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2 years ago

How many moles of oxygen are consumed in the complete combustion of 1.60 moles of benzene, c6h6?

Chemistry

1 answer:

My name is Ann [436]2 years ago

3 0

In the complete combustion of 1.60 moles of benzene, C6H6, 12 moles of oxygen, O2, is consumed.

Combustion is defined as the process of burning something. In chemistry, combustion refers to the chemical process between a fuel and an oxidant, usually oxygen to produce heat and light in the form of flame.

In a complete combustion, oxygen is sufficient to react with any hydrocarbons to produce carbon dioxide and water.

Balancing the combustion reaction of benzene, we have:

2C6H6 + 15 O2 = 12CO2 + 6H2O

Based on the balanced combustion reaction above, 2 moles of benzene requires 15 moles of oxygen to have a complete combustion.

If we have 1.60 moles C6H6,

moles O2 = mole ratio x mole of benzene

moles O2 = (15 moles O2/2 moles C6H6) x 1.60 moles C6H6

moles O2 = 12

To learn more about combustion: brainly.com/question/9913173

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Enter your answer in the provided box.S(rhombic) + O2(g) → SO2(g) ΔHo rxn= −296.06 kJ/molS(monoclinic) + O2(g) → SO2(g) ΔHo rxn=

IgorC [24]

Answer: $\Delta H^0=+0.3kJ/mol$ .

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$S_{rhombic}+O_2(g)\rightarrow SO_2(g)$ $\Delta H^0_1=-296.06kJ$ (1)

$S_{monoclinic}+O_2(g)\rightarrow SO_2(g)/tex] [tex]\Delta H^0_2=-296.36kJ$ (2)

The final reaction is:

$S_{rhombic}\rightarrow S_{monoclinic}$ $\Delta H^0_3=?$ (3)

By subtracting (1) and (2)

$\Delta H^0_3=\Delta H^0_1-\Delta H^0_2=-296.06kJ-(-296.36kJ)=0.3kJ$

Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ

3 0

3 years ago

The organic compounds that are the starting materials for compounds like fats and amino acids are called

alisha [4.7K]

The answer is; acetyl-CoA

This organic compound is an intermediate of the TCA/Citric/Krebs cycle. To make fats, the coenzyme is carboxylated to manolyCoA. This becomes a precursor fo palmitate and other lipids in the body by the addition of more manolyCoA.

The coenzyme is also a precursor to the formation of alpha-ketoglutarate, another intermediate of the TCA cycle, which is also a precursor in the formation of proteins.

5 0

3 years ago

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A 1.0 L buffer solution is 0.300 M HC2H3O2 and 0.045 M LiC2H3O2. Which of the following actions will destroy the buffer?

zheka24 [161]

Answer:

b) Adding 0.075 moles of HCl

Explanation:

A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa (Weak base with its conjugate acid).

The buffer of the problem is the acetic acid / lithium acetate.

The addition of any moles of the acid and the conjugate base will not destroy the buffer, just would change the pH of the buffer. Thus, a and c will not destroy the buffer.

The addition of an acid (HCl) or a base (NaOH), produce the following reactions:

HCl + LiC₂H₃O₂ → HC₂H₃O₂ + LiCl

The acid reacts with the conjugate base to produce the weak acid.

And:

NaOH + HC₂H₃O₂ →NaC₂H₃O₂ + H₂O

The base reacts with the weak acid to produce conjugate base.

As the buffer is 1.0L, the moles of the species of the buffer are:

HC₂H₃O₂ = 0.300 moles

LiC₂H₃O₂ = 0.045 moles

The reaction of HCl with LiC₂H₃O₂ consume all LiC₂H₃O₂ -because there are an excess of moles of HCl that react with all LiC₂H₃O₂-

As you will have just HC₂H₃O₂ after the reaction, the addition of b destroy the buffer.

In the other way, 0.0500 moles of NaOH react with the HC₂H₃O₂ but not consuming all HC₂H₃O₂, thus d doesn't destroy the buffer.

5 0

3 years ago

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $2.3\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$2.3\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.28\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.28\times 10^{-5}M$

2) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.28\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.28\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.034\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.034\times 10^{-5}M$

3) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.034\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.034\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=4.0\times 10^{-8}M$

Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

5 0

3 years ago

Suggest a way that chemistry knowledge could help you make decisions about a claim

Tanya [424]

Answer and explanation;

-Knowledge of chemistry allows the public to make informed decisions.

-Since chemistry is the basis of everything in life, knowing the basics allows us to make smarter decisions. With a knowledge of chemistry, one can form better opinions about the day’s news. An individual will be able to distinguish the truth from the lies in environmental reporting, and better understand the meaning of a chemical spill.

-One can even better decide what your opinion on public policy should be by using your own knowledge of science rather than trying to figure out which politician or pundit is the least untrustworthy.

4 0

3 years ago

Read 2 more answers