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3 years ago

How much of a 400g sample remains after 4 years if a radioactive isotope has a half-life of 2 years?

Chemistry

1 answer:

timama [110]3 years ago

3 0

Answer:

100 g

Explanation:

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

Original amount (N₀) = 400 g

Time (t) = 4 years

Half-life (t½) = 2 years

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Time (t) = 4 years

Half-life (t½) = 2 years

Number of half-lives (n) =?

n = t / t½

n = 4 / 2

n = 2

Thus, 2 half-lives has elapsed.

Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:

Original amount (N₀) = 400 g

Number of half-lives (n) = 2

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

N = 1/4 × 400

N = 0.25 × 400

N = 100 g

Thus, the amount of the radioactive isotope remaing is the 100 g.

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Chlorine is the most widely used disinfectant for killing pathogens during water treatment. Determine the kilograms of chlorine

makkiz [27]

Explanation:

The given data is as follows.

Flow of chlorine (Q) = $10,000 m^{3}/day$

Amount of liter present per day is as follows.

$10,000 \times 10^{3} l/day$

It is given that dosage of chlorine will be as follows.

10 mg/l = $10 \times 10^{-6}$ kg/l

Therefore, total chlorine requirement is as follows.

Total chlorine requirement = $(10,000 \times 10^{3}) \times (10 \times 10^{-6})$ kg/day

= 100 kg/day

Thus, we can conclude that the kilograms of chlorine used daily at the given water treatment plant is 100 kg/day.

4 0

3 years ago

Enthalpies of reaction calculated from bond energies and from enthalpies of formation are often, but not always, close to each o

jolli1 [7]

The enthalpy change in a reaction is given by-

ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants

This can be expressed in terms of bond energy as-

ΔH°rxn = BEreactants - BEproducts

Therefore, the calculated bond energy according to the above equation will be-

ΔH°rxn = [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H) ] - [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H) = 0 kJ/mol

<h3>What is enthalpy change?</h3>

Enthalpy change is a measure of the energy emitted or consumed in a reaction. This can be determined using the following equation which involves standard enthalpy of reactant and product formation:

ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants

<h3>What is bond energy?</h3>

Bond energy is defined as the amount of energy needed to dissociate a mole of molecules into their individual atoms.

Learn more about the Enthalpy Change here:

brainly.com/question/14047927

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2 years ago

write the balanced chemical equation of the reaction between strong acid and weak base also mention the types of salt obtained i

steposvetlana [31]

Answer:

write the balanced chemical equation of the reaction between strong acid and weak base also mention the types of salt obtained in the reaction

Explanation:

HCl → H+(aq) + Cl−(aq) (100%)

Any acid that dissociates 100% into ions is called a strong acid. If it does not dissociate 100%, it is a weak acid. HC2H3O2 is an example of a weak acid:

HC2H3O2→ H+(aq) + C2H3O2−(aq) (~5%)

Because this reaction does not go 100% to completion, it is more appropriate to write it as an equilibrium:

HC2H3O2 ⇄ H+(aq) + C2H3O2−(aq)

As it turns out, there are very few strong acids, which are given in Table 12.2 “Strong Acids and Bases”. If an acid is not listed here, it is a weak acid. It may be 1% ionized or 99% ionized, but it is still classified as a weak acid.

8 0

3 years ago

Calculate the radius of tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6 g/cm, and an atomic wei

Ivahew [28]

Answer:

The radius of tantalum (Ta) atom is $R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

Explanation:

From the Body-centered cubic (BBC) crystal structure we know that a unit cell length <em>a </em>and atomic radius <em>R </em>are related through

$a=\frac{4R}{\sqrt{3} }$

So the volume of the unit cell $V_{c}$ is

$V_{c}= a^3=(\frac{4R}{\sqrt{3} } )^3=\frac{64\sqrt{3}R^3}{9}$

We can compute the theoretical density ρ through the following relationship

$\rho=\frac{nA}{V_{c}N_{a}}$

where

n = number of atoms associated with each unit cell

A = atomic weight

$V_{c}$ = volume of the unit cell

$N_{a}$ = Avogadro’s number ( $6.023 \times 10^{23}$ atoms/mol)

From the information given:

A = 180.9 g/mol

ρ = 16.6 g/cm^3

Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.

We can use the theoretical density ρ to find the radio <em>R</em> as follows:

$\rho=\frac{nA}{V_{c}N_{a}}\\\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}$

Solving for <em>R</em>

$\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}$