All categories

2 years ago

How many moles are in 1.22 * 10^20 atoms of Si

Chemistry

1 answer:

mezya [45]2 years ago

4 0

Answer:

2.03 × 10⁻⁴ moles

Explanation:

To convert atoms to moles, use Avogadro's number (6.022 × 10²³).

(1.22 × 10²⁰ atoms)/(6.022 × 10²³ atoms/mol) = 2.03 × 10⁻⁴ mol

You might be interested in

What is HC2H3O2 in science

Leona [35]

The chemical name of Hc2h3o2 is Acetic Acid.

3 0

3 years ago

Read 2 more answers

Balance the following Chemical Equation: NaBr +CaCl2-> NaCl+ CaBr2

frozen [14]

First write all of the compounds/atoms in either side then fill in existing values and balance

Na- 1
Br- 1
Ca- 1
Cl- 2

Na- 1
Cl- 1
Ca-1
Br-2

Balance to get

2NaBr+CaCl2=2NaCl+CaBr2

7 0

3 years ago

4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o

Vilka [71]

Answer: The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol$

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

$\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}$

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = $\frac{1}{1}\times 0.0711=0.0711mol$ of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

$0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g$

To calculate the percentage yield of 3-hexanol, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

$\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%$

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

4 0

3 years ago

Water vapor is not usually seen.

xeze [42]

Answer:

True

Explanation:

Water vapor is totally invisible.

Without it, there would be no clouds or rain or snow, since all of these require water vapor in order to form. All of the water vapor that evaporates from the surface of the Earth eventually returns as precipitation - rain or snow.

Helped by the One & Only #QUEEN aka #DRIPPQUEENMO

Hoped this helped! :)

5 0

3 years ago

If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of

Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.1\ g}{186.04\ g/mol}$

$Moles\ of\ ferrocene= 0.0005375\ mol$

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

$\rho=\frac{Mass}{Volume}$

or,

$Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g$

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.044\ g}{78.49\ g/mol}$

$Moles\ of\ acetyl\ chloride= 0.0005606\ mol$

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms 0.0005375 mole of monoacetylferrocene

Moles of product formed = 0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %

5 0

3 years ago