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77julia77 [94]
3 years ago
8

List some non-examples of Convection​

Chemistry
1 answer:
dimulka [17.4K]3 years ago
7 0

Non Example: It is not convection because heat is not transferred through a fluid (gas or liquid).

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In the laboratory a student determines the specific heat of a metal. He heats 19.5 grams of copper to 98.27 °C and then drops it
siniylev [52]

Answer:

The specific heat of copper is 0.37 J/g°C

Explanation:

<u>Step 1: </u>Data given

Mass of copper = 19.5 grams

Initial temperature of copper = 98.27 °C

Mass of water = 76.3 grams

Initial temperature of water = 24.05 °C

Final temperature of water and copper = 25.69 °C

<u>Step 2:</u> Calculate specific heat of copper

Qgained = -Qlost

Q = m*c*ΔT

Qwater = -Qcopper

m(water) * c(water) * ΔT(water) = - m(copper) * c(copper) *ΔT(copper)

⇒ with m(water) = 76.3 grams

⇒ with c(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2-T1 = 25.69 - 24.05 = 1.64

⇒ with m(copper) = 19.5 grams

⇒ with c(copper) = TO BE DETERMINED

⇒ with ΔT(copper) = T2-T1 = 25.69 - 98.27 = -72.58

76.3 * 4.184 * 1.64 = - 19.5 * c(copper) * -72.58

523.552 = 1415.31 * c(copper)

c(copper) = 0.37 J/g°C

The specific heat of copper is 0.37 J/g°C

3 0
4 years ago
Most chloride salts are soluble. Which of the following is an exception to this generalization?
siniylev [52]

The exception to the rule concerning the solubility of chlorides in water is PbCl2.

The solubility rules give us an idea of which substances are soluble in water and what substances are not soluble in water. According to the solubility rules, chlorides are soluble in water.

However, chlorides of lead are not soluble in water hence, the exception to the rule is PbCl2.

Learn more: brainly.com/question/6505878

7 0
3 years ago
What is the layer of the sun from which energy escapes into space?
Jet001 [13]
The core is the layer of the sun from which energy escapes into space.
8 0
3 years ago
Phosphoglucoisomerase interconverts glucose 6‑phosphate to, and from, glucose 1‑phosphate.
lawyer [7]

Answer:

Keq = 0.053

7.3 kJ/mol

Explanation:

Let's consider the following isomerization reaction.

glucose 6‑phosphate ⇄ glucose 1 - phosphate

The concentrations at equilibrium are:

[G6P] = 0.19 M

[G1P] = 0.01 M

The concentration equilibrium constant (Keq) is:

Keq = [G1P] / [G6P]

Keq = 0.01 / 0.19

Keq = 0.053

We can find the standard free energy change, ΔG°, of the reaction mixture using the following expression.

ΔG° = -R × T × lnKeq

ΔG° = -8.314 J/mol.K × 298 K × ln0.053

ΔG° = 7.3 × 10³ J/mol = 7.3 kJ/mol

7 0
4 years ago
Which element has the larger Zell
Eddi Din [679]

Answer:

a

Explanation:

because of potassium and lithium belongs to

alkali metals

6 0
3 years ago
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