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1 year ago

Write out the mechanism for the reaction of the alkene side product with concentrated hydrochloric acid.

1 answer:

6 0

The mechanism for the reaction of the alkene side product with concentrated hydrochloric acid cannot easily abstract proton to form alkene.

We do not observe alkene side product because Cl- (chloride) is a poor base which cannot easily abstract proton to form alkene.

For obtaining alkene as product, Sulphuric acid H₂SO₄ is used.

Hydrochloric acid, additionally called muriatic acid, is an aqueous answer of hydrogen chloride. it is a drab solution with a specific pungent smell. Hydrochloric acid, additionally known as muriatic acid, is an aqueous solution of hydrogen chloride. Hydrochloric acid is corrosive to the eyes, skin, and mucous membranes. Acute (quick-term) inhalation publicity may also purpose eye, nostril, and breathing tract irritation and infection and pulmonary edema in humans.

A reaction mechanism is the collection of standard steps with the aid of which a chemical response occurs. A reaction that occurs in or greater elementary steps is known as a multistep or complex response. A response intermediate is a chemical species that is fashioned in a single essential step and ate up in a subsequent step.

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Part G

notka56 [123]

Answer:

Therefore the theoretical density of iron is 7.877 g/cm³ .

Therefore the number of vacancy per cm³ is

Explanation:

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2 years ago

The source of alpha particles in a smoke detector is the

pantera1 [17]

The source of alpha particles in a smoke detector is the Americium.

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2 years ago

predict the product that forms in the synthesis reaction between potassium metal and iodine gas. provide the balanced chemical e

Doss [256]

K + I - > KI
Potassium (needs to lose 1 electron) responds with Iodine (needs to pick up 1 electron) to fulfill both component's octet, shaping a salt, potassium iodide
This is a similar case for NaCl, simply unique components. Trust this made a difference.

5 0

3 years ago

Read 2 more answers

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

2 years ago

How many moles of NaCl must be dissolved in 0.5 L of water to make a 4 mole/L

kvv77 [185]

C = 4 mol/l
v = 0.5 l

n(NaCl)=cv

n(NaCl) = 4 mol/l · 0.5 l = 2 mol

2 moles of NaCl must be dissolved

5 0

3 years ago