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2 years ago

Why are the three atoms are isotopes of magnesium

Chemistry

1 answer:

LenKa [72]2 years ago

6 0

Answer:

Magnesium is a naturally ubiquitous; (appearing & found evrywhere) element and has three naturally occurring stable isotopes, 24Mg, 25Mg and 26Mg, with relative abundance of 78.99%, 10.00% and 11.01%, respectively.

However, they differ only because a 24Mg atom has 12 neutrons in its nucleus, a 25Mg atom has 13 neutrons, and a 26Mg has 14 neutrons.

Explanation:

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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]

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

<u>Explanation:</u>

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

Draw structures from the following names, and determine which compounds are optically active:(c) 1,2-dibromo-2-methylbutane

dolphi86 [110]

The given compound 1,2-dibromo-2-methylbutane is an optically active compound .

Because this compound does not have plane of symmetry (POS) and center of symmetry (COS) i.e. does not have di-symmetry . And also forms non superimposable mirror image . the compound is optically active .

It has chiral center.

Here , the chiral carbon is attached to the 4 distinct groups such as : methyl , ethyl , bromine , bromomethane .

<h3>What is di-symmetry?</h3>

Di-symmetry is that which have no center of symmetry and plane of symmetry and alternate axis of symmetry .

<h3>Chiral center :</h3>

Have Sp3 hybridized center (4sigma bond ) .

4 distinct group is attached to the chiral atom. form non -superimposable mirror image .

<h3>What is optical isomerism ?</h3>

Same molecular formula and same structural formula . also have same physical and chemical properties .

They differ in their behavior towards plane polarized light (ppl) .

Learn more about chiral center here:

brainly.com/question/9522537

#SPJ4

8 0

1 year ago

Indicate the number of unpaired electrons for following: [noble gas]ns2np5

Fantom [35]

1 unpaired electron.

4 0

3 years ago

The National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. The units the NWS

pashok25 [27]

Answer:

d. 103.3

Explanation:

In the given question, the National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. And the units of atmospheric pressure used for reporting the atmospheric pressure data are inches of mercury. For a barometric pressure of 30.51 inches of mercury, we can calculate the pressure in kPa as follow:

In principle, 3.386 kPa is equivalent to the atmospheric pressure of 1 inch of mercury. Thus, 30.51 inches of mercury is equivalent to 30.51 in *(3.386 kPa/1 in) = 103.307 kPa.

Therefore, a barometric pressure of 30.51 inches of mercury corresponds to _____103.3_____ kPa.

4 0

2 years ago

9.8 x 10^-6 regular notation

topjm [15]

Answer:

0.0000098 should be the answer

Explanation:

3 0

3 years ago