Swim ability is it materials ability to burn in the presence of

Chemistry

1 answer:

eduard2 years ago

6 0

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.

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Vanillin (used to flavor vanilla ice cream and other foods) is the substance whose aroma the human nose detects in the smallest

balu736 [363]

Answer:

Cost to supply enough vanillin is $3.2\$$

Explanation:

Threshold limit of vanillin in air is $2.0\times 10^{-11}g$ per litre means there should be $2.0\times 10^{-11}g$ of vanillin in 1L of air to detect aroma of vanillin.

$1ft^{3}=28.32L$

So, $5.0\times 10^{7}ft^{3}=(5.0\times 10^{7}\times 28.32)L$

So amount of vanillin should be present to detect = $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g$

As cost of 50 g vanillin is $112\$$ therefore cost of $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g$ vanillin = $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32\times 112)\$ = 3.2\$$