calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass

Answers

Answer 1

Answer:

Answer:

We are given that there is 95% ethanol by mass in rectified spirit

so, we can say that in a 100g sample, we have 95 grams of ethanol and 5 grams of water

we will find the number of moles of ethanol and water in 100g solution of rectified spirit and use that to calculate the mole fraction

Moles of Ethanol:

Molar mass of ethanol = 46 grams / mol

Number of moles = Given mass / molar mass

Number of moles = 95 / 46

Moles of Ethanol = 2 moles (approx)

Moles of Water:

