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shepuryov [24]
3 years ago
8

The Canadian lynx is a large wild cat that lives in the cold regions of Canada. The snowshoe hare is the main prey species that

is hunted by the Canadian lynx. The snowshoe hare eats grass, fern, and leaves. If the population of Canadian lynx were to completely disappear, what would most likely happen to the food supply of the snowshoe hare?
Chemistry
2 answers:
g100num [7]3 years ago
6 0
If the population of the Canadian lynx will decline and completely disappear, the event will directly affect the population of the snowshoe hare. The snowshoe hare population will rapidly increase due to the absence of their predator. Also, their food supply will rapidly decrease due to the increase in their population. 
zmey [24]3 years ago
3 0
<span>The population of snowshoe hare would outgrow the food supply, so there would be less food available.
</span>
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#5.) Explain why the rock cycle is referred to as a "cycle".
Natalka [10]
The rock cycle is referred to as cycle because it repeats over and over again. The definition of cycle is a series of events that are regularly related in the same order. So with the rock cycle they stay in the same order and repeat.
Hope this helps :)
5 0
3 years ago
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How many moles of water were lost if the amount of water lost was 0.456 grams? Do not include units and assume three significant
nika2105 [10]
<h3>Answer:</h3>

0.0253 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 0.456 g H₂O (water)

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 0.456 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 0.025305 \ mol \ H_2O

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.025305 mol H₂O ≈ 0.0253 mol H₂O

3 0
2 years ago
Calculate the partial pressure of oxygen given the total barometric pressure of 515 mmHg. Round to the nearest hundredth.
Eva8 [605]

The partial pressure of oxygen given the total barometric pressure is : 108.15 mmHg

<u>Given data : </u>

Total barometric pressure = 515 mmHg

Assuming oxygen percentage = 21%

Barometric pressure dry at 37°C

<h3 /><h3>Determine the partial pressure of oxygen </h3>

Applying  the relation below

Partial pressure = oxygen percentage * Barometric pressure

                          = 21% * 515 mmHg

                          = 108.15 mmHg

Hence we can conclude that the partial pressure of oxygen is 108.15 mmHg.

Learn more about Partial pressure : brainly.com/question/1835226

8 0
2 years ago
A 25.0g sample of brass, which has a specific heat capacity of 0.375·J·g−1°C−1, is dropped into an insulated container containin
Angelina_Jolie [31]

Answer:

The equilibrium temperature of water is 25.6 °C

Explanation:

Step 1: Data given

Mass of the sample of brass = 25.0 grams

The specific heat capacity = 0.375 J/g°C

Mass of water = 250.0 grams

Temperature of water = 25.0 °C

The initial temperature of the brass is 96.7°C

Step 2: Calculate the equilibrium temperature

Heat lost = heat gained

Q(sample) = -Q(water)

Q = m*C* ΔT

m(sample)*c(sample)*ΔT(sample) = - m(water)*c(water)*ΔT(water)

⇒m(sample) = the mass of the sample of brass = 25.0 grams

⇒with c(sample) =The specific heat capacity = 0.375 J/g°C  

⇒with ΔT = the change of temperature = T2 - T1 =T2 - 96.7 °C

⇒with m(water) = the mass of the water = 250.0 grams

⇒with c(water) = the specific heat capacity = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = T2 - 25.0°C

25.0 * 0.375 * (T2 - 96.7) = - 250.0 * 4.184 J/g°C * (T2 - 25.0°C)

9.375T2 - 906.56 = -1046T2 + 26150

9.375T2 + 1046T2 = 26150 + 906.56

1055.375T2 = 27056.26

T2 = 25.6 °C

The equilibrium temperature of water is 25.6 °C

4 0
3 years ago
Which of these is an environmental effect of building dams?
nadezda [96]
Hmm... I'm unsure but its either B or D.
7 0
3 years ago
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