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artcher [175]
3 years ago
6

The initial temperature of three moles of oxygen gas is 32.5°C, and its pressure is 6.60 atm. (a) What will its final temperatur

e be when heated at constant volume so the pressure is three times its initial value? °C (b) Now the volume of the gas is also allowed to change. Determine the final temperature if the gas is heated until the pressure and the volume are "quadrupled".
Chemistry
1 answer:
Anastasy [175]3 years ago
8 0

Answer:

a. 916.95K is final temperature of the gas

b. 3667.8K

Explanation:

a. We can solve the temperature of a gas when the volume is expanded three times using Gay-Lussac's law:

P₁T₂ = P₂T₁

<em>Where P is pressure,</em>

<em>And T is absolute temperature of 1, initial states and 2, final states of the gas.</em>

<em />

Initial pressure is 6.60 atm, initial absolute temperature is 32.5°C + 273.15K = 305.65K. Final pressure is three times initial pressure = 6.60atm*3 = 19.8atm.

Solving for final temperature:

P₁T₂ = P₂T₁

6.60atm*T₂ = 19.8atm*305.65K

T₂ =

916.95K is final temperature of the gas

<em />

b.  Charle's law is the gas law that relates changes in volume and temperature of a gas:

V₁T₂ = V₂*T₁

If the volume is quadrupled: V₂/V₁ = 4:

T₂ = 4*916.95K

T₂ = 3667.8K

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When 1.045 g of CaO is added to 50.0 mL of water at 25.0 °C in a calorimeter, the temperature of the water increases to 32.3 °C.
Rus_ich [418]

Answer:

1.71 kJ/mol

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

\Delta H=m\times C\times \Delta T

Where,  

\Delta H  is the enthalpy change

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass of CaO = 1.045 g

Specific heat = 4.18 J/g°C

\Delta T=32.3-25.0\ ^0C=7.3\ ^0C

So,  

\Delta H=1.045\times 4.18\times 7.3\ J=31.88713\ J

Also, 1 J = 0.001 kJ

So,  

\Delta H=0.03189\ kJ

Also, Molar mass of CaO = 56.0774 g/mol

Moles=\frac{Mass}{Molar\ mass}=\frac{1.045}{56.0774}\ mol=0.01863\ mol

Thus, Enthalpy change in kJ/mol is:-

\Delta H=\frac{0.03189}{0.01863}\ kJ/mol=1.71\ kJ/mol

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3 years ago
A bottle sitting on a table has no forces at all acting upon it.
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Answer:

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Explanation:

5 0
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(a) if a sample containing 2.00 ml of nitroglycerin is detonated, how many total moles of gas are produced? (b) if each mole of
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Select the reason why two molecules of benzaldehyde cannot react via the aldol condensation.
Lesechka [4]
Benzaldehyde or C6H5CHO would not undergo the aldol condensation because it does not contain an alpha-hydrogen in its structure. Aldol condensation is a type of reaction that happens between an enolate and an aldehyde or ketone leading to a alkene that has a planar structure. The lack of an alpha-hydrogen would not allow for it to undergo such process since it cannot enolize. Benzaldehyde undergoes a nucleophilic reaction known as Claisen-Schmidt condensation. It has somehow same mechanism of the aldol reaction however, the nucleophilic attack on the carbonyl happens even without the alpha-hydrogen but with an enolate that is from a ketone.
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3 years ago
It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta
Zigmanuir [339]

Answer:

\lambda=241.9\ nm

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that  

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,  

1 mole = 6.023\times 10^{23}\ electrons

So,  

6.023\times 10^{23} electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of \frac {495.0}{6.023\times 10^{23}}\ kJ of energy.

Energy required = 82.18\times 10^{-23}\ kJ

Also,  

1 kJ = 1000 J

So,  

Energy required = 82.18\times 10^{-20}\ J

Also, E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

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\lambda=2.4188\times 10^{-7}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=241.9\ nm

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