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Goryan [66]
3 years ago
7

Explain whether or not you expect the chaparral biome to be sensitive to the loss of a single species.

Chemistry
1 answer:
aleksley [76]3 years ago
3 0

Answer:

See the answer below

Explanation:

The chaparral biome is a temperate biome with a characteristic high temperature and dryness during summer and mild rainy winters and springs. The biome can be found in relatively small amounts in the major continents of the world with its rich plant and animal diversity who have successfully adapted to the conditions of the biome.

Due to the high biodiversity of the chaparral biome, <u>one would expect it to be resilient to the loss of a single species.</u> <em>The more the biodiversity of a biome or community, the more resilient such biome or community would be to the loss of species and lower the biodiversity, the more sensitive the community would be to the loss of species. </em>

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26.76 miles

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How many significant figures in 0.002030?
scoundrel [369]
Answer:
Three.
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8 0
3 years ago
Read 2 more answers
how does the total mass of each object increases the amount of force that is needed to get them moving at 5 m/s increase by abou
irakobra [83]

Answer:

The answer is below

Explanation:

Newton's second law of motion states that the force applied to an object is directly proportional to the rate of change of momentum with respect to time, going in the same direction as the force.

Let F = force, m = mass of object, v = velocity of object, mv = momentum.

F = d/dt(mv) = m(dv / dt) = ma; a = acceleration.

Let us assume that the object starts from rest to 5 m/s within 1 seconds, hence:

F = m(dv / dt)

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7 0
2 years ago
50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the
trapecia [35]

Answer:

-21 kJ·mol⁻¹  

Explanation:

Data:

                    H₃O⁺ +  OH⁻ ⟶ 2H₂O

       V/mL:    50         50  

c/mol·dm⁻³:   1.0         1.0

     

ΔT = 4.5 °C  

       C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

     nΔH   +         mCΔT       + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 +   50×4.5  = 0

0.050ΔH +          1883        +      225    = 0

                                  0.050ΔH + 2108 = 0

                                              0.050ΔH = -2108

                                                        ΔH = -2108/0.0500

                                                              = -42 000 J/mol

                                                              = -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

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