Why do organisms evolve?Organisms make changes to help them survive and reproduce in their environment

Organisms with traits that help them survive and reproduce in their environment pass on those traits.

Changes in the environment cause organisms to change.
Organisms naturally select traits that help them survive and reproduce

Answers

Answer 1
Answer:

In an ecosystem, organisms evolve to make changes to help them survive and reproduce in their environment.

What is an ecosystem?

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.

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Related Questions

What is the empirical formula of a compound composed of 36.7 g 36.7 g of potassium ( K K ) and 7.51 g 7.51 g of oxygen ( O O )? Insert subscripts as needed.
Is A and C correct? I'm a little unsure of this question, but is it right?
What is the mass of 0.73 moles of AgNO3?
The solubility of glucose at 30°C is125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.
HELP Which type of light is stored energy?A. Kinetic B. Potential C. Thermal D. Field

Transpiration is the loss of water from the leaves of plants. The stomata of leaves must open to allowcarbon dioxide to enter the leaf for photosynthesis, but when they are open, water vapor escapes into
the atmosphere.
HYPOTHESIS: As the intensity of light is increased, the rate of transpiration will increase, as
measured in by the loss of mass of the plant.
Independent variable:
Dependent variable:

Answers

Independent variable are the light intensity that what we change and dependent variable are the rate of transpiration that what we measure.

What is photosynthesis ?

Photosynthesis is the process that can be used by the plants and other animals to covert light energy into the chemical energy. Photosynthesis is the main source of food in the earth.

Photosynthesis is the process in which oxygen is released. For survival oxygen is very important, from this process we obtain sufficient amount of oxygen.

The dependent variable is the rate of transpiration and the independent variable is time.The dependent variable is the rate of transportation because it is depends on the environmental factor the plant is placed.

Thus,Independent variable are the light intensity that what we change and dependent variable are the rate of transpiration that what we measure.

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Answer:

Independent variable: The light intensity (what YOU CHANGE)

Dependent variable: The rate of transpiration (what YOU MEASURE)

During an experiment, a student adds 2.90 g CaO to 400.0 mL of 1.500 M HCl . The student observes a temperature increase of 6.00 °C . Assuming that the solution's final volume is 400.0 mL , the density is 1.00 g/mL , and the heat capacity is 4.184 J/g⋅°C , calculate the heat of the reaction, ΔHrxn .

Answers

Answer:

ΔHrxn = 193107.69 J/mol

Explanation:

ΔHrxn = mcΔT

m = mass

c = heat capacity

ΔT = temperature variation

density = m/V

m = density x V

m = 1.00 g/mL x 400.0 mL

m = 400.0 g

ΔHrxn = mcΔT

ΔHrxn = 400 g x 4.184 J/g°C x 6.00 °C

ΔHrxn = 10041.6 J

CaO + 2HCl  → CaCl₂ + H₂O

CaO = 56.0774 g/mol

2.90 g CaO = 0.052 mol

400.0 mL of 1.500 mol/L HCl = 0.6 mol HCl

ΔHrxn = 10041.6 J is for 0.052 mol of CaO

ΔHrxn = 193107.69 J is for 1 mol of CaO

I have a cup of hot coffee at 140 oC but I want to cool it to 110 oC. My cup holds about 0.3 kg of coffee. Fortunately, I have a bunch of aluminum cubes in the freezer that I can drop into my hot coffee to cool it down. If each aluminum cube has a mass of 1 g (not 1 kg!) and my freezer keeps its contents at a temperature of –10 oC, how many cubes do I have to drop into my coffee? The specific heat of water is around 4000 joules/kg/oC and aluminum is about 900 joules/kg/oC. (Pick the answer closest to the true value and ignore any thermal losses to surroundings.)

A. 200
B. 330
C. 400
D. 110
E. 88

Answers

Answer:

The correct answer is option B.

Explanation:

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)

where,

C_1 = specific heat of metal = 900 J/kg^oC

C_2 = specific heat of coffee= 4000 J/kg^oC

m_1 = mass of metal = x

m_2 = mass of coffee = 0.3 kg

T_f = final temperature of aluminum metal= 110^oC

T_1 = initial temperature of aluminum metal = -10^oC

T_2 = initial temperature of coffee= 140^oC

Now put all the given values in the above formula, we get

x* 900 J/kg^oC* (110-(-10))^oC=-(0.3 kg* 4000 J/kg^oC* (110-140)^oC

x=0.333 kg

Mass of aluminum cubes = 0.3333 kg = 333.3 g

If mass of 1 cube is 1 gram, then numbers of cubes in 333.3 grams will be:

=(333.3 g)/(1 g)=333.3\approx 330

330 cubes of aluminum cubes will be required.

