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Tems11 [23]
2 years ago
11

The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h

alf-life of this reaction at 1000°C?
Chemistry
1 answer:
RoseWind [281]2 years ago
6 0

Answer:

2.5×10⁶ s

Explanation:

From the question given above, the following data were obtained:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 2.8×10¯⁷

t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

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Answer:

Density independent factor

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in dense areas where people livein  very close and tight spaces

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3 years ago
Is an apple a pure substance or a mixture?
guajiro [1.7K]
An apple would be a mixture. While a mixture is a combination of elements and substances, a pure substance acts as an element itself, therefore, having the properties of only THAT certain element. 


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3 years ago
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What is produced when calcium reacts with fluorine in a synthesis reaction?
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C is the correct answer (CaF2) (sorry dont have subscript)

Explanation: synthesis reaction forms a compound and calcium reacting with fluorine produces Calcium Fluoride (CaF2) chemical name
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3 years ago
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Calculate the percent by volume of a solution that has 75 mL of solute dissolved in 375 mL of solution. Show your work and round
Mandarinka [93]

Answer:

The answer is 20 % V/V

Explanation:

We use this formula for calculate the %V/V:

%V/V= (ml solute/ml solution) x 100= (75ml/375 ml)x 100 = 20 % V/V

<em>The% V / V represents the amount of ml of solute dissolved in 100 ml of solution</em>

7 0
3 years ago
The standard reduction potentials of lithium metal and chlorine gas are as follows:Reaction Reduction potential(V)Li+(aq)+e−→Li(
meriva

Answer:

A) E° = 4.40 V

B) ΔG° = -8.49 × 10⁵ J

Explanation:

Let's consider the following redox reaction.

2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)

We can write the corresponding half-reactions.

Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq)      E°red = 1.36 V

Anode (oxidation):  2 Li(s) → 2 Li⁺(aq) + 2 e⁻         E°red = -3.04

<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>

The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.

E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V

<em>B) Calculate the free energy ΔG° of the reaction.</em>

We can calculate Gibbs free energy (ΔG°) using the following expression.

ΔG° = -n.F.E°

where,

n are the moles of electrons transferred

F is Faraday's constant

ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J

8 0
3 years ago
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