistry
1 answer:
nadya68 [22]3 years ago
5 0

Answer: 1709.4 Joules

Explanation:

The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since Q = ?

M = 18.5 grams

Recall that the specific heat capacity of copper C = 0.385 J/g.C

Φ = 285°C - 45°C = 240°C

Then, Q = MCΦ

Q = 18.5grams x 0.385 J/g.C x 240°C

Q = 1709.4 Joules

Thus, 1709.4 Joules is released when copper is cooled.

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The state of matter in which a material has a definite shape and a definite volume is the _____ state.
Scilla [17]

Answer:

solid

Explanation:

in the solid state the material will has a fixed shape and volume whatever the container that contains it

where in liquid the shape will be different depending on the container

and in gas state the shape and volume are not definite

3 0
2 years ago
How many moles are there in 3.9 grams of potassium
zimovet [89]
1.77 so A. I don’t need to explain this just helping out haha
7 0
2 years ago
Read 2 more answers
Façam 3 perguntas sobre os oxidos (quimica) (socorro)
nataly862011 [7]

Answer:

Pergunta 1: Como se forma um óxido?

Pergunta 2: O que acontece com uma moeda quando escurece?

Pergunta 3: Quais outros compostos podem ser formados a partir de óxidos?

4 0
3 years ago
A cylinder of O2 gas occupies a volume of 60.50 L at STP. How many moles of oxygen
Dahasolnce [82]

Answer:

The answer to your question is 2.52 moles

Explanation:

Data

Volume = 60.5 l

Temperature = 20°C

Pressure = 1 atm

Constant of ideal gases = R = 0.082 atm l/mol°K

Formula

PV = nRT

-Solve for n

   n = PV / RT

-Convert temperature to °K

Temperature = 20 + 273

                      = 293°K

-Substitution

   n = (1 x 60.5) / (0.082 x 293)

-Simplification

    n = 60.5 / 24.026

-Result

    n = 2.52 moles

3 0
3 years ago
The enthalpies of formation of the compounds in the combustion of methane, , are CH4 (g): Hf = –74.6 kJ/mol; CO2 (g): Hf = –393.
iris [78.8K]

Answer:

ΔH°c = - 1605.1 KJ

Explanation:

  • CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

∴ ΔHf CH4(g) = - 74.6 KJ/mol

∴ ΔHf CO2(g) = - 393.5bKJ/mol

∴ ΔHfH2O(g) = - 241.82 KJ/mol

standard enthalpy of combustion (ΔH°c):

⇒ ΔH°c = (2)(ΔHf H2O) + ΔHfCO2 - ΔHfCH4

⇒ ΔH°c = (2)(- 241.82) + ( - 393.5 ) - ( - 74.6 )

⇒ ΔH°c = - 802.54 KJ/mol

⇒ ΔH°c = ( - 802.54 KJ/mol )(  2 mol CH4 )

⇒ ΔH°c = - 1605.08 KJ

6 0
3 years ago
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