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Illusion [34]
2 years ago
6

If you attempt to produce work using the formation reaction of methanol (CH3OH), and the reaction is carried out reversibly, wha

t is the minimum amount of heat wasted?
Chemistry
1 answer:
Veronika [31]2 years ago
7 0

The minimum amount of heat wasted in the production of methanol is 239kJ/mol of heat.

<h3>What is the heat wasted?</h3>

Now we know that the enthalpy is the heat that is released or absorbed in a reaction. The reaction involves the combination of reactants that leads to the formation of the product(s).

Now we know that the production of methanol leads to the evolution of about 239kJ/mol of heat. This heat is wasted as it goes to no good use but is just lost to the environment.

As such, the minimum amount of heat wasted in the production of methanol is 239kJ/mol of heat.

Learn more about methanol:brainly.com/question/24077457

#SPJ1

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Determine the mmol of both starting materials (factoring in that formic acid is not pure, but rather 88% weight/volume, or 88g/1
Oxana [17]

Solution :

Molecular      Molar Mass       Volume      Density       Mass      Moles      nmoles

formula            (g/mol)               (mL)          (g/mL)           (g)

$C_6H_8N_2$            108.14                                                    0.108      0.001          1

HCOOH           46.02                0.064          1.22     0.07808     0.0017       1.7

mmoles of o-phenylenediamine = 1 mmoles

mmoles of formic acid = 1.7 \approx 2 mmoles

From the reaction of o-phenylenediamine and formic acid, we see,

1 mmole of o-phenylenediamine reacts with 1 mmole of formic acid.

But here, 2 mmoles of the formic acid , this means that the formic acid is an excess reagent and the o-phenylenediamine is the limiting reagent here.

The amount of product depends on the limiting reagent that is o-phenylenediamine. So, 1mmole of o-phenylenediamine will give 1mmole of product.

molar mass of Benzimidazole = 118.14 g/mol

mmoles of Benzimidazole formed = 1 mmol

Mass of benzimidazole formed = molar mass x \frac{nmoles}{1000}

                                                    $=\frac{118.14 \times 1}{1000}$

                                                     = 0.11814 g

So the theoretical yield of Benzimidazole is = 0.118 g = 118mg

5 0
3 years ago
What is the atomic number of a sodium atom that has 11 protons and 12 nucleus
Pani-rosa [81]

Answer:

The Atomic Number is 11

Explanation:

7 0
3 years ago
What is true of electrons?
Julli [10]

Answer:

electrons are negatively charged and are located in the electron cloud (outside the nucleus)

6 0
3 years ago
Read 2 more answers
How much heat is absorbed when 90.5 g of ice is heated from -11.0 °C to 145.0 °C?
Nadusha1986 [10]

Answer:

Q(total) = 283Kj

Explanation:

5 Heat Transitions …

Specific Heats => c(s) = 0.50cal/g∙⁰C,  c(l) = 1.0 cal/g∙⁰C, c(g) = 0.48 cal/g∙⁰C

Phase Transition Constants => ΔHᵪ = Heat of Fusion = 80 cal/g; ΔHᵥ = Heat of Vaporization = 540cal/g

Note => Phase change regions => no temp. change occurs when 2 phases are in contact (melting and evaporation). Only when single phase substance exists (s, l or g) does temperature change occur. See heating curve for water diagram. The increasing slopes are temperature change regions and heat flow is given by Q =mcΔT. The horizontal slopes are phase changes ( melting & evaporation) and heat flow for each of those regions is given by Q = m·ΔH. Each transition energy is calculated individually (see below) and added to obtain the total heat flow needed.

Q = mcΔT for temperature change regions of the heating curve (single phase only)

Q = m∙ΔH for phase transition regions of the heating curve (2 phases in contact)

Solid (ice) => Melting Pt  => Q(s) = mcΔT = (90.5g)(0.50cal/g∙⁰C)(11⁰C) = 478 cal

Melting (s/l) => Liquid (water) =>   Q(s/l) = m∙ΔHᵪ = (90.5g)(80cal/g) = 7240 cal

Liquid (water) => Boiling Pt => Q(l) = mcΔT = (90.5g)(1.0cal/g∙⁰C)(100⁰C) = 9050 cal

Boiling (l/g) => Gas (steam) => Q(l/g) = m∙ΔHᵥ = (90.5g)(540cal/g) = 48,870 cal

Gas (steam) => Steam @ 145⁰C => Q(g = mcΔT = (90.5g)(0.48cal/g∙⁰C)(45⁰C) = 2036 cal

Total Heat Transfer (Qᵤ) = Q(s) + Q(s/l) + Q(l) + Q(l/g) + Q(g)  

                                 = 478cal +7240cal + 9050 cal + 48,870cal + 2036cal

                                 = 67,674 cal x 4.184 j/cal = 283,148 joules = 283 Kj

4 0
4 years ago
What is the number of electrons shared in the multiple carbon-carbon bond in one molecule of 1-pentyne?
Vesnalui [34]
The correct answer for the given question above would be option A. The number of electrons shared <span> in the multiple carbon-carbon bond in one molecule of 1-pentyne is 6. -YNE is considered as a triple bond. This would mean 3 pairs of electrons of 6 electrons. Hope this answers your question. Have a great day!</span>
8 0
3 years ago
Read 2 more answers
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