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.
Complete balanced equation: 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O
Ionized equation (with spectator ions):
2H⁺ + 2NO₃⁻ + Ca²⁺ + 2OH⁻ → Ca²⁺ + 2NO₃⁻ + 2H₂O
By eliminating the ions that are the same of both sides of the equation (spectator ions):
2H⁺ + 2OH⁻ → 2H₂O [Net Ionic Equation]
Answer:
This element is Rubidium (Rb) and has an average atomic mass of 85.468 u
Explanation:
The average mass of an element is calculated by taking the average of the atomic masses of its stable isotopes.
The enitre atomic mass = 100 % or 1
⇒ this consists of X-85 with 72.17 % abundance with atomic massof 84.9118 g/mol
72.17 % = 0.7217
⇒ this consists of X-87 with 27.83 % abundance with atomic mass of 86.9092 g/mol
27.83 % = 0.2783
To calculate the mass of this isotope we use the following:
0.7271 * 84.9118 + 0.2783 * 86.9092 =85.468 g/mol
This element is Rubidium(Rb) and has an average atomic mass of 85.468 u
Answer:
isotope 2
Explanation:
it has the highest percentage abundance
Answer:
The answer to your question is 25.9 g of KCl
Explanation:
Data
Grams of KCl = ?
Volume = 0.75 l
Molarity = 1 M
Formula
Solve for number of moles
Substitution
Number of moles = 1 x 0.75
Simplification
Number of moles = 0.75 moles
Molecular mass KCl = 39 + 35.5 = 34.5
Use proportions to find the grams of KCl
34.5 g of KCl ---------------- 1 mol
x ---------------- 0.75 moles
x = (0.75 x 34.5) / 1
x = 25.9 g of KCl
