Answer: Flammability is a material's ability to burn in the presence of oxygen.
Explanation: Chemical properties can be observed only when the substance changes into one or more different substances through chemical reactions or transformations. One of the chemical properties is flammability.
Flammability is a material's ability to burn in the presence of oxygen.
Remember, oxygen doesn't burn. Precisely flammable substances obtain substances that burn. Oxygen remains an oxidizing agent, which means it supports the combustion process. Oxygen causes other objects to catch fire at low temperatures and burns hotter and faster. But oxygen itself does not burn. Consequently, if you at present deliver fuel and fire, adding oxygen will provide the fire.
Carbon dioxide is the result of combustion. An example can be seen in firewood in a fireplace. One of the chemical properties of carbon-based wood is having the ability to burn. Chemically the wood turns into carbon dioxide when it burns and leaves a residue of ash. Furthermore, this ash residue cannot be turned back into the wood. Chemical changes result in new substances.
Consider an example of a combustion reaction to methane gas:
Our balanced equation for methane combustion implies that every one CH₄ molecule reacts with two O₂ molecules. The product of combustion is one carbon dioxide molecule and two steam or water vapor molecules.
ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K
<h3>
Answer:</h3>
495 g K₃N
<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>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.77 mol K₃N
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
495.039 g K₃N ≈ 495 g K₃N
Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours
Explanation:
We are given:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains
number of particles.
We know that:
Charge on 1 electron = 
Charge on 1 mole of electrons = 

is passed to deposit = 1 mole of copper
63.5 g of copper is deposited by = 193000 C
of copper is deposited by =
To calculate the time required, we use the equation:

where,
I = current passed = 40.0 A
q = total charge = 42551181 C
t = time required = ?
Putting values in above equation, we get:

Converting this into hours, we use the conversion factor:
1 hr = 3600 seconds
So, 
Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours