All categories

3 years ago

What major product is formed when heptyne is treated with Br2 in CCl4?

Chemistry

2 answers:

melisa1 [442]3 years ago

8 0

Answer:

Option A is the Answer. 

Explanation:

Treatment of alkyne with Bromine in the presence of $CCl_4$ produces a trans product that is, a alkenyl bromide.

$CH_3 \ CH_2 \ CH_2 \ CH_2 \ CH_2 - C = CH \frac {(CCl_4)}{(Br_2 )} > CH_3 \ CH_2 \ CH_2\ CH_2 \CH_2\ -CBr = CHBr$

E-1, 2-dibromoheptene

It is a trans intermediate on further halogenations produces a tetrabromopentane.

It is an electrophilic expansion response in which the triple bond breaks to turn into a twofold bond and creates a dibromoalkene (E-1,2 dibromoheptene), and on further expansion, the twofold bond turns into a solitary bond and delivers a tetrabromoalkane.

ANTONII [103]3 years ago

4 0

Answer:A 1,2-dibromoheptene

Explanation:

You might be interested in

3. Given the reaction:

RideAnS [48]

A balanced chemical equation is a ratio: this one shows that for every 5 moles O2, 3 moles CO2 are produced. Since there's a 5:3 ratio of O2/CO2.

((5 m O2)/(3 m CO2)) * 1.5 m CO2 = 2.5 m
O2.

Answer B.

8 0

2 years ago

What is an Atom? What are the 5 most common elements that make up the human body?.

dalvyx [7]

An atom is the smallest unit of ordinary matter that forms a chemical element.

oxygen, carbon, hydrogen, nitrogen, calcium, and phosphorus.

6 0

3 years ago

Please answer both questions Thankyou

erma4kov [3.2K]

Answer:

The limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water

Explanation:

Step 1: Convert everything into moles

nH2(l) = 1.06 x 10^8 g / 2.016 g/mol = 5.26 x 10^7 mols

nO2(l) = 6.29 x 10^8 g / 32.00 g/ mol = 1.966 x 10^7 mols

Step 2: Find the limiting reagent

The limiting reagent would be oxygen gas from

the balanced equation because we have less moles of oxygen gas needed to fully combust with the hydrant gas

Step 3: Stoichiometry time

The mole ratio from oxygen gas to water is 1:2

This means that for every mole of oxygen gas two moles of water is produced

We need to multiply the moles of oxygen gas by two to find out how many moles of water has been produced

nH2O = nO2 x 2

nH2O = 1.966x10^7 x 2

nH2O = 3.932x10^7

Step 4: Therefore statement

Therefore the limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water

7 0

3 years ago

11. Propane (C3Hg) is a fuel commonly used in gas grills.

Mrac [35]

Answer:

81.71%

Explanation:

One mole of propane contains 3 moles of carbon atoms and 8 moles of hydrogen atoms, as seen from the molecular formula of $C_3H_8$ . In order to calculate the percent of carbon in propane by mass, we need to remember that %w/w (or percent mass) formula states that:

$\omega=\frac{m_{component}}{m_{total}}\cdot100\%$

That is, we need to divide the mass of the component of interest by the total mass of the compound and multiply by 100 to obtain the percentage.

For simplicity, let's take 1 mole of propane and find the mass of 1 mole (hence, we'll be finding the molar mass of propane). To do that, we add the 3 molar masses of carbon and 8 molar masses of hydrogen to obtain a total of:

$M_{C_3H_8}=3M_C+8M_H=3\cdot12.011 \frac{g}{mol}+8\cdot1.00784\frac{g}{mol}=44.096 \frac{g}{mol}$

Now that we have the molar mass of propane, we also need to find the total mass of carbon in 1 mole of propane. We know that we have a total of 3 moles of carbon which corresponds to:

$M_C=3\cdot12.011 \frac{g}{mol}=36.033 \frac{g}{mol}$

Dividing the mass of carbon present by the total mass of the compound will yield the mass percentage as defined by the formula we introduced:

$\omega_C=\frac{36.033\frac{g}{mol}}{44.096 \frac{g}{mol}}\cdot 100\%=81.71\%$

5 0

3 years ago

What is happening in this picture?

True [87]

Answer:

in first picture pressure is low and in second picture the pressure is high

2ans in first picture air is

less amount

in second picture air is high amount

3 0

3 years ago