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PolarNik
2 months ago
8

Why would it be better to be an r-selected species if the water resources in an area were to become more limited over a short pe

riod of time?
Also, sorry if this is the wrong subject, it was hard to find a fitting one for Environmental Science.
Chemistry
2 answers:
castortr0y [3K]2 months ago
7 0
Hi there!

In situations where water resources become scarce, being an r-selected species is advantageous because they typically reproduce faster than other species.

Hope this helps!:)
lions [2.9K]2 months ago
6 0
An r-selected species has a significantly faster reproductive rate compared to K-selected species.

The focus of r-selected species is on quick maturation and reproduction. They are likely to breed during short periods when water supply is available, thus enhancing their survival chances.

Conversely, K-selected species prioritize nurturing their young and tend to reproduce later. Due to the longer maturation time before breeding, by the time K-selected species are ready, the water supply may be depleted, leading to lower survival odds.

Hope this clarifies!
If you have more questions, don’t hesitate to ask! Have a fantastic day!:)

~Collinjun0827, Junior Moderator
You might be interested in
You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di
lorasvet [2795]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

500mM = 500mM \times (\frac{1M}{1000M})= 0.5M

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M

10 mL of this solution is further diluted to 250 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M

10 mL of this solution is further diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M

Convert μM:

2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M

5 mL of this solution is diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M

Convert μM:

0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M

10 mL of this solution is further diluted to 500 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M

10 mL of this solution is further diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

7 0
2 months ago
Liquid nitrogen has a density of 0.807 g/ml at –195.8 °c. if 1.00 l of n2(l) is allowed to warm to 25°c at a pressure of 1.00 at
castortr0y [3046]
Step 1: Convert density from g/mL to g/L; 0.807 g/mL is equivalent to 807 g/L. Step 2: Calculate Moles of N₂; Density = Mass / Volume, or Mass = Density × Volume. Plugging in values, Mass = 807 g/L × 1 L gives us Mass = 807 g. Similarly, Moles = Mass / M.mass, which leads to Moles = 807 g / 28 g.mol⁻¹, giving us Moles = 28.82 moles. Step 3: Apply the Ideal Gas Law to determine Volume of gas occupied; P V = n R T, thus V = n R T / P. Remember to convert temperature to Kelvin (25 °C + 273 = 298 K). Hence, V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm, resulting in V = 704.76 L.
8 0
2 months ago
A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man
castortr0y [3046]

Answer:

0.400 moles of Oxygen

Explanation:

By employing the equation PV = nRT, the initial pressure of the flask can be calculated prior to the reaction, which leads to:

P = nRT/V

Where:

n signifies moles (4,000 moles: 2,000 moles of CO and 2,000 moles of H₂O)

R represents the gas constant (0.082 atm·L/mol·K)

T is the temperature (300.0 K)

V denotes volume (0.2000 L)

Substituting values results in P = 492.0 atm

To achieve a pressure reduction of 10.00%, the resulting pressure should be:

492.0 atm - 49.2 atm = 442.8 atm

Calculating with the new pressure under the same conditions gives the moles as:

n = PV/RT

n = 3,600 total moles

In the reaction:

2CO(g) + O₂(g) ⟶ 2CO₂(g)

The resulting moles are:

CO: 2,000 moles - 2X

O₂: 2,000 moles - X

CO₂: 2X

Where X accounts for the moles that react

Consequently, the total moles are:

4,000 moles - X = 3,600 moles

X = 0.400 moles

This indicates that the amount of oxygen needed for the reaction is 0.400 moles of Oxygen

I hope this is useful!

7 0
2 months ago
A 1.20 g sample of water is injected into an evacuated 5.00 l flask at 65°c. part of the water vaporizes and creates a pressure
alisha [2963]
Assuming the water vapor behaves as an ideal gas,


PV = nRT

For conversions, 760 mmHg = 101325 Pa and 1,000 L = 1 m³
(187.5 mmHg)(101325 Pa/760 mmHg)(5 L)(1 m³/1,000 L) = n(8.314 m³Pa/molK)(65+273 K)
Calculating for n,
n = 0.0445 mole of water

Considering the molar mass of water is 18 g/mol,
The mass of the vaporized water = 0.0445 * 18 = 0.8 g of water evaporated

Therefore,
The fraction of water that vaporized = 0.8/1.2 * 100 = 66.7%
3 0
2 months ago
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