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gogolik [260]
3 years ago
11

Please can I have help on question e as I’m really stuck.I promise I’ll give out brainiest

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
5 0

The relative atomic mass of element A : 6.075 u

<h3>Further explanation </h3>

The elements in nature have several types of isotopes

Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

Atomic mass is the average atomic mass of all its isotopes

Mass atom X = mass isotope 1 . % + mass isotope 2.% ...

Element A

mass number of isotope 1 = 6 , percent : 92.5%

mass number of isotope 2 = 7, percent : 7.5

Relative atomic mass of element A

\tt = 92.5\%\times 6+7.5\%\times 7\\\\=5.55+0.525=6.075~u

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Answer:

C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}

Explanation:

The differential equation is given as:

\frac{dC}{dt} = r- kC

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Taking integral of both sides; we have:

\int\limits  \frac{dC}{r- kC} = \int\limits dt

-\frac{1}{k} In(r-kC) = t+D\\In(r-kC)=-kt-kD

r-kC=e^{-kt-kD}

r-kC=e^{-kt}e^{-kD}

r-kC=Ae^{-kt}

kC=r-Ae^{-kt}

C=\frac{r}{k}-\frac{A}{k}e^{-kt}

C(t)=\frac{r}{k}-\frac{A}{k}e^{-kt}     ------- equation (1)

If C(0)= C_o ; we have:

C(0)= \frac{r}{k}-\frac{A}{k}e^0         (where; A is an integration constant)

C_o = \frac{r}{k}- \frac{A}{k}

C_o=\frac{r-A}{k}

kC_o=r-A

A=r-kC_o

Substituting A=r-kC_o into equation (1); we have;

C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}

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