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4 years ago

11

The first order reaction A → B, has k = 2.10. If the initial concentration of reactant A is 1.500 M, how long will it take [A] =

0.600 M?

1 answer:

Semenov [28]4 years ago

6 0

Answer:

$ln( \binom{at}{ao}{} ) = - kt$

Explanation:

$ln( \binom{0.6}{1.5} ) = - 2.1t$

$ln(0.4) = - 2.1t$

T = 0.436.

I used the formula for first order reaction.

I hope you get this.

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With regards to Lewis-electron diagrams, which of the following compounds has resonance forms?

andre [41]

Explanation:

The resonance compounds are compounds that have the same position of the atoms (same quantity and elements) but the position of the electrons arround them is different in each resonance compound.

In reality, the compound switches between all the resonance structures it has.

One example of resonance Lewis structures is the Ozone's as can be seen in the figure.

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3 years ago

In a similar experiment, unlabeled 4-butanolide was allowed to stand in an acidic solution in which the water had been labeled w

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Answer:

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Explanation:

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3 years ago

What is radioactive decay?

swat32

Answer:

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3 years ago

Carbon tetrachloride, CCl4, once used as a cleaning fluid and as a fire extinguisher, is produced by heating methane and chlorin

Elena-2011 [213]

Answer:

1.882 g

Explanation:

Data Given

mass of Cl₂ = 33.4 g

mass of CH₄ = ?

Reaction Given:

CH₄+ 4Cl₂ --------→ CCl₄ + HCl

Solution:

First find the mass of CH₄ from the reaction that it combine with how many grams of Chlorin.

Look at the balanced reaction

CH₄ + 4Cl₂ --------→ CCl₄ + 4HCl

1 mol 4 mol

So 1 mole of CH₄ combine with 4 moles of Cl₂

Now

convert the moles into mass for which we have to know molar mass of CH₄ and Cl₂

Molar mass of Cl₂ = 2 (35.5)

Molar mass of Cl₂ = 71 g/mol

mass of Cl₂

mass in grams = no. of moles x molar mass

mass of Cl₂ = 4 mol x 71 g/mol

mass of Cl₂ = 284 g

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Molar mass of CH₄= 16 g/mol

mass of CH₄

mass in grams = no. of moles x molar mass

mass of CH₄= 1 mol x 16 g/mol

mass of CH₄ = 16 g

So,

284 g of Cl₂ combine with 16 g of methane ( CH₄ ) then how many grams of CH₄ is needed to combine with 33.4 g of Cl₂

Apply unity Formula

284 g of Cl₂ ≅ 16 g of methane ( CH₄ )

33.4 g of Cl₂ ≅ X g of methane ( CH₄ )

By cross multiplication

X g of methane ( CH₄ ) = 16 g x 33.4 g / 284 g

X g of methane ( CH₄ ) = 1.88 g

1.882 g of methane (CH₄) will needed to combine with 33.4 g of Cl₂

So

methane (CH₄) = 1.882 g

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3 years ago

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Licemer1 [7]

Answer:

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