Metal hydrides react with water to form hydrogen gas and the metal hydroxide. srh2(s) + 2 h2o(l) sr(oh)2(s) + 2 h2(g) you wish to calculate the mass of hydrogen gas that can be prepared from 5.00 g of srh2 and 5.47 g of h2o. (a) how many moles of h2 can be produced from the given mass of srh2?

Answers

The reaction forms 0.112 mol H_2.  

We have the masses of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them.  

M_r: __89.64 ___18.02 ___________2.016

______SrH_2 + 2H_2O → Sr(OH)_2 + 2H_2

Mass/g: 5.00 ___5.47

Step 2. Calculate the moles of each reactant  

Moles of SrH_2 = 5.00 g SrH_2 × (1 mol SrH_2 /89.64 g SrH_2)

= 0.055 77 mol SrH_2

Moles of H_2O = 5.47 g H_2O × (1 mol H_2O/18.02 g H_2O)

= 0.3036 mol H_2O

Step 3. Identify the limiting reactant

Calculate the moles of H_2 we can obtain from each reactant.  

From SrH_2: Moles of H_2 = 0.055 77 mol SrH_2 × (2 mol H_2 /1 mol SrH_2) = 0.112 mol H_2

From H_2O: Moles of H_2 = 0.3036 mol H_2O × (2 mol H_2/2 mol H_2O)

= 0.3036 mol H_2

SrH_2 is the limiting reactant because it gives the smaller amount of H_2.

It produces 0.112 mol H_2.

2.67 grams of butane (C4H10) is combusted in a bomb calorimeter. The temperature increases from 25.68 C to 36.2C. What is the change in the internal energy (deltaE) in KJ/mol for the reaction if the heat capacity of the bomb calorimeter is 5.73 kJ/C?

Answers

The internal energy : 1310.43 kJ/mol

Further  explanation

Internal energy (ΔE) can be formulated for Calorimeter :

\tt \Delta E=C*.\Delta t

C= the heat capacity of the calorimeter

Δt=36.2-25.68=10.52°C

\tt \Delta E=5.73 kJ/^oC* 10.52^oC=60.2796~kJ

  • mol butane(MW=58,12 g/mol)

\tt (2.67)/(58,12 )=0.046

  • the internal energy (ΔE) in KJ/mol

\tt (60.2796)/(0.046)=1310.43~kJ/mol

Final answer:

The change in internal energy when 2.67 grams of butane is combusted in a bomb calorimeter, given a temperature increase from 25.68 C to 36.2C and a heat capacity of 5.73 kJ/C for the calorimeter, is approximately 1308 kJ/mol.

Explanation:

To solve the problem of calculating the changes in internal energy when 2.67 grams of butane (C4H10) is combusted in a bomb calorimeter, it is necessary to understand calorimeter's heat capacity and how a bomb calorimeter works.

The first step will be to calculate the change in temperature which here is the final temperature subtracted from the initial temperature: 36.2 C - 25.68 C = 10.52 C.

Then, we multiply this temperature change by the heat capacity of the calorimeter to find the total heat produced by the reaction in kJ: 10.52 C * 5.73 kJ/C = 60.18 kJ.

The final step is to convert grams of butane to moles, because we are asked to find the energy change in kJ/mol. The molar mass of butane (C4H10) is approximately 58.12 g/mol. So we have approximately 2.67 g / 58.12 g/mol = 0.046 mol.

Finally, we divide the heat produced by the number of moles to get the energy change per mole of butane: 60.18 kJ / 0.046 mol = approximately 1308 kJ/mol.

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In our modern view of matter and energy, is the law of mass conservation still relevant to chemical reactions? Explains

Answers

Explanation:

Law of conservation of mass states that mass can neither be created nor it can be destroyed but it can be transformed into one form to another.

Similarly, law of conservation of energy states that energy can neither be created nor it can be destroyed as it can only be transformed from one form to another.

In modern view of matter and energy, is the law of mass conservation still relevant to chemical reactions as follows.

For example, 2Na + Cl_(2) \rightarrow 2NaCl

Atomic mass of Na = 23

Atomic mass of Cl = 35

Hence, mass of total number of reactants is calculated as follows.

             [(2 \times 23) + (35 \times 2)] g/mol = 116 g/mol

Mass of total number of products is calculated as follows.

             [2 \times (23 + 35)] = 116 g/mol

Thus, it is proved that in our modern view of matter and energy, is the law of mass conservation still relevant to chemical reactions.