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1 year ago

5

the enthalpy of vaporization of an roganic alchol is 35.3 kJ/mol at the boiling pointof 64.2 C calculate the entropy change for

this alchol going forma liquid to vapor

1 answer:

bonufazy [111]1 year ago

3 0

Answer:

Explanation:

s=h/t

=35.3/64.2(c)

=35.3/337.2(k)

=0.104

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Calculate the value of E°cell for the following reaction:2Au(s) + 3Ca2+(aq) → 2Au3+(aq) + 3Ca(s)Au3+(aq) + 3e- → Au(s) E° = 1.50

uranmaximum [27]

<u>Answer:</u> The standard electrode potential of the cell is -4.37 V

<u>Explanation:</u>

For the given cell reaction:

$2Au(s)+3Ca^{2+}(aq.)\rightarrow 2Au^{3+}(aq.)+3Ca(s)$

The half reactions follows:

<u>Oxidation half reaction:</u> $Au(s)\rightarrow Au^{3+}(aq.)+3e^-;E^o_{Au^{3+}/Au}=1.50V$ ( × 2 )

<u>Reduction half reaction:</u> $Ca^{2+}(aq.)+2e^-\rightarrow Ca(s);E^o_{Ca^{2+}/Ca}=-2.87V$ ( × 3 )

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-2.87-(1.50)=-4.37V$

Hence, the standard electrode potential of the cell is -4.37 V

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3 years ago

Consider the reaction of C3H8 with O2 to form CO2 and H2O. If 5.11 g O2 is reacted with excess C3H8 and 3.35 g of CO2 is ultimat

JulsSmile [24]

Answer : The percent yield of the reaction is, 79.8 %

Explanation : Given,

Mass of $O_2$ = 5.11 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $O_2$ .

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{5.11g}{32g/mole}=0.159mole$

Now we have to calculate the moles of $CO_2$ .

The balanced chemical reaction will be,

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From the balanced reaction, we conclude that

As, 5 moles of $O_2$ react to give 3 moles of $CO_2$

So, 0.159 moles of $O_2$ react to give $\frac{3}{5}\times 0.159=0.0954$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(0.0954mole)\times (44g/mole)=4.1976g$

The theoretical yield of $CO_2$ = 4.1976 g

The actual yield of $CO_2$ = 3.35 g

Now we have to calculate the percent yield of $CO_2$

$\%\text{ yield of }CO_2=\frac{\text{Actual yield of }CO_2}{\text{Theoretical yield of }CO_2}\times 100=\frac{3.35g}{4.1976g}\times 100=79.8\%$

Therefore, the percent yield of the reaction is, 79.8 %

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3 years ago

An exothermic reaction has A) a negative ΔH, absorbs heat from the surroundings, and feels cold to the touch. B) a positive ΔH,

Strike441 [17]

Answer:

Option C: a negative ΔH, gives off heat to the surroundings, and feels warm to the touch

Explanation:

Step 1:

An exothermic reaction = a reaction that gives off heat to the surroundings.

There is a temperature raise.

<em>A) a negative ΔH, absorbs heat from the surroundings, and feels cold to the touch.</em>

This is false. since there is a temperature raise, it doesn't feel cold to the touch, and gives off heat to the surroundings.

<em>B) a positive ΔH, gives off heat to the surroundings, and feels warm to the touch.</em>

This is false. An exothermic reaction has a negative ΔH.

<em>C) a negative ΔH, gives off heat to the surroundings, and feels warm to the touch. </em>This is correct

<em>D) a positive ΔH, absorbs heat from the surroundings, and feels cold to the touch</em>.

This is false. An exothermic reaction doesn't absorb heat from the surroundings, it doesn't feel cold to the touch.

<em>E) a positive ΔH, absorbs heat from the surroundings, and feels warm to the touch.</em>

This is false. An exothermic reaction doesn't have a positive ΔH.

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3 years ago

Examples of laboratory accidents

Anika [276]

Answer:

Explanation:

1. Spilling off a chemical on you(

2. High voltages can harm you

3. biohazards including infective organisms and their toxins

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2 years ago

An element has a half-life of 5.0 days. How many days would

Olin [163]

Answer:

15 days

Explanation:

Given half-life = 5 days => k = 0.693/half-life = 0.693/5 days⁻¹

A = A₀e^-k·t => t = ln(A/A₀) / -k = ln(1/8) / -0.1386 days⁻¹ = 15 days

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2 years ago