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Afina-wow [57]
3 years ago
8

Hydrogen gas explosive within range of 4%-75% v/v. Assuming that each student in your class produces 6 L of H2 and that the lab

is not ventilated, is danger of an explosion? Show your calculations to back up your answer.
Chemistry
1 answer:
SpyIntel [72]3 years ago
3 0
To determine whether the amount of H2 in the lab is dangerous, we first need to know how much hydrogen gas is present in the room in units of percent by volume. For this particular problem, we cannot exactly determine since we do not know the total volume of the room. Hope this answers the question.
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How much energy is required to vaporize 155 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol
netineya [11]

The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ

<h3>Further explanation</h3>

Enthalpy is the amount of system heat at constant pressure.

The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.

\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The standard unit is kilojoules (kJ)

The enthalpy change symbol (ΔH) is usually written behind the reaction equation.

Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.

Examples of water evaporation:

 H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ

The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid

  • 155 g of butane

relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol

tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]

\large mole\:=\:\large \frac{155}{58}

mole = 2,672

Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:

23.1 kJ / mol x 2,672 mol = 61,723 kJ

<h3>Learn more</h3>

the heat of vaporization

brainly.com/question/11475740

The latent heat of vaporization

brainly.com/question/10555500

brainly.com/question/4176497

Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic

4 0
4 years ago
Read 2 more answers
You're legally obligated to yield the right-of-way:
kenny6666 [7]
It is most likely B.
3 0
3 years ago
Ca(s)+2hno3(aq)→ca(no3)2(aq)+h2(g) identify the oxidizing agent.
aleksandr82 [10.1K]

To find for the oxidizing agent, first let us write the half reactions of this complete chemical reaction:

Ca = Ca2+ + 2e- <span>
2 H+ + 2e- = H2</span>

 

The oxidizing agent would be the substance of the element that is reduced. We know that an element is reduced when an electron is added to it. In this case, the element being reduced is H. Therefore the oxidizing agent is HNO3.

 

Answer:

<span>HNO3</span>

8 0
3 years ago
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Why is the structure of a molecule important to its function
Scilla [17]

Answer:

It determines how biological molecules recognize and respond to one another with specificity.

Explanation:

A molecule has a characteristic size and shape. The precise shape of a molecule is usually very important to its function in the living cell. Molecular shape is crucial in biology because it determines how biological molecules recognize and respond to one another with specificity.

3 0
3 years ago
Draw the structure of the product of each step in the following three-step synthesis. Show the formal charges, if applicable. As
Dvinal [7]

Answer:

Second step: 4-bromo-1-methyl-2-nitrobenzene.

Third step: 1.5-dibromo-2-methyl-3-nitrobenzene.

Explanation:

To solve this exercise I will use the concepts of electrophilic substitution. In these reactions, a functional group is displaced by an electrophile. In the attached image are the two main products.

5 0
3 years ago
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