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Nataly [62]
3 years ago
9

During the winter in grassland ecosystems, rabbits and ground hogs both hibernate in holes. Due to an extremely wet summer much

of the ground cannot be used for winter hibernation burrows. This causes an increase in competition between rabbits and ground hogs. Explain a structural and a behavior adaptation that the rabbits could adapt to deal with these new pressures.
Chemistry
1 answer:
vovangra [49]3 years ago
3 0

Explanation:

They have to battle them dang hogs and get them burrows before them.

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If 22.5L of nitrogen at 0.98 atm is compressed to 0.95 atm,what is the new volume?
Ksenya-84 [330]

At constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.

<h3>What is Boyle's law?</h3>

Boyle's law simply states that "the volume of any given quantity of gas is inversely proportional to its pressure as long as temperature remains constant.

Boyle's law is expressed as;

P₁V₁ = P₂V₂

Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.

Given that;

  • Initial volume of the gas V₁ = 22.5L
  • Initial pressure of the gas P₁ = 0.98atm
  • Final pressure of the gas P₂ = 0.95atm
  • Final volume of the gas V₂ = ?

P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

V₂ = (0.98atm × 22.5L) / 0.95atm

V₂ = 22.05Latm / 0.95atm

V₂ = 23.2L

Therefore, at constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.

Learn more about Boyle's law here: brainly.com/question/1437490

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6 0
2 years ago
Show
SIZIF [17.4K]

Answer: 0.9375 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}     .....(1)

Molarity of HCl solution = 0.75 M

Volume of HCl solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

\text{Moles of} HCl={0.75}\times{0.025}=0.01875moles  

CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)  

According to stoichiometry :

2 moles of HCl require = 1 mole of CaCO_3

Thus 0.01875 moles of HCl will require=\frac{1}{2}\times 0.01875=0.009375moles  of CaCO_3

Mass of CaCO_3=moles\times {\text {Molar mass}}=0.009375moles\times 100g/mol=0.9375g

Thus 0.9375 g of CaCO_3 is required to react with 25.0 ml of 0.75 M HCl

6 0
3 years ago
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Answer:

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A flask holds 3.01 x 1023 molecules of carbon dioxide. The container is holding ________
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