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3 years ago

How many moles of C2H2 are needed to react completely with 84.0 mol O2?

Chemistry

2 answers:

docker41 [41]3 years ago

5 0

Answer: 33.6 moles

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Combustion of ethylene is represented by following balanced equation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

According to stoichiometry:

5 moles of $O_2$ react completely with = 2 moles of $C_2H_2$

Thus 84.0 moles of $O_2$ react completely with = $\frac{2}{5}\times 84.0=33.6$ moles of $C_2H_2$

Thus 33.6 moles of $C_2H_2$ are needed to react completely with 84.0 mol $O_2$

Airida [17]3 years ago

3 0

33.6 moles are needed to completely react with 84.0 moles of O2

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The vapor pressure of water and the partial pressure of hydrogen contribute to the total pressure of 715 torr. What is the parti

Lostsunrise [7]

Answer:

0.91 atm is the partial pressure of just hydrogen gas.

Explanation:

Vapor pressure of water , p= 0.0313 atm

Partial pressure of hydrogen gas = $p_{H_2}$

Total pressure of the water vapors and hydrogen gas = P = 715 Torr

1 atm = 715 Torr

$715 Torr=\frac{715}{760} atm=0.94 atm$

According Dalton's law of partial pressure:

$P=p+p_{H_2}$

$0.94 atm=0.0313 atm+p_{H_2}$

$p_{H_2}=0.94 atm - 0.0313 atm =0.9087 atm \approx 0.91 atm$

0.91 atm is the partial pressure of just hydrogen gas.

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3 years ago

When a piece of metal A is placed in a solution containing ions of metal B, metal B plates out on the piece of A.(a) Which metal

zmey [24]

While metal B is being reduced from an ion to a solid metal, metal A is being oxidized.

Metal A is being displaced because, through oxidation, it transforms from a solid metal to ions in a solution.

What is Oxidation ?

Redox reactions include a change in the oxidation state of the substrate. Loss of electrons or a rise in an element's oxidation state are both considered to be oxidation.

Gaining electrons or lowering the oxidation state of an element or its constituent atoms are both examples of reduction.

In general, the term "reduction" refers to a reaction in which a chemical receives additional electrons; the compound that gets electrons is referred to as being reduced. We can think of the transformation of a ketone or an aldehyde into an alcohol as a two-electron reduction because hydride can be thought of as a proton plus two electrons.

So finally we can say that Metal A is being oxidized.

To know more about Oxidation please click here : brainly.com/question/25886015

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2 years ago

Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.

bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr (1)

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = 181,66 kJ/mol

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

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ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

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