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White raven [17]
4 years ago
6

One problem with organic chemistry is that we tend to focus on one molecule at a time, we forget that there are many, many molec

ules undergoing the reaction. Assume you started with exactly 1 g of bromobenzene, how many MOLECULES OF bromobenzene is this?
Chemistry
1 answer:
dangina [55]4 years ago
7 0

Answer:

1g of bromobenzene = 3.84× 10^21 molecules

Explanation:

Molar mass of bromobenzene (C6H5Br)= 79.9+(12×6)+ 5= 156.9

156.9g(1mole) has 6.023×10^23 molecules

This implies that 1g will contain 6.023×10^23/156.9 = 3.84×10^21 molecules

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3 years ago
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A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is conve
Helen [10]

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

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

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}

3 0
3 years ago
Write a balanced equation for the following:
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(ANS1)— P4 + 5O2 ---> 2P2O5

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7 0
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xenn [34]

Answer: a

Explanation:

this is common sense

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sdas [7]

The four metals that will not replace hydrogen in an acid are Cu, Ag, Au, Hg.

Explanation:

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