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3 years ago

Describe how different environments support different varieties of organisms

Chemistry

1 answer:

vladimir2022 [97]3 years ago

3 0

Answer:

How is an environment able to support different varieties of organisms? ... It provides all the necessary conditions, such as temperature, space, water and food resources, oxygen and carbon dioxide, that suit the needs of the organisms. Examples of habitat include forests, mountains, bodies of water, and even puddles.

Explanation:

You might be interested in

How many moles of oxygen must be placed

umka21 [38]

Answer: 0.245 moles of oxygen must be placed in the container to exert the given pressure at the given temperature. The Ideal Gas Law equation gives the relationship among the pressure, volume, temperature, and moles of gas.

Further Explanation:

The Ideal Gas Equation is:

$PV = nRT$

where:

P - pressure (in atm)

V - volume (in L)

n - amount of gas (in moles)

R - universal gas constant $0.08206 \frac{L-atm}{mol-K}$

T - temperature (in K)

In the problem, we are given the values:

P = 2.00 atm (3 significant figures)

V = 3.00 L (3 significant figures)

n = ?

T = 25.0 degrees Celsius (3 significant figures)

We need to convert the temperature to Kelvin before we can use the Ideal Gas Equation. The formula to convert from degree Celsius to Kelvin is:

$Temperature \ in \ Kelvin = Temperature\ in \ Celsius \ + \ 273.15$

Therefore, for this problem,

$Temperature\ in \ K = 25.0 +273.15\\Temperature\ in \ K = 298.15$

Solving for n using the Ideal Gas Equation:

$n \ = \frac{PV}{RT}\\n \ = \frac{(2.00 \ atm) \ (3.00 \ L)}{(0.08206 \ \frac{L-atm}{mol-K})( 298.15 \ K)} \\n \ = 0.245 \ mol$

The least number of significant figures is 3, therefore, the final answer must have only 3 significant figures.

Learn More

1. Learn more about Boyle's Law brainly.com/question/1437490

2. Learn more about Charles' Law brainly.com/question/1421697

3. Learn more about Gay-Lussac's Law brainly.com/question/6534668

Keywords: Ideal Gas Law, Volume, Pressure

4 0

3 years ago

You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p

Elenna [48]

Answer:

<u>1.29 mol</u>

Explanation:

This is a direct application of the equation for ideal gases.

$PV=nRT$

Where:

P = pressure = 1.25 atm
V = volume = 25.2 liter
R = Universal constant of gases = 0.08206 atm-liter/K-mol
T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
n = number of moles

Solving for n:

$n=\frac{PV}{RT}$

Substituting:

$n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol$

8 0

3 years ago

Which of the following groups in the periodic

Dimas [21]

Elements of Group 1 and group 2 in the periodic table contain elements so reactive that they are never found in the free state

<u>Explanation</u>:

The metals in group 1 of periodic table consisting of 'alkali metals' which include lithium, potassium, sodium, rubidium, Francium and caesium. They are highly reactive because they have low ionisation energy and larger radius. The group 2 metals consist of 'alkaline earth metals' which include calcium, strontium, barium, beryllium, radium and magnesium. These alkaline earth metal have +2 oxidation number, hence are highly reactive.

These both group metals are mostly reactive and so are never found in a free state. When they are exposed to air they would immediately react with oxygen. Hence, are stored in oils to avoid oxidation.

7 0

3 years ago

compose time graph that contains learning and insight that contains social science disiplins that led to its emergences

iren2701 [21]

Answer:

The major social sciences are:

Anthropology.

Economics & Management.

History.

Human Services.

Political Science.

Psychology.

Sociology

3 0

3 years ago

Methanol, ethanol, and n-propanol are three common alcohols. When 1.00 g of n-propanol (C3H7OH)is burned in air, -33.4 kJ of hea

Inga [223]

Answer : The heat of combustion of n-propanol is 0.554 kJ/mol

Explanation :

First we have to calculate the moles of n-propanol.

$\text{Moles of n-propanol}=\frac{\text{Mass of n-propanol}}{\text{Molar mass of n-propanol}}$

Molar mass of n-propanol = 60.09 g/mole

$\text{Moles of n-propanol}=\frac{1.00g}{60.09g/mole}=0.0166mole$

Now we have to calculate the heat of combustion of n-propanol.

As, 0.0166 mole of n-propanol liberated heat of combustion = -33.4 kJ

So, 1 mole of n-propanol liberated heat of combustion = 0.0166 × (-33.4 kJ)

= 0.554 kJ/mol

Therefore, the heat of combustion of n-propanol is 0.554 kJ/mol

5 0

3 years ago