What statement most accurately classifies this cell?

The production of heat by metabolic processes takes place throughout the volume of an animal, but loss of heat takes place only

Sever21 [200]

To solve this problem we will apply the concepts related to the change in length in proportion to the area and volume. We will define the states of the lengths in their final and initial state and later with the given relationship, we will extrapolate these measures to the area and volume

The initial measures,

$\text{Initial Length} = L$

$\text{Initial surface Area} = 6L^2$ (Surface of a Cube)

$\text{Initial Volume} = L^3$

The final measures

$\text{Final Length} = L_f$

$\text{Final surface area} = 6L_f^2$

$\text{Final Volume} = L_f^3$

Given,

$\frac{(SA)_f}{(SA)_i} = 2$

Now applying the same relation we have that

$(\frac{L_f}{L_i})^2 = 2$

$\frac{L_f}{L_i} = \sqrt{2}$

The relation with volume would be

$\frac{(Volume)_f}{(Volume)_i} = (\frac{L_f}{L_i})^3$

$\frac{(Volume)_f}{(Volume)_i} = (\sqrt{2})^3$

$\frac{(Volume)_f}{(Volume)_i} = (2\sqrt{2})$

$\frac{(Volume)_f}{(Volume)_i} = 2.83$

Volume of the cube change by a factor of 2.83

6 0

4 years ago

An ideal gas is brought through an isothermal compression process. The 3.00 mol 3.00 mol of gas goes from an initial volume of 2

ozzi

Answer:

The answers can be found by considering the isothermal expansion equation as well as the ideal gas equation from where we have

The temperature T = 602.64K and the final pressure P = 110.24MPa

Explanation:

Numbeer of moles of gas = 3.00 mol

initial volume = 230.8×10−6 m3

final volume = 133.4×10−6 m3 .

released energy = 8240 J

Temperature = Constant = T

Pressure = $p_{f}$ =unknown

From the relation the combined ideal gas law, PV = nRT

Where R = 8.314 4621.JK−1mol−1

we have The release energy from compression P1V1 -P2V2

-qrev = -nRTln( $\frac{V_{2} }{V_{1} }$ ) = 8240J

n = 3

Hence -nRTln( $\frac{V_{2} }{V_{1} }$ ) = 3×8.314 462×ln $(\frac{133.4}{230.8})$ × T= -8240 J

or -13.67×T = -8240J, thus T = -8240/-13.67 = 602.64K

The Final pressure is given by

PV = n×R×T from where we have V = final volume thus

P = (n×R×T)/V = (3×8.134×602.64)÷(133.4× $10^{-6}$ ) = 110237041.1 N/ $m^{2}$ = 110.237MPa

8 0

3 years ago

Do only number 3 and thank

IrinaK [193]

Charge = 0.2 Ah is the correct answer...

3 0

3 years ago

Accorrding to Kepler's third law, a planet whose distance from the Sun is 2 A.U. would have an orbital period of how many Earth-

Lilit [14]

Answer:

hi

Explanation:

4 0

3 years ago

gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an

Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0

3 years ago