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Andrews [41]
3 years ago
8

In the presence of sulfuric acid, this alcohol is dehydrated to form an alkene through an E1 mechanism. In the box, draw the maj

or alkene product of this reaction.

Chemistry
1 answer:
Sophie [7]3 years ago
5 0

Answer:

figure is attached

Explanation:

When we treat alcohol with H₂SO₄ we get elimination as the major product.

As we can see in the given reaction that in step 1 the lone pair of electrons of oxygen attached to the alcohol make a bond with the hydrogen of H₂SO₄.

In the 2nd step H₂O gets detached from the parent ring which generated a positive charge on the ring.

In the 3rd step elimination of hydrogen from the carbon next to the carbonium carbon results into formation of an alkene.

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Agata [3.3K]

Answer:

Nose

Explanation:

Because every time I touch it's nose, it will hide itself into the shell faster than other places

8 0
3 years ago
During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of
almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

  • Ca: 40.078;
  • C: 12.011;
  • O: 15.999.

Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

  • Ca: 40.078;
  • Cl: 35.45.

M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

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4 years ago
What element is 1s2 2s22p6 3s23p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f10
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3 years ago
Read 2 more answers
PLEASE HELP!!!!!!!!!
andreev551 [17]

The average atomic mass of the imaginary element : 47.255 amu

<h3>Further explanation  </h3>

The elements in nature have several types of isotopes  

Isotopes are elements that have the same Atomic Number (Proton)  

Atomic mass is the average atomic mass of all its isotopes  

Mass atom X = mass isotope 1 . % + mass isotope 2.% ..

isotope E-47 47.011 amu, 87.34%

isotope E-48 48.008 amu, 6.895

isotope E-49 50.009 amu, 5.77%

The average atomic mass :

\tt avg~mass=0.8734\times 47.011+0.06895\times 48.008+0.0577\times 50.009\\\\avg~mass=41.059+3.310+2.886\\\\avg~mass=47.255~amu

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3 years ago
What type of compound is fe3n2
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Fe3N2, also known as Iron (II) nitride, is an ionic compound.


Ionic compounds are compounds that consists of metals and non-metals bonded with ionic bonds. The metal ion gives up electron(s) to the non-metals.


Since iron is a metal and nitrogen is an non-metal, the bond they would form would be an ionic bond. Iron gives up 2 electrons to form iron(II) ion, while nitrogen gains 3 electrons to form nitride ion. Since one iron cannot let a nitrogen gain 3 electrons, so in the compound, there would be 3 iron (ii) ions that has given up 6 electrons in total while 2 nitride ions have gained 6 electrons in total.

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