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3 years ago

Which answer provides the correct name for the following hydrocarbon?

Chemistry

2 answers:

motikmotik3 years ago

7 0

Answer: Option (b) is the correct answer.

Explanation:

In the given molecule, there are five carbon atoms attached linearly to each other will single bonds only.

So, it is known that when a compound contains only carbon and hydrogen atoms attached single bondedly to each other then this type of hydrocarbon is known as alkane.

Their general formula is $C_{n}H_{2n+2}$ , where n is the number of carbon atoms present. Suffine "ane" is added to the name of alkane.

Thus, we can conclude that name of the given hydrocarbon is pentane.

ivanzaharov [21]3 years ago

4 0

1) Hydrocarbon: CH3 - CH2 - CH2 - CH2 - CH3

2) Only single bonds => alkane => sufix ane

3) no substitutions

4) 5 carbons = > prefix penta.

Therefore, the name is pentane.

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8. Why would it be a bad idea to make coins made of sodium metal?

Elina [12.6K]

Answer:

Sodium reacts with the oxygen in air. It reacts vigorously with oxygen and the moisture that is already present in the air and thus catches fire.

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2 years ago

2. How many grams of zinc will be formed if 32.0 g of copper reacts with zinc nitrate? Copper(I)nitrate is the other product.

Yakvenalex [24]

Answer:

16.46 g.

Explanation:

It is a stichiometry problem.

We should write the balance equation of the mentioned chemical reaction:

2Cu + Zn(NO₃)₂ → Zn + 2Cu(NO₃).

It is clear that 2.0 moles of Cu reacts with 1.0 mole of Zn(NO₃)₂ to produce 1.0 mole of Zn and 2.0 moles of Cu(NO₃).

We need to calculate the number of moles of the reacted Cu (32.0 g) using the relation:

n = mass / molar mass

The no. of moles of Cu = mass / atomic mass = (32.0 g) / (63.546 g/mol) = 0.503 mol.

Using cross multiplication:

2.0 moles of Cu produces → 1.0 mole of Zn, from the stichiometry.

0.503 mole of Cu produces → ??? mole of Zn.

The no. of moles of Zn produced = (1.0 mol)(0.503 mol) / (2.0 mol) = 0.2517 mol.

∴ The grams of Zn produced = no. of moles x atomic mass of Zn = (0.2517 mol)(65.38 g/mol) = 16.46 g.

3 0

2 years ago

How might the yield of 1-bromobutane be affected if water was not added, and what product(s) would be favored?

jek_recluse [69]

Due to the inability of the reaction to take place, the yield of 1-Bromobutane would drop.

Since 1-Butanol won't react with the additional sodium bromide, bromination won't happen.

If water had been supplied, the equilibrium would have shifted extremely far to the left, preventing the reactants from interacting with the acid and favoring the yield of 1-bromobutane instead.

<h3>What is Bromination?</h3>

When a substance undergoes bromination, bromine is added to the compound as a result of the chemical reaction.

After bromination, the result will have different properties from the initial reactant.
For example, an alkene is brominated by electrophilic addition of $Br_{2}$ .
Benzene ring bromination by electrophilic aromatic substitution.

Learn more about Bromine here:

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5 0

1 year ago

Pb + Fe(SO4) =Pb(SO4) + Fe what type of reaction is this ?

VikaD [51]

Answer:

Single replacement

Explanation:

3 0

2 years ago

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (M) is calculated from the following equation:

M = n ÷ V

where n is the number of moles of the solute in the solution, and V is the volume of the solution.

Accordingly, the number of moles of the solute is given by

n = M x V

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

n (1) = M (1) x V (1)

n (2) = M (2) x V (2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

n (1) = n (2)

and

M (1) x V (1) = M (2) x V (2)

For the stock solution, M (1) = 1.244 M, and V (1) is the volume needed. For the dilute solution, M (2) = 0,1281 M, and V (2) = 250.0 mL.

The volume of the stock solution needed, V (1), is calculated as follows:

V (1) = M (2) x V (2) ÷ M (1)

V (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

V (1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0

2 years ago