Which three are advantages of asexual reproduction?A:Offspring are more likely to survive environmental changes.

B:Some offspring are more likely to survive a disease.

C:Less energy is required to reproduce.

D:The population can increase from only one parent.

E:The population can increase quickly.

Answers

Answer 1

Answer: C, D, and E

A and B cannot be true because asexual reproduction means the parent organism is essentially creating clones of itself, providing no variation in DNA and making all offspring vulnerable to the same environmental changes and diseases as the parent.

Answer 2

Answer:

Answer:

sorry if I get this wrong I think it is C

Explanation:

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A chemist designs a galvanic cell that uses these two half-reactions:O2 (g) + 4H+(aq) + 4e− → 2H2O (l) Eo =+1.23VZn+2 (aq) + 2e− → Zn(s) Eo=−0.763VAnswer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode. Write a balanced equation for the half-reaction that happens at the anode. Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions
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PLEASE HELP ASAP‼️ Which of the following elements has the highest metallic characterA. Zt
B. Rh
C. Hf
D. Co

Answers

Answer: C

Explanation:

The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen, if the rate of the formation of carbon monoxide is 0.35 M/s ? g

Answers

Answer : The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-(1)/(a)(d[A])/(dt)$

$\text{Rate of disappearance of B}=-(1)/(b)(d[B])/(dt)$

$\text{Rate of formation of C}=+(1)/(c)(d[C])/(dt)$

$\text{Rate of formation of D}=+(1)/(d)(d[D])/(dt)$

$Rate=-(1)/(a)(d[A])/(dt)=-(1)/(b)(d[B])/(dt)=+(1)/(c)(d[C])/(dt)=+(1)/(d)(d[D])/(dt)$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$CH_4+H_2O\rightarrow CO+3H_2$

The expression for rate of reaction :

$\text{Rate of disappearance of }CH_4=-(d[CH_4])/(dt)$

$\text{Rate of disappearance of }H_2O=-(d[H_2O])/(dt)$

$\text{Rate of formation of }CO=+(d[CO])/(dt)$

$\text{Rate of formation of }H_2=+(1)/(3)(d[H_2])/(dt)$

The rate of reaction expression is:

$\text{Rate of reaction}=-(d[CH_4])/(dt)=-(d[H_2O])/(dt)=+(d[CO])/(dt)=+(1)/(3)(d[H_2])/(dt)$

As we are given that:

$+(d[CO])/(dt)=0.35M/s$

Now we to determine the rate for the formation of hydrogen.

$+(1)/(3)(d[H_2])/(dt)=+(d[CO])/(dt)$

$+(1)/(3)(d[H_2])/(dt)=0.35M/s$

$(d[H_2])/(dt)=3* 0.35M/s$

$(d[H_2])/(dt)=1.05M/s$

Thus, the rate for the formation of hydrogen is, 1.05 M/s

A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74 atm. When the experiment is repeated using 0.020 mol of ethanoic acid instead of methyl methanoate, the measured pressure is lower than 0.74 atm. The lower pressure for ethanoic acid is due to the following reversible reaction.CH3COOH(g)+CH3COOH(g) ⇋ (CH3COOH)2(g)+Assume that when equilibrium has been reached, 50 percent of the ethanoic acid molecules have reacted.i. Calculate the total pressure in the vessel at equilibrium at 450 K.ii. Calculate the value of the equilibrium constant, Kp, for the reaction at 450 K

Answers

Explanation:

Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

= 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product $(CH3COOH)^(2)$ gas formed are calculated as

0.010 mol CH3COOH * 1 mol $(CH3COOH)^(2)$ / 2 mol CH3COOH

= 0.005 mol $(CH3COOH)^(2)$

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^(2)$

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

= 0.74 atm * 0.015 mol / 0.020 mol

= 0.555 atm

Answer True or False for each of the following statements. (a) The carburization surface was maintained at slightly less than 0.25 wt% carbon for each specimen. (b) Comparing the finished specimens at a depth of 0.20 mm, specimen A features the lowest carbon concentration. (c) Comparing the finished specimens as a whole, specimen D features the lowest overall amount of carbon.

