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Mariulka [41]
2 years ago
15

when solid pellets of sodium hydroxide (naoh) dissolve in water, the temperature of the water can rise dramatically. taking naoh

as the system, what can you deduce about the signs of the entropy change of the system (δssys) and surroundings (δssurr) from this?
Chemistry
1 answer:
irga5000 [103]2 years ago
8 0

ΔSsys  and ΔSsurr both have values larger than 0.

<h3>Entropy Change: What Is It?</h3>
  • Entropy change is a phenomena that measures the evolution of randomness or disorder in a thermodynamic system.
  • It has to do with how heat or enthalpy is converted during work.
  • More unpredictability in a thermodynamic system indicates high entropy.
  • Heat transport (delta Q) divided by temperature equals the change in entropy (delta S).
<h3>What causes variations in entropy?</h3>
  • When a substance is divided into several pieces, entropy rises.
  • Because the solute particles are split apart when a solution is generated, the dissolving process increases entropy.
  • As the temperature rises, entropy increases.

learn more about entropy change here

brainly.com/question/6364271

#SPJ4

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The enthalpy of combustion for octane (C8H18(l)), a key component of gasoline, is -5,074 kJ/mol. This value is the Delta. Hrxn f
Inessa [10]

Combustion can be defined as the reaction of a compound with oxygen. The enthalpy of combustion of octane is \Delta H_{\rm rxn} for \rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O.

<h3>What is the enthalpy of reaction?</h3>

The enthalpy of reaction is the amount of heat energy absorbed or lost by the molecules in the chemical reaction.

The enthalpy of combustion is the amount of heat energy released by the compound in the reaction with oxygen.

The reaction in which heat is liberated with the reaction of a compound with oxygen has an enthalpy of combustion, equivalent to the enthalpy of reaction.

The combustion of octane can be given as:

\rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O

Thus, the reaction has combustion energy equivalent to the enthalpy of the reaction is \rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O. Thus, option B is correct.

Learn more about enthalpy of reaction, here:

brainly.com/question/1657608

6 0
3 years ago
A solution in a dish contains 4.0 grams of salt dissolved in 100 grams of water. If 50 grams of the water evaporates, this is ev
MaRussiya [10]
D. A mixture

If the water is evaporating while the salt remains, it means the two are not chemically bonded and therefore are not a compound.
6 0
3 years ago
Which statements accurately describe cells? Check all that apply.
Wewaii [24]

Answer: 1, 3, and 4

Explanation: i just did it

3 0
3 years ago
Read 2 more answers
What is the frequency of a radio wave with a a wavelength of 3 m? (Hint: MHz
Ostrovityanka [42]

Answer:
a. 1 x 10^8

Explanation:

100 MHz = 100,000,000 Hz = 10^8 Hz

4 0
2 years ago
The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
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