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Nina [5.8K]
3 years ago
5

Which of the following displays the correct change in enthalpy and best describes the reaction below? 2CsCl(aq) + Na2SO4(aq) 2Na

Cl(aq) + Cs2SO4(aq) Given: Cs2SO4: ∆H= -1400 kJ CsCl: ∆H= -415 kJ Na2SO4: ∆H= -1380 kJ NaCl: ∆H = -411 kJ
https://gyazo.com/bb924c15fe3b2c341d56dcb96b91658c
Chemistry
1 answer:
Bond [772]3 years ago
3 0
We subtract the enthalpies of the reactants from that of the products:
2\Delta
 H(NaCl)+\Delta H(Cs_2 SO_4)-2\Delta H(CsCl)-\Delta H(Na_2 SO_4) \\ 
=2(-411)+(-1400) -2(-415)-(-1380) \\ = -12 kJ
Since this is < 0, this is an exothermic reaction.

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in a typical person, the level of glucose is about 85 mg/ 100 mL of blood. if an average body contains about 11 pt of blood, how
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Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.

We make a proportion out of the word problem

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One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
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Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

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Can you be more specific

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