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True [87]
4 years ago
10

What’s the connection between population and limiting factors?

Chemistry
1 answer:
irakobra [83]4 years ago
4 0

Answer:

Limiting factors are resources or other factors in the environment that can lower the population growth rate. Limiting factors include a low food supply and lack of space. Limiting factors can lower birth rates, increase death rates, or lead to emigration.

Explanation:

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2. A pure substance has constant physical and chemical properties, while mixtures have varying physical and chemical properties (i.e., boiling point and melting point).

3. A pure substance is pure, while a mixture is impure.

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What is FeCl3+NaOH=Fe(OH)3+NaCl
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Hydrogen fuel cells are a promising alternative to gasoline engines in part because they release water instead of carbon dioxide
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Answer:

The correct answer is option (B) Two molecules of H2 and a molecule of O2 must collide at one time

Explanation:

2H2(g) + O2(g) → 2 H2O(g)

   2             1            2

Considering the given reaction, for water molecule to be form, two molecules of hydrogen and one molecule of oxygen must collide at the same time. The product obtained from the collision is two molecules of water as steam. 2 H2O(g)

6 0
3 years ago
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For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl
Usimov [2.4K]

Answer: The approximate equilibrium partial pressure of H_2S is 3.92  atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

      2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)

K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

On reversing the reaction:

     2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)

initial pressure  4.00atm    2.00 atm       0

eqm          (4.00-2x)atm      (2.00-x) atm      2x atm

K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}

K_p'=\frac{1}{K_p}=0.67\times 10^5

2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)

0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}

x=1.96

[H_2S]=2x=2\times 1.96=3.92 atm

Thus approximate equilibrium partial pressure of H_2S is 3.92 atm

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