I believe that the best definition of the word common in environmental science is a typical resource. A typical resource is a resource that is not hard to find and is very common. The answer to the question will hence be A.
The answer could not be the other three choices because a shared resource is a resource that is used by more than one organism in the environment. A worthless resource is one that has no use in the environment while a rare resource is one that is hard to find.
The ball travels 10m/s
Speed = distance/time
Answer :
is the oxidizing agent and Fe is the reducing agent.
Explanation :
Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.
Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.
The balanced redox reaction is :

The half oxidation-reduction reactions are:
Oxidation reaction : 
Reduction reaction : 
In order to balance the electrons, we multiply the oxidation reaction by 4 and reduction reaction by 3 then added both equation, we get the balanced redox reaction.
Oxidation reaction : 
Reduction reaction : 

In this reaction,
is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and
is the oxidizing agent that gain an electron to another chemical species in a redox chemical reaction and itself gets reduced.
Thus,
is the oxidizing agent and Fe is the reducing agent.
Answer: the percent composition of carbon in heptane is 83.9%
Explanation:
<u>1) Atomic masses of the atoms:</u>
<u>2) Molar mass of heptane:</u>
- C₇H₁₆: 7 × 12.01 g/mol + 16×1.008 g/mol = 100.2 g/mol
<u>3) Mass of carbon in one mole of heptane:</u>
- C₇: 7 × 12.01 g/mol = 84.07 g/mol
<u>3) Percent composition of carbon:</u>
- % = (mass in grams of C) / (mass in grams of C₇H₁₆) × 100 =
= (84.07 g/ 100.2 g) × 100 = 83.9% ← answer
<h3>
Answer:</h3>
0.024 kg CaO
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Aqueous Solutions</u>
- Molarity = moles of solute / liters of solution
<u>Atomic Structure</u>
- Reading a Periodic Tables
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
0.41 mol CaO
2.5 M Solution
<u>Step 2: Identify Conversions</u>
1000 g = 1 kg
Molar Mass of Ca - 40.08 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of CaO - 40.08 + 16.00 = 56.08 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>
0.024114 kg CaO ≈ 0.024 kg CaO