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3 years ago

What is the change in entropy of the lead when 2.0 kg of molten lead at its melting point temperature solidifies.

1 answer:

7 0

Answer: The change in entropy of the lead when 2.0 kg of molten lead at its melting point temperature solidifies is 20 cal/K.

Explanation:

$Pb(l)\rightleftharpoons Pb(s)$

$\Delta S=\frac{q}{T}$

${\text{Change in entropy}}=\frac{heat}{Temperature}$

T is the temperature at which the process is carried out, here the temperature is constant, 327.5°C = $(273+327.5)K=600.5K$ ,because only phase change occurs.

q = heat given in the process = $m\times L_f$

m = mass of lead = 2.0 kg

$L_f$ = latent heat of fusion of lead = 5.9 kcal/kg

$\Delta S=\frac{2kg\times 5.9kcal/kg}{600.5K}$

$\Delta S=0.02 kcal/K=20cal/K$

Thus the change in entropy of the lead when 2.0 kg of molten lead at its melting point temperature solidifies is 20 cal/K.

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If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$ = $10^{-2.88}$ = 0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

  CH3COOH H+ CH3COOH
Initial   $moldm^{-3}$      $x$ 0 0

Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

8 0

3 years ago

Answer the ones please get right

Natali [406]

Answer:

Hope this helps!

Explanation:

I would help you, but I won't. Because you take others people points without helping them and but random words. So, since you are doing that I won't be helping you <3. Thanks for the points idiot.

6 0

3 years ago

A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached

lara [203]

Answer:

a) cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

b) 0.068 V.

Explanation:

A) Cu2+ + 2e- euilibrium cu (s)

Hg2Cl2 + 2e- equilibrium 2Hg (l) + 1cl-

Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

B) To calculate the cell voltage

E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+

putting values we get

= 0.339V + (90.05916V/2)log(0.100) = 0.309V

E_cell = E Cu2+/Cu - E SCE = 0.309 V - 0.241 V = 0.068V.

6 0

3 years ago

What is the density of an object that has a mass of 1.08 kg and displaces 50.50 cm3 of water? (Please report your answer in g/cm

masha68 [24]

Answer:

Explanation:

1.08kg/50.5 cm^3(1000g/kg)=21.386 g/cm^3

3 0

2 years ago

Read 2 more answers

What molality of a nonvolatile, nonelectrolyte solute is needed to raise the boiling point of water by 7.10°c (kb = 0.520°c/m)?

faltersainse [42]

According to boiling point elevation equation:
Δ T = i Kb m
when ΔT (change in boiling point) = 7.10 C°
and i (van't Hoff factor)= 1
and Kb = 0.520
so, by substitution:
7.10 = 1*0.520 *m
m = 7.1 / 0.52 = 13.65 m

7 0

3 years ago