A first-order decomposition reaction has a rate constant of 0.00140 yr−1. How long does it take for [reactant] to reach 12.5% of its original value? Be sure to report your answer to the correct number of significant figures.

Answers

Answer 1

Answer:

Reactants take 504.87 yr to reach 12.5% of their original value in first-order decomposition reaction.

Equation for the first-order decomposition reaction:-

$A_(t) =A_(0) e^(-kt)$ ....(1)

Here, $A_(t)$ is the final concentration, t is the time, $A_(0)$ is the initial concentration, and k is the rate constant.

Given:-

$A_(t) =0.125A_(0)$

k= $0.00140yr^(-1)$

Substitute the above value in equation (1) as follows:-

$0.125A_(0) =A_(0) e^(-kt) \n0.125A_(0) =A_(0) e^{-k*0.00140 yr^(-1) }\nln(0.125)/(-0.00140)=t\nt=504.87 year$

So, 504.87 yr does it take for the reactant to reach 12.5% of its original value.

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Answer 2

Answer:

Final answer:

The time required for a reactant to reach 12.5% of its original value in a first-order reaction is approximately 1482 years, obtained by applying the formula for the half-life of a first-order reaction and multiplying by 3.

Explanation:

In a first-order reaction, the half-life of the reaction, which is the time it takes for half of the reactant to be consumed, is independent of the concentration of the reactant. Also, for a first-order reaction, it would take approximately 3 half-lives for the reactant to be reduced to 12.5% of its original value. The Integrated Rate Law for a First-Order Reaction can be applied to determine the time it will take.

Given the rate constant (k) is 0.00140 yr¯¹, we will use the formula for the half-life of a first-order reaction: t₁/₂ = 0.693 / k. After calculating the half-life (t₁/₂), multiply it by 3 to determine the time for the reactant concentration to reach 12.5% of its original value. Hence, it would take approximately 1482 years to reach 12.5% of the original value when rounded to the correct number of significant figures.

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Ammonium nitrate dissociates in water according to the following equation:43() = 4+()+03−()When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.1) Calculate q for the reaction. You must show your work.2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.3) Calculate ΔH for the reaction in kJ/mol. You must show your work.

Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21 ∘C∘C. Express the pressures in atmospheres to three significant digits separated by commas.

Answers

The question is incomplete, complete question is ;

A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of $O_2$ and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.

Answer:

Partial pressure of the oxygen gas is 3.91 atm.

Partial pressure of the helium gas is 20.0 atm

Total pressure of the gases is 24.0 atm

Explanation:

Moles of oxygen gas = $n_1=(51.8)/(32 g/mol)=1.619 mol$

Moles of helium gas = $n_2=(33.1 g)/(4 g/mol)=8.275 mol$

Total moles of gas = $n_1+n_2=(1.619 +8.275 ) mole=9.894 mol$

Volume of the cylinder = V = 10.0 L

Total pressure in the cylinder = P = ?

Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K

PV = nRT ( ideal gas equation )

$P=(nRT)/(V)$

$=(9.894 mol* 0.0821 atm L/mol K* 294 K)/(10.0 L)$

P = 23.88 atm ≈ 23.9

Partial pressure of the individual gas will be determined by the help of Dalton's law:

partial pressure = Total pressure × mole fraction of gas

Partial pressure of the oxygen gas

$p_(1)=P* \chi_(1)=P* (n_1)/(n_1+n_2)$

$p_1=23.88 atm* (1.619 mol)/(9.894 mol)=3.91 atm$

Partial pressure of the helium gas

$p_(2)=P* \chi_(2)=P* (n_2)/(n_1+n_2)$

$p_2=23.88 atm* (8.275 mol)/(9.894 mol)=19.97 atm\approx 20.0 atm$

Find the mole ratio of H2SO4 and H20 in the equation Fe2O3 + H2SO4 → Fe2(SO4)3 + H20.

Answers

Explanation:

$Fe2O3 +3 H2SO4 → Fe2(SO4)3 + 3H20.$

Therefore the ratio is 3:3

In each of the three reactions between NaOH and HCl, the sign of q for the water was positive. This means the the sign of q for the reaction was ______ and the reaction was ______.

Answers

Answer:

This means the the sign of q for the reaction was _NEGATIVE _____ and the reaction was _EXOTHERMIC_____.

Explanation:

In calorimetry, when heat is absorbed by the solution, the q-value of the solution will have a positive value. This means that the reaction will produce heat for the solution to absorb and thus the q-value for the reaction will be negative. This is an exothermic reaction.

Whereas, when heat is absorbed from the solution, the q-value for the solution will have a negative value. This means that the reaction will absorb heat from the solution and so the reaction is endothermic, and q value for the reaction is positive.

So, from the question, since the q-value of water is positive, it means that heat is absorbed by the solution and the reaction will produce a negative value of q and it's an exothermic reaction because the reaction produces heat for the solution.

A fruit fly is considered a consumer rather than a producer why

Answers

In an ecosystem, the only true producers are autotrophic organisms like plants and bacteria. These organisms produce energy by converting the energy from the sun into simple sugars.
All of the other organisms in the food chain, like the fruit fly, are simply consuming the energy produced by the plants/bacteria and not actually making/producing energy from a new source.

What formula was used to find the answer

Answers

Can you give more information?

A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of H2O. Determine the masses of C and H in the sample and the percentages of these elements in this hydrocarbon.

Answers

Answer: The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=0.3986g$

Mass of $H_2O=0.0578g$

Mass of sample = 0.1153 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.3986 g of carbon dioxide, $(12)/(44)* 0.3986=0.1087g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, $(2)/(18)* 0.0578=0.0066g$ of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

$\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}* 100$ ......(1)

For Carbon:

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=(0.1087g)/(0.1153g)* 100=94.27\%$

For Hydrogen:

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

$\%\text{ composition of hydrogen}=(0.0066g)/(0.1153g)* 100=5.72\%$

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

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