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anyanavicka [17]
3 years ago
13

Xavier was conducting a long term study of two populations of butterflies. Both populations had the same number of individuals.

Both populations lived in forests that were the same size. Population A lived in a forest that was full of milkweed, the favorite food of this butterfly species. Population B lived in a forest that had limited amounts of milkweed and other foods. Over many years, Xavier found that one of the populations was showing signs of evolutionary change. Their mouth parts were changing and they could get food from new kinds of plants. Which statement explains why one population would be more likely to evolve in response to natural selection? *
a. Population B would be more likely to change because it has a ready supply of food, and only certain butterflies would be able to survive
b. Population A would be more likely to change because it has a ready supply of food and only certain butterflies would be able to survive
c. Population A would be more likely to change because it has a limited amount of food and some butterflies might be better able to get enough food, survive, and reproduce
d. Population B would be more likely to change because those butterflies have a limited supply of food, and butterflies with certain traits might be better able to get enough food, survive, and reproduce.
Chemistry
1 answer:
nata0808 [166]3 years ago
8 0
B. I had that same question!!
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A 34.0 g piece of metal is heated to 92.0°C then placed in a beaker of water containing 22.0 g of water at 19.0°C. The temperatu
lapo4ka [179]

Answer:

0.1988 J/g°C

Explanation:

-Qmetal = Qwater

Q = mc∆T

Where;

Q = amount of heat

m = mass of substance

c = specific heat of substance

∆T = change in temperature

Hence;

-{mc∆T} of metal = {mc∆T} of water

From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.

For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?

Note that, the final temperature of water and the metal = 24°C

-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)

-{34 × c × (-68°C)} = 459.8

-{34 × c × -68} = 459.8

-{-2312c} = 459.8

+2312c = 459.8

c = 459.8/2312

c = 0.1988

The specific heat capacity of the metal is 0.1988 J/g°C

6 0
3 years ago
The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k
Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})

We have given in the question

P_1=102\ mmHg

T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol

And R is the Universal Gas Constant.

R=0.008 314 kJ/Kmol

ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165

Taking inverse log both side we get,

\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg

8 0
3 years ago
Which is a negative result of eating too many lipids plz help
fenix001 [56]
Eating too much lipids is hazardous to the health. It has a lot of negative effects. One is cardiovascular complications, too much of lipids can cause high blood cholesterol, arterial hardening, heart diseases and stroke. Also, obesity is possible. Eating too much lipids is associated with metabolism abnormalities.
5 0
3 years ago
Read 2 more answers
Calculate the value of the diffusion coefficient D (in m2/s) at 547°C for the diffusion of some species in a metal; assume that
svetlana [45]

Answer : The value of diffusion coefficient is, 2.97\times 10^{-16}m^2/s

Explanation :

Formula used :

D=D_o\times \exp \left(-\frac{Q_d}{RT}\right )

where,

D = diffusion coefficient = ?

D_o = 5.6\times 10^{-5}m^2/s

Q_d = 177kJ/mol

R = gas constant = 8.314 J/mol.K

T = temperature = 547^oC=273+547=820K

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

D=5.6\times 10^{-5}m^2/s\times \exp \left(-\frac{177kJ/mol}{(8.314J/mol.K)\times (820K)}\right )

D=2.97\times 10^{-16}m^2/s

Thus, the value of diffusion coefficient is, 2.97\times 10^{-16}m^2/s

6 0
3 years ago
Help me out hear ples
hoa [83]
Number 6 is D the mitochondrion
3 0
2 years ago
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