PLEASE HELP ASAP.... Part Two

Radioactive decay can be described by the following equation where is the original amount of the substance, is the amount of the

soldi70 [24.7K]

Answer:

Iron remains = 17.49 mg

Explanation:

Half life of iron -55 = 2.737 years (Source)

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{2.737}\ year^{-1}$

The rate constant, k = 0.2533 year⁻¹

Time = 2.41 years

$[A_0]$ = 32.2 mg

Using integrated rate law for first order kinetics as:

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

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$[A_t]=32.2\times e^{-0.2533\times 2.41}\ mg$

$[A_t]=32.2\times e^{-0.610453}\ mg$

$[A_t]=17.49\ mg$

Iron remains = 17.49 mg

8 0

3 years ago

Do cells divide twice in asexual reproduction, sexual reproduction, or both?

igor_vitrenko [27]

Cells divide in asexual reproduction!

I hope this helps

please mark me as brainliestt!!

8 0

3 years ago

Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine

Lilit [14]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 34.9689 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 36.9659 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7577\times 100=75.77\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.7577)=0.2423\times 100=24.23\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

6 0

3 years ago

Write the overall molecular equation for the reaction of hydroiodic acid ( HI ) and potassium hydroxide. Include physical states

Setler [38]

Answer:

HI(aq) + NaOH(aq)-------------> NaI(aq) + H2O(l)

Explanation:

The molecular reaction between hydroiodic acid and aqueous sodium hydroxide is shown above. It is a reaction of one mole of hydroiodic acid with one mole of sodium hydroxide to yield salt and water only. It is a neutralization reaction. Hydrogen iodide dissolves in water to produce hydronium ions which agrees with Arrhenius description of acids. Hydroiodic acid is a strong acid with PKa of -9.3

3 0

3 years ago

What is the solution to the equation below, rounded to the correct number of Significant figures? 4700 L - 281.4 L = ___ L

Tems11 [23]

Answer:

46

Explanation:

3 0

3 years ago