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1 year ago

In general ethers have a much lower boiling point than their isomeric alcohols. why?

1 answer:

4 0

Ethers have a much lower boiling point than their isomeric alcohols because ether have not hydrogen bond that why ethers have lower boiling point .

Ethers will not have a hydrogen atom attached to the oxygen atom as alcohols do. Because ether molecules can not form intermolecular hydrogen bonds, their boiling points are much shorter than those of alcohol with such a comparable mass.

Owing to the combination of intermolecular H-bonding in alcohols though not in ethers, alcohols possess higher boiling temperatures than isomeric ethers.

Therefore, ethers have a much lower boiling point than their isomeric alcohols because ether have not hydrogen bond that why ethers have lower boiling point .

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A sample was prepared by mixing 18. ml of 3.00 x 10^-3 m crystal violet (cv) with 2.00 ml of 0.250 m naoh. calculate the resulti

Aleks [24]

Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.

Explanation :

Step 1 : Find moles of crystal violet and NaOH.

The molarity formula is

$Molarity = \frac{mol}{L}$

Molarity of crystal violet = $3.00 \times 10^{-3} = \frac{mol (CrystalViolet)}{L}$

The volume of crystal violet solution is 18 mL which is 0.018 L.

Moles of crystal violet = $3.00 \times 10^{-3} \times 0.018 = 5.4 \times 10^{-5}$

Moles of crystal violet = 5.4 x 10⁻⁵

Moles of NaOH = $Molarity \times L = 0.250 \times 0.00200 = 5.00 \times 10^{-4}$

Moles of NaOH = 5.00 x 10⁻⁴

Step 2 : Find total volume of the solution

The total volume of the solution after mixing NaOH and crystal violet is

0.018 L + 0.00200 = 0.020 L

Step 3 : Use molarity formula to find final concentrations

Molarity of crystal violet = $\frac{mol(CrystalViolet)}{Total Volume(L) } = \frac{5.4 \times 10^{-5}}{0.020} = 2.7 \times 10^{-3}$

Final concentration of CV = 0.0027 M

Molarity of NaOH= $\frac{mol(NaOH)}{Total Volume(L) } = \frac{5.00 \times 10^{-4}}{0.020} = 0.025 \times 10^{-3}$

NaOH is a strong base and dissociates completely as follows.

$NaOH (aq) \rightarrow Na^{+} (aq) + OH^{-} (aq)$

The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.

Concentration of OH⁻ = 0.025 M

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3 years ago

An unknown solution has a pH of 7. How would you classify this solution?

xenn [34]

Solution with a ph of 7 Is neutral

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3 years ago

fig. 15 shows a simplified diagram of the electrolysis of a molten electrolyte containing lithium chloride. b. Describe how the

melisa1 [442]

During the electrolysis of the molten lithium chloride, the Lithium ions (Li⁺) at the cathode undergoes reduction, and the electron configuration of lithium becomes 1s²2s¹.

<h3>What is electrolysis?</h3>

Electrolysis can be described as the process in which the electric current is passed through the chemical compound to break them. In this process, the atoms and ions are interchanged by the addition or removal of electrons.

The ions are allowed to move freely in this process. When an ionic compound is melted or dissolved in water then ions are produced which can move freely.

During the electrolysis of molten lithium chloride, the lithium ions reach the cathode and accept the electrons while chloride ions reach at anode and loss electrons to become chlorine gas.

At anode : 2 Cl⁻ → Cl₂ + 2e⁻

At cathode: 2 Li⁺ + 2e⁻ → Li

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1 year ago

The ability of a substance to be hammered or rolled into thin sheets is called ________

zaharov [31]

Answer: Malleability

Explanation: is a physical property of metals that defines their ability to be hammered, pressed, or rolled into thin sheets without breaking. In other words, it is the property of a metal to deform under compression and take on a new shape.

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3 years ago

Calculate the mass percent of 0.485 g of H, which reacts with oxygen to form 2.32 g H2O

Nataly [62]

Answer:

Mass % of hydrogen = 20.9 %

Explanation:

Given data:

Mass of hydrogen = 0.485 g

Mass of water = 2.32 g

Mass percent of hydrogen = ?

Solution:

Mass % of hydrogen = mass of hydrogen / mass of water × 100

Mass % of hydrogen = 0.485 g/ 2.32 g × 100

Mass % of hydrogen = 0.209 × 100

Mass % of hydrogen = 20.9 %

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3 years ago