Answers

Answer:

verdadero/a

falso/b

verdadero/c

Explanation:

A mass of 34.05 g of H2O(s) at 273 K is dropped into 185 g of H2O(l) at 310. K in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equi- librium has been reached. Assume that CP, m for H2O is con- stant at its values for 298 K throughout the temperature range of interest.

Answers

Answer:

The temperature of the system once equilibrium is reached, is 292 Kelvin

Explanation:

Step 1: Data given

Mass of H2O = 34.05 grams

⇒ temperature = 273 K

Mass of H2O at 310 K = 185 grams

Pressure = 1 bar = 0.9869 atm

Step 2: Calculate the final temperature

n(ice)*ΔH(ice fusion) + n(ice)*CP(H2O)(Tfinal- Ti,ice) + n(H20)*CP(H2O)*(Tfinal-Ti,H2O) = 0

Tfinal = [n(ice)*CP(ice)*Ti(ice) + n(H2O)*CP(H2O)*Ti(H20) - n(ice)*ΔH(ice fusion)] / [n(ice)*CP(ice) +n(H2O)*CP(H2O)]

⇒ with n(ice) = moles of ice = 34.05 grams / 18.02 g/mol = 1.890 moles

⇒ with CP(ice) = 75.3 J/K*mol

⇒ with Ti(ice) = the initial temperature of ice = 273 K

⇒ with n(H2O) = the moles of water = 185.0 grams / 18.02 g/mol = 10.27 moles

⇒ with CP(H2O) = CP(ice) = 75.3 J/K*mol

⇒ with Ti(H2O) = the initial temperature of the water = 310 K

⇒ with ΔH(ice, fusion) = 6010 J/mol

Tfinal = [1.890 moles * 75.3 J/K*mol * 273 + 10.27 mol * 75.3 J/K*mol * 310 K - 1.890 moles * 6010 J/mol] / [1.890 moles *75.3J/k*mol + 10.27 mol * 75.3 J/K*mol]

38852.541 + 239732.61 - 11358.9 = 267226.251

Tfinal= 291.8 ≈ 292 Kelvin

The temperature of the system once equilibrium is reached, is 292 Kelvin

Final answer:

We must use the principle of conservation of energy to equate the heat gained by the ice to the heat lost by the water, in a given equation, to calculate the final equilibrium temperature.

Explanation:

The topic here is thermodynamics, specifically calculating the final equilibrium temperature when two substances are mixed. Given the information, we can apply the principle of conservation of energy, which in this context is the heat gained by one substance is equal to the heat lost by the other. In this case, the heat gained by the ice (H2O(s)) as it melts and increases in temperature is equal to the heat lost by the water (H2O(l)). Therefore, we have the equation 34.05 g * 1 kcal/kg * K *(T - 273 K) + 34.05 g * 80 Cal/g = 185 g * 1 kcal/kg*K *(310 K - T) where T is the final temperature to be solved.

Learn more about heat transfer here:

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If you mix different amounts of two ideal gases that are originally at different temperatures, what must be true of the final state after the temperature stabilizes? (There may be more than one correct choice.) a) Both gases will reach the same final temperature.

b) The final rms molecular speed will be the same for both gases.

c) The final average kinetic energy of a molecule will be the same for both gases.

Answers

Answer:

a,c are correct

Explanation:

a) On mixing two gases the final temperature of both the gases becomes the same. The heat will flow from high temp. gas to lower temp gas till the temp of both gases become equal (Thermal equilibrium). This is correct.

b) The rms speed of the molecule is inversely proportional to its molar mass so the final rsm will not be the same. This is incorrect.

c) The average kinetic energy of the system will remain the same. Hence this is also correct.

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The destruction of the ozone layer by chlorofluorocarbons (CFC’s) can be described by the following reactions:ClO(g) + O3(g) ? Cl(g) + 2 O2(g) ?H°rxn = –29.90 kJ2 O3(g) ? 3 O2(g) ?H°rxn = 24.18 kJDetermine the value of heat of reaction for the following:Cl(g) + O3(g) ? ClO(g) + O2(g) ?H°=_____________?

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