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2 years ago

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

Chemistry

1 answer:

liubo4ka [24]2 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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Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (

aleksley [76]

Answer: The amount of heat produced by the reaction is -21.36 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol$

To calculate the number of moles, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = $0.0831\text{ L. bar }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol$

To calculate the heat released of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

$-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ$

Hence, the amount of heat produced by the reaction is -21.36 kJ

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3 years ago

What ion depolarizes the membrane when it diffuses into the axon of a neuron?

Agata [3.3K]

Answer:Sodium ions

Explanation:

The membrane potential of most cells is negative, when sodium ions enters the cells, it can reverse the polarity and makes it positively charge.

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3 years ago

The average dosage of oxcarbazepine for an epileptic child between the ages of 4 and 16 is 9.00 mg per 1 kg of body weight (9.00

melamori03 [73]

Convert the child weight (37.3 pounds) to kilograms

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multiply the dose (9.00mg/kg) by the weight of the child to find how much you need to give him

A kg * 9.00 mg/1kg = "B mg"

calculate the mL of suspension dividing the "B mg" by the concentration of the suspension 60.0 mg/mL

B mg * 1mL/ 60.0 mg = C mL oxcarbazepine

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3 years ago

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What causes an ice cube to melt when removed from a freezer?

OverLord2011 [107]

Answer:

the melting process begins right away because the air temperature around the ice cubes is warmer than the temperature in the freezer

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3 years ago

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Gaseous hydrogen iodide is placed in a closed container at 425∘C, where it partially decomposes to hydrogen and iodine: 2HI(g)⇌H

murzikaleks [220]

Answer:

0.0184

Explanation:

Let's consider the following reaction at equilibrium.

2 HI(g) ⇌ H₂(g) + I₂(g)

The concentration equilibrium constant (Kc) is equal to the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

Kc = [H₂] × [I₂] / [HI]²

Kc = (4.78 × 10⁻⁴) × (4.78 × 10⁻⁴) / (3.52 × 10⁻³)²

Kc = 0.0184

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3 years ago