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2 years ago

Find the amount in grams of 2.0 x 10^23 atoms Cu

Chemistry

1 answer:

-Dominant- [34]2 years ago

5 0

Answer:

21.182 g

Explanation:

There are about (6.0)(10^23) atoms in one mole of a substance, so the given sample has about 0.333 mol of Cu.

The atomic mass of Cu is 63.546 g/mol, meaning that the answer is about 21.182 g

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A sample of carbon monoxide gas is initially in a 5858 mL container. The gas is then moved to 3.29 L container at a temperature

Doss [256]

Answer:

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Explanation:

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3 years ago

Please hep its my last question

atroni [7]

Answer:Noble gases:

are highly reactive.

react only with other gases.

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are not very reactive with other elements.

Explanation:Noble gases:

are highly reactive.

react only with other gases.

do not appear in the periodic table.

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2 years ago

If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution

Shtirlitz [24]

Explanation:

a) Using Beer-Lambert's law :

Formula used :

$A=\epsilon \times c\times l$

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

$\epsilon$ = molar absorptivity of this solution = $5.56\times 10^4 M^{-1} cm^{-1}$

$0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c$

$c=1.4163\times 10^{-5} M=14.16 \mu M$

( $1\mu M=10^{-6} M$ )

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) $c=1.4163\times 10^{-5} mol/L$

1 L of solution contains $1.4163\times 10^{-5}$ moles of red dye.

Mass of $1.4163\times 10^{-5}$ moles of red dye:

$1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g$

$(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100$

$red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%$

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = $c=1.4163\times 10^{-5} M$

Concentration of red solution after dilution = c'

$c=c'\times 5$

$1.4163\times 10^{-5} M=c'\times 5$

$c'=2.83\times 10^{-6} M$

The final concentration of the diluted solution is $2.83\times 10^{-6} M$

8 0

3 years ago

How many cyanides are needed to bond with Strontium?

Gala2k [10]

Answer:

Explanation:

6 0

3 years ago

The exhaust gas from an automobile contains 3% by volume of carbon monoxide (CO). Express this concentration in mg/m3 at 25oC an

ser-zykov [4K]

Answer:

$24540\frac{mg}{m^3}$

Explanation:

Hello,

In this case, since the 3% by volume is represented as:

$\frac{3L\ CO}{L\ gas}$

By using the ideal gas equation we compute the density of CO:

$\rho =\frac{MP}{RT} =\frac{28g/mol*1atm}{0.082\frac{atm*L}{mol*K}*298K}= 0.818g/L$

Then we apply the conversion factors as follows:

$=\frac{3L\ CO}{100L\ gas}*\frac{0.818g\ CO}{1L\ CO} *\frac{1000mg\ CO}{1g\ CO} *\frac{1000L\ gas}{1m^3\ gas} \\\\=24540\frac{mg}{m^3}$

Regards.

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3 years ago