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3 years ago

8

What does the second law of throdynamics state?

2 answers:

Anna71 [15]3 years ago

5 0

Explanation:

what does the second law of throdynamics state?

A. The amount of molecular disorder in the universe will always increase

B. The amount of molecular disorder in the universe will always decrease

C. The amount of molecular disorder in the universe will stay constant

Elena L [17]3 years ago

4 0

Answer:

<h2>A.</h2>

The second law of thermodynamics states that the total entropy of a system either increases or remains constant in any spontaneous process; it never decreases. ... This is because entropy increases for heat transfer of energy from hot to cold (Figure 12.9).

Explanation:

Hope It Helps

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Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction

lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Cu^{2+}/Cu]}=0.34V$

$E^o_{[Ag^{+}/Ag]}=0.80V$

$E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=0.80V-(0.34V)=0.46V$

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}$

where,

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

$E_{cell}$ = ?

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

$E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}$

$E_{cell}=0.434V$

Therefore, the cell potential for this cell 0.434 V

8 0

3 years ago

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

gayaneshka [121]

Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law, PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

$V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}$

Given;

$\frac{dP}{dt}$ (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) = 600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( $\frac{dv}{dt}$ );

(600 x 40) + (150 x $\frac{dv}{dt}$ ) = 0

$\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min$

Therefore, the volume is decreasing at 160 cm³/min

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3 years ago

A forward horizontal force of 3 lb is used to pull a 60 lb sled at constant velocity on a frozen pond. the coefficent of (kineti

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It’s either 0.05 or 20. Assuming that the coefficient friction is a damping factor, I feel like 0.05 would be correct m

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When a Lunar Module landed on the Moon, it used thrusters to slow its descent to the surface. When other spacecraft are returned

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3 years ago

A weather balloon is partially inflated with helium gas to a volume of 2.0 m³. The pressure was measured at 101 kPa and the temp

Yanka [14]

Answer:

4.7 m³

Explanation:

We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2

* Givens :

P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K

( We must always convert the temperature unit to Kelvin "K")

* What we want to find :

V2 = ?

* Solution :

101 × 2 / 300.15 = 40 × V2 / 283.15

V2 × 40 / 283.15 ≈ 0.67

V2 = 0.67 × 283.15 / 40

V2 ≈ 4.7 m³

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2 years ago