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3 years ago

If you have 3.45 × 1015 atoms of iron, how many moles of iron do you have?

Chemistry

2 answers:

kumpel [21]3 years ago

7 0

3501.75

is the answer

i think

(your welcome)

Vinil7 [7]3 years ago

4 0

It's 5.73x10^-9 moles of iron***

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Draw the most stable resonance structure for the intermediate in the electrophilic aromatic bromination of aniline, anisole, and

ASHA 777 [7]

Answer:

Here's what I get

Explanation:

(a) Intermediates

The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).

(b) Relative Stabilities

The relative stabilities decrease in the order shown.

N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.

(c) Relative reactivities

The relative reactivities would be

C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃

4 0

3 years ago

By friends leaving this app

Lunna [17]

Answer:

Whyyyyyyyyyy

Don't please...

3 0

3 years ago

Read 2 more answers

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .