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1 year ago

Predict the sign of ΔSsys for each process: (b) Hot air expands.

Chemistry

1 answer:

mars1129 [50]1 year ago

8 0

The sign of ΔSsys for hot air expands is positive.

Entropy (S) is the degree of randomness or disorderliness in a thermodynamic system. The randomness increases when the motion of molecules increases due to the addition of heat. It also increases when the distance between the molecules increases.

ΔS is a measurement of the change in entropy in a thermodynamic system. It is positive when entropy increases and negative when it decreases.

When the hot air expands the distance between the molecules of the gas increases and thus the randomness also increases. Entropy increases and the of ΔS becomes positive.

If you need to learn more about ΔSsys signs, click here

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C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati

hammer [34]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. Therefore, the given reaction is exothermic.

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. Therefore, the given combustion reaction is an oxidation-reduction reaction.

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.

5 0

2 years ago

Explain the process of how a star starts as a cloud of dust and gas that then forms into a star.

Makovka662 [10]

Answer:

Stars form from an accumulation of gas and dust, which collapses due to gravity and starts to form stars. The process of star formation takes around a million years from the time the initial gas cloud starts to collapse until the star is created and shines like the Sun.

7 0

2 years ago

Read 2 more answers

Given the following information: Mass of proton = 1.00728 amu Mass of neutron = 1.00866 amu Mass of electron = 5.486 × 10^-4 amu

lesya692 [45]

Answer: The nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

Explanation:

For the given element $_3^6\textrm{Li}$

Number of protons = 3

Number of neutrons = (6 - 3) = 3

We are given:

$m_p=1.00728amu\\m_n=1.00866amu\\A=6.015126amu$

M = mass of nucleus = $(n_p\times m_p)+(n_n\times m_n)$

$M=[(3\times 1.00728)+(3\times 1.00866)]=6.04782amu$

Calculating mass defect of the nucleus:

$\Delta m=M-A\\\Delta m=[6.04782-6.015126)]=0.032694amu=0.032694g/mol$

Converting this quantity into kg/mol, we use the conversion factor:

1 kg = 1000 g

So, $0.032694g/mol=0.032694\times 10^{-3}kg/mol$

To calculate the nuclear binding energy, we use Einstein equation, which is:

$E=\Delta mc^2$

where,

E = Nuclear binding energy = ? J/mol

$\Delta m$ = Mass defect = $0.032694\times 10^{-3}kg/mol$

c = Speed of light = $2.9979\times 10^8m/s$

Putting values in above equation, we get:

$E=0.032694\times 10^{-3}kg/mol\times (2.9979\times 10^8m/s)^2\\\\E=2.938\times 10^{12}J/mol$

Hence, the nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

3 0

3 years ago

Given the balanced equation representing a reaction:

Hunter-Best [27]

The answer is (4) a substitution reaction. Notice that propane lost an H and gained a Br. This constitutes substitution.

8 0

3 years ago

How many liters of oxygen are required to completely react with 2.0 liters of CH4 at30 °C and 3.0 atm?CH4(g) + 2O2(g) → CO2(g) +

Dominik [7]

1) Write the chemical equation.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

2) List the known and unknown quantities.

Sample: CH4.

Volume: 2.0 L.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Moles: unknown.

3) Moles of CH4.

3.1- Set the equation.

$PV=nRT$

3.2- Plug in the known values and solve for n (moles).

$(3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$ $n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=$ $n=0.24\text{ }mol\text{ }CH_4$

4) Moles of oxygen that reacted.

The molar ratio between CH4 and O2 is 1 mol CH4: 2 mol O2.

$mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2$

5) Volume of oxygen required.

Sample: O2.

Moles: 0.48 mol.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Volume: unknown.

5.1- Set the equation.

$PV=nRT$

5.2- Plug in the known values and solve for V (liters).

$(3.0\text{ }atm)(V)=0.48\text{ }O_2*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$ $V=\frac{(0.48\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(303.15\text{ }K)}{3.0\text{ }atm}$ $V=3.98\text{ }L$

3.98 L of O2 is required to react with 2.0 L CH4.

6 0

1 year ago