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adelina 88 [10]
3 years ago
5

The first-order decomposition of cyclopropane has a rate constant of 6.7 * 10-4 5-1. If the initial concentration of cyclopropan

e is 1.33 M, what is the concentration of cyclopropane after 644 s? OA) 0.43 M B) 0.15 M C) 0.94 M D) 0.86 M
Chemistry
1 answer:
Sati [7]3 years ago
7 0

Answer:

D) 0.86 M

Explanation:

Given that:

The rate constant, k = 6.7×10⁻⁴ s⁻¹

Initial concentration [A₀] = 1.33 M

Time = 644 s

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,

[A_t] is the concentration at time t

So,

[A_t]=1.33\times e^{-6.7\times 10^{-4}\times 644}

[A_t]=0.86 M

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How many moles are equal to 89.23g of calcium oxide, CaO?
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1.59moles

Explanation:

Mass  of CaO = 89.23g

Unknown

Number of moles = ?

Solution:

The mole is a unit of measurement in chemistry used to delineate the number of particles an atom contains.

A mole of a substance contains the avogadro's number of particles.

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2 years ago
Solid-solid homogeneous mixture give one example
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2 years ago
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Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO
sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂   → CaCO₃

Number of moles of CaO:

Number of moles  = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles  = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

                  CO₂         :                CaCO₃  

                  1               :                 1

                 0.31           :              0.31

                CaO           :               CaCO₃  

                 1                :                 1

                 0.26         :              0.26

The number of moles of  CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ =  26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of  CaCO₃ produced by CaO are less it will be limiting reactant.

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The tractor trailer will push the car when they hit so therefore the car will have to greatest change

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