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Liula [17]
3 years ago
6

Thermodynamic PropertiesProperty Value

Chemistry
1 answer:
igomit [66]3 years ago
8 0

Answer:

1,620 J.

Explanation:

  • The amount of heat added to a substance (Q) can be calculated from the relation:

<em>Q = m.c.ΔT.</em>

where, Q is the amount of heat released from ethanol cooling,

m is the mass of ethanol (m = 60.0 g),

c is the specific heat of ethanol in the liquid phase, since the T is cooled below the boiling point and above the melting point (c = 1.0 J/g °C),

ΔT is the temperature difference (final T - initial T) (ΔT = 43.0 °C – 70.0 °C = - 27.0 °C).

<em>∴ Q = m.c.ΔT</em> = (60.0 g)(1.0 J/g °C)(- 27.0 °C) = - 1620 J.

<em>The system releases 1620 J.</em>

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Each nucleus of a cell contains:
Jobisdone [24]

Hi there!

<u>Answer</u>:

Genes and chromosomes

Explanation:

All cells contain Genetic information (or Dna) which is organised into structures called chromosomes.

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5 0
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Hydrogen sulfide,H2S, is a very toxic gas with a smell of rotten eggs. using the following: H2S+3/2 O2=SO2+H2O H2+1/2O2=H2O S+O2
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Answer:

ΔH = -20kJ

Explanation:

The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

H₂(g) + S(g) → H₂S(g)

Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>

<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>

<em />

I hope it helps!

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3 years ago
What property do all of the group 18 elements have that make them stand out from other elements?
matrenka [14]
They are very stable (not reactive).

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In defining work, we awareness at the outcomes that the system (e.g. an engine) has on its surroundings. for this reason we outline work as being fine when the gadget does work at the environment (strength leaves the machine). If work is completed at the gadget (energy added to the machine), the work is bad.

The classical sign convention states that warmth switch into a device and work produced with the aid of it are tremendous, while heat transfer faraway from a device and work produced on it is negative.

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