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3 years ago

What is the lower concentration limit (vol%) at which a mixture of ethanol in air can explode?

Chemistry

1 answer:

liubo4ka [24]3 years ago

4 0

Answer:

Lower explosive limit (LEL) of ethanol = 3.3%

Explanation:

In the case of alcohol, ethanol presents certain fire hazards. Its momentary flash point is 55ºF (12.9ºC), while the momentary flash point of gasoline is -45ºF (-42.8ºC), and the E85 mixture ranges between -20ºF and -4ºF (between -28 , 9ºC and -20ºC), and has a wider range of flammability limits than gasoline. For emergency response teams, this implies that during a release of the typical ethanol / gasoline mixture, the fuel can be expected to behave like gasoline: It is heavier than air - as we mentioned earlier - and can produce vapors and form flammable mixtures in the air, under most environmental conditions.

General properties and comparison with other inflambles products:

Flash point momentary Gasoline = -45 ° F

Ethanol = 55 ° F

E 85 = between -20º and -4º F

Flammability limits

Lower explosive limit (LEL) of ethanol = 3.3%

Upper Explosive Limit (UEL) = 19%

Lower explosive limit (LEL) of the mixture E 85 = 1.4%

Upper Explosive Limit (UEL) 85 = 19%

Lower explosive limit (LEL) of gasoline = 1.4%

Upper Explosive Limit (UEL) = 7.6%

They have a wider range than gasoline

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25cc of 5 % NaOH solution neutralized 30cc of h2sO4 solution. Whatis normality of H2SO4?

MAVERICK [17]

The normality of the H₂SO₄ that reacted with 25cc of 5 % NaOH solution is 1.1 N.

<h3>What is the molarity of 5% NaOH?</h3>

The molarity of 5% NaOH is 1.32 M

25 cc of NaOH neutralized 30cc of H₂SO₄ solution.

Equation of reaction is given below:

2 NaOH + H₂SO₄ ---> Na₂SO₄ + 2 H₂O

Molarity of H₂SO₄ = 1.32 x 1 x 25/(30 x 2) = 0.55 M

Normality = Molarity × moles of H⁺ ions per mole of acid

moles of H⁺ ions per mole of H₂SO₄ = 2

Normality of H₂SO₄ = 0.55 x 2 = 1.1 N

In conclusion, the normality of an acid is determined from the molarity and the moles of H⁺ ions per mole of acid.

Learn more about normality at: brainly.com/question/22817773

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1 year ago

select from this list the one metal or alloy that is best suited for each of the following applications, and cite at least one r

Brut [27]

List of metal or alloy that is best-suited include :

1) Block of an internal combustion engine is made of grey cast iron because it is ductile and hard enough to bear the loads.

2) Steam heat exchanger is made up of stainless steel material as it has high resistance to corrosion and has the capacity to bear high-temperature differences.

3) For jet engine turbofan blades, titanium alloys are being used because they can able to bear high pressure and high energy loads.

4) Tool steel is used to make drill bit material as it has high tensile and point strength.

5) Aluminum alloys are used for cryogenic applications as it is the best suited for work at low temperatures.

6) Magnesium is used for pyrotechnics as it gives works on the effect of heat, light, and gas.

Alloy, a metallic substance composed of two or more elements, as both a compound or a solution. The components of alloys are ordinarily themselves metals, though carbon, a nonmetal, is an essential constituent of metal.

Alloys, in trendy, are more potent and harder, less malleable, much less ductile, and greater corrosion-resistant than the alloy's predominant metallic.

An alloy is created with the aid of mixing metallic with another issue, both any other metal or a nonmetal substance. Metallic alloys are commonly made by means of melting the substances, mixing them collectively, after which letting them cool to room temperature, ensuing in a strong material.

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1 year ago

If a substance has a half life of 58 years and starts with 500 g radioactive, how much remains radioactive after 30 years?

Vilka [71]

Answer:

A = 349 g.

Explanation:

Hello there!

In this case, since the radioactive decay kinetic model is based on the first-order kinetics whose integrated rate law is:

$A=Ao*exp(-kt)$

We can firstly calculate the rate constant given the half-life as shown below:

$k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{58year}=0.012year^{-1}$

Therefore, we can next plug in the rate constant, elapsed time and initial mass of the radioactive to obtain:

$A=500g*exp(-0.012year^{-1} *30year)\\\\A=349g$

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