All categories

3 years ago

Predict the products when cyclohexanol is dehydrated.

1 answer:

5 0

<span>When cyclohexanol is dehydrated using various means it creates a new hydrocarbon. This new hydrocarbon is called cyclohexene. Cyclohexene is a liquid which has no color, yet has a very strong smell to it. It is used in common industrial processes, but is also considered somewhat unstable due to it's tendency to form peroxides when exposed to light.</span>

You might be interested in

The movement of organisms into a range is called

Brrunno [24]

A. immigration, how is this chemistry?

8 0

3 years ago

write a balanced chemical equation for solid copper reacting with aqueous silver nitrate to produce aqueous copper (II) nitrate

sashaice [31]

Make sure have same amounts of species on both sides
Cu (s) + 2 AgNO3 (aq) -> Cu(NO3)2 (aq) + 2 Ag (s)

7 0

3 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

3 years ago

What is Markanikov rule?<br>

lana [24]

Markovnikov rule, in organic chemistry, a generalization, formulated by Vladimir Vasilyevich Markovnikov in 1869, stating that in addition reactions to unsymmetrical alkenes, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component ...

5 0

2 years ago

Read 2 more answers

What is the percent composition of dinotrogen pentoxide? With work shown please

iris [78.8K]

Answer:

%N = 25.94%

%O = 74.06%

Explanation:

Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅

We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.

m(N): 2 × 14.01 g = 28.02 g

Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅

We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.

m(O): 5 × 16.00 g = 80.00 g

Step 3: Calculate the mass of 1 mole of N₂O₅

We will sum the masses of N and O.

m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g

Step 4: Calculate the percent composition of N₂O₅

We will use the following expression.

%Element = m(Element)/m(Compound) × 100%

%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%

%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%

7 0

3 years ago