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Elena-2011 [213]
3 years ago
6

What are the two different Energetics that relate to Environmental Science

Chemistry
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:

Organisms inhabit nearly every environment on Earth, from hot vents deep in the ocean floor to the icy reaches of the Arctic. Each environment offers both resources and constraints that shape the appearance of the species that inhabit it, and the strategies these species use to survive and reproduce. Some of the broadest patterns of environmental difference arise from the way our planet orbits the Sun and the resulting global distribution of sunlight (Chapin et al. 2002).

Explanation:

In the tropics, where solar radiation is plentiful year-round, temperatures are warm, and plants may photosynthesize continuously as long as water and nutrients are available. In polar regions, where solar radiation is seasonally limited, mean temperatures are much lower, and organisms must cope with extended periods when photosynthesis ceases.

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How much heat (kJ) is absorbed by 229.1 g of water in order for the temperature to increase from 25.00∘C to 32.50∘C?
hammer [34]

Answer:

(Q1) 9.42 kJ.

(Q2) 1.999 kJ

Explanation:

Heat: This is a form of Energy that brings about the sensation of warmth.

The S.I unit of Heat is Joules (J).

The heat of a body depend on the mass of the body, specific heat capacity, and temperature difference. as shown below

Q = cm(t₂-t₁) ........................ Equation 1

(Q1)

Q = cm(t₂-t₁)

Where Q = amount of heat absorbed, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 229.1 g = 0.2991 kg, t₁ = 25.0 °C, 32.50 °C

Constant: c = 4200 J/kg.°C

Substituting into equation 1

Q = 0.2991×4200(32.5-25)

Q = 1256.22(7.5)

Q = 9421.65 J

Q = 9.42 kJ.

Hence the heat absorbed = 9.42 kJ

(Q2)

Q = cm(t₂-t₁)

Where Q = amount of heat required, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 34 g = 0.034 kg, t₁ = 9 °C, t₂ = 23 °C

Constant: c = 4200 J/kg.°C

Q = 0.034×4200(23-9)

Q = 142.8(14)

Q = 1999.2 J

Q = 1.999 kJ.

Thus the Heat required = 1.999 kJ

4 0
3 years ago
Water can be formed in the following reaction:
just olya [345]

Answer

12

Explanation

We have a balanced chemical equation from the question that depicts the formation of water.

                                                  2H2+O2-->2H2O,


We can clearly see from the equation that, the formation of 2 moles of water molecules requires the input of 2 moles of hydrogen and 1 mole of Oxygen.

So indirectly, it tells that the moles of water molecules will be double of the moles of Oxygen molecules used in the reaction.

So if we say that 6 moles of oxygen is used and the reaction is going in such a way that hydrogen is not a limiting reactant, then 12 moles of water will be produced.


Hope it help!

8 0
3 years ago
What is the pressure in a 5.00 L tank with 49.00 grams of oxygen gas at 350 K? ___atm
ziro4ka [17]

Answer:

The right answer is "8.81 atm".

Explanation:

Given:

V = 5.00 L

Mass = 4900 g

MW = 32 g/mol

T = 350 K

Now,

Number of moles will be:

n = \frac{Mass}{MW}

   =\frac{49.00}{32}

   =1.53125 \ mole

By using the ideal gas equation, we get

⇒ PV=nRT

or,

⇒    P=\frac{nRT}{V}

By substituting the values, we get

           =\frac{1.53125\times 0.0821\times 350}{5.00}

           =\frac{44.1}{5.00}

           =8.81 \ atm  

4 0
3 years ago
What is one property that is different between water and oil? *
Arlecino [84]
Oil is less dense than water, so the difference would be its density. Water is a good solvent, which means It can dissolve other substances.
3 0
3 years ago
A chemistry student is given 700. mL of a clear aqueous solution at 26.° C. He is told an unknown amount of a certain compound
Alex73 [517]

Answer:

The correct answer is - yes, 4.57 g of solute per 100 ml of solution

Explanation:

The correct answer is yes we can calculate the solubility of X in the water at 22.0°C. The salt will remain after the evaporate from the dissolved and cooled down at 26°C.

Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.

Then solublity will be :

32. g solute / 700 ml solution = y / 100 ml solution

⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.

Thus, the answer is 4.57 g of solute per 100 ml of solution.

5 0
3 years ago
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