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3 years ago

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Earthquake damage causes two rabbits to be separated from the rest of the rabbits in their large habitat. They have no way to ge

t back to their original habitat. The two rabbits mate with each other. Over time, all the offspring in the new habitat are descendants of the original two rabbits. What are the outcomes of this situation?

1 answer:

Sati [7]3 years ago

5 0

The animals have to adapt to their new environment meant and continue mating so they do not go extinct

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In a sample containing a mixture of only these gases at exactly one atmosphere pressure, the partial pressures of carbon dioxide

Black_prince [1.1K]

Answer:

Explanation:

The pressure of a gaseous mixture is equal to the sum of the partial pressures of the individual gases:

Σ $P_g_a_s$ $= P_1+P_2+P_3+...+P_n$

The prompt is trying to confuse you, but it actually tells us the pressure of the mixture to be 1 atm, but this can be converted to torr. Furthermore, we are informed only three gases are in the mixture: diatomic nitrogen, diatomic oxygen, and carbon dioxide:

$P_g_a_s=1 \ atm = 760 \ torr= P_N_2+P_O_2+P_C_O_2\\760 \ torr = 582.008 \ torr + P_O_2 \ + 0.285 \ torr$

Solve for Po2:

$P_o_2=(760-582.008-0.285) \ torr = 177.707 \ torr$

Thus, the partial pressure of diatomic oxygen is 177.707 torr.

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3 years ago

The volume of a gas-filled balloon is 30.0 L at 313 K and 1520 mm Hg pressure. What would the volume be at standard temperature

pashok25 [27]

313 k
step-by-step-explanation

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3 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago

Calculate the value deltaG°

atroni [7]

Answer:

ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.

Explanation:

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3 years ago

Plsss help!!! ASAP!!!

Verizon [17]

The electronic configuration is for iron (Fe) because if you add all those power up it will give you 26 and it’s the atomic number of Fe

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3 years ago