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valkas [14]
2 years ago
12

A student wants to form 2‑hexanol using acidic hydration. He finds four alkenes in the inventory cabinet that could be possible

reagents for this reaction. Which of the alkene starting material(s) will generate high yields of 2‑hexanol (or hexan‑2‑ol) after acidic hydration? An unknown alkene is treated with H 3 O plus to give 2 hexanol, a 6 carbon chain with a hydroxyl group on cabron 2.

Chemistry
1 answer:
julia-pushkina [17]2 years ago
8 0

Answer:

Look on the picture.

Explanation:

He could find only 2 isomers of n-hexane alkenes for this reaction. Other two could be marked from other direction.

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A sample of an unknown compound was decomposed and found to be composed of 1.36 mol oxygen, 4.10 mol hydrogen, and 2.05 mol carb
Flura [38]

Answer:

C3H6O2

Explanation:

To find the empirical formula of the compound, we divide the amount in moles of each of the elements by the amount in mole of the element with the smallest number of mole. In this question, the element with the smallest number of moles is oxygen with 1.36 mole. Hence, we divide the number of moles of each element by this.

H = 4.10/1.36 = 3

O = 1.36/1.36 = 1

C = 2.05/1.36 = 1.5

We then multiply through by 2 to yield the compound with the empirical formula C3H6O2

7 0
2 years ago
a block of iron has a mass of 826g. what is the mass of a block of magnesium that has the same volume as the block of iron? the
fiasKO [112]

Answer is: the mass of a block of magnesium is 177.75 grams.

m(Fe) = 826 g.

d(Fe) = 7.9 g/cm³.

1) Calculate volume of iron and magnesium:

d(Fe) = m(Fe) ÷ V(Fe).

V(Fe) = m(Fe) ÷ d(Fe).

V(Fe) = 826 g ÷ 7.9 g/cm³.

V(Fe) = V(Mg) = 104.56 cm³.

2) Calculate mass of magnesium:

m(Mg) = V(Mg) · d(Mg).

m(Mg) = 104.56 g/cm³ · 1.7 g/cm³.

m(Mg) = 177.75 g.

8 0
2 years ago
A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr
rusak2 [61]

Answer:

Explanation:

Given parameters:

Initial temperature T₁  = 25.2°C  = 25.2 + 273  = 298.2K

Initial pressure  = P₁  = 0.6atm

Final temperature = 72.4°C   = 72.4 + 273  = 345.4K

Unknown:

Final pressure = ?

Solution:

To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

                  \frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

where P and T are temperatures, 1 and 2 are initial and final temperatures.

 Input the parameters and solve;

          \frac{0.6}{298.2}   = \frac{P_{2} }{345.4}  

          P₂   = 0.7atm

         

3 0
2 years ago
Sodium thiosulfate (Na2S2O3), photographer’s
vova2212 [387]

3Na2S2O3 + AgBr ------>Na5[Ag(S2O3) 3] +NaBr

from equation 3 mol 1 mol

given x mol 0.10 mol


x= (3*0.10)/1=0.30 mol Na2S2O3


Answer: 0.30 mol Na2S2O3

3 0
2 years ago
Read 2 more answers
A 15.0-L rigid container was charged with 0.500 atm of kryp‑ ton gas and 1.50 atm of chlorine gas at 350.8C. The krypton and chl
Alecsey [184]

Answer: 32.94 g

Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

Kr+2Cl_2\rightarrow KrCl_4

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.

for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm

V = 15.0 L

T = 350.8 + 273 = 623.8 K

For krypton, n=\frac{0.500*15.0}{0.0821*623.8}

n = 0.146 moles

for chlorine, n=\frac{1.50*15.0}{0.0821*623.8}

n = 0.439

From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.

Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.

So, the amount of product formed is calculated from moles of krypton.

Molar mass of krypton tetrachloride is 225.61 gram per mol.

There is 1:1 mol ratio between krypton and krypton tetrachloride.

0.146molKr(\frac{1molKrCl_4}{molKr})(\frac{225.61gKrCl_4}{1molKrCl_4})

= 32.94 g of KrCl_4

So, 32.94 g of the product will form.

5 0
2 years ago
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