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uranmaximum [27]
3 years ago
6

Which of the following statements best describes the relationships between carrying captivity in population size

Chemistry
1 answer:
Ierofanga [76]3 years ago
4 0

The correct answer is letter a, whereas, carrying capacity in captivity increases population size. It is because the carrying captivity is the one responsible for having a maximum population size in which will help in sustaining the necessities that the species need in the environment in which makes it responsible for the population size to increase depending on its capacity. The correct answer is letter a.

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Please help with that problem <br> Important İ have quiz tomorrow
iogann1982 [59]

Answer:

Final pressure of the gas remaining in the first container is 3.5 atm

Explanation:

Since it is given that temperature is constant , we can apply -

PV=constant , where

P = Pressure of the gas

V = Volume of the container in which the gas is contained

Initially,

For container 1 -

P_{i} = 4 atm

V_{1} = 6 L

Finally,

For container 2 -

P_{f,2} = 3 atm

V_{2} = 1 L

For container 1 -

P_{f,1} = ?

V_{1} = 6 L

∴ P_{i}V_{1} = P_{f,1}V_{1} + P_{f,2}V_{2}

∴ 4×6 =  (P_{f,1}×6) +(3×1)

∴ P_{f,1} = 3.5 atm

6 0
3 years ago
[Send a message] as you move across a row from left to right in the period table, atomic numbers of the elements
SashulF [63]
I think it might be D or B 
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6 0
3 years ago
Suppose 215 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?
charle [14.2K]

Answer:

The answer is "41.23 \ L\  N_2"

Explanation:

2 NO_3^{-} + 10 e^{-} + 12 H^{+} \longrightarrow N_2 + 6 H_2O\\\\= \frac{( 215 \ g \ NO_3^{-})}{(62.0049  \frac{\ g NO_3^{-}}{mol})} \times  \frac{(1 \ mol \ N_2}{ 2 \ mol \ NO_3^{-})}\\\\

=3.46746789 \times 0.5\\\\= 1.733 \ mol \ N_2 \\\\\to V = \frac{nRT}{P} \\\\= (1.733 \ mol) \times (0.08205746 \frac{L\ atm}{Kmol}) \times \frac{ (17 + 273) K}{(1.00 atm)}\\\\= 41.23

8 0
3 years ago
Does anyone know what order?
mamaluj [8]

I believe it is decomposition, neutralization, combustion

4 0
3 years ago
Read 2 more answers
What volume of a 6.67 M NaCl solution contains 3.12 mol NaCl? L
harkovskaia [24]

Answer:

0.47dm³

Explanation

Given parameters :

Molarity of NaCl = 6.67M

Number of moles = 3.12mol

Volume of NaCl =?

Volume of NaCl = number of moles/Molarity

Volume of NaCl = 3.12mol/6.67M

Volume of NaCl = 0.47dm³

4 0
4 years ago
Read 2 more answers
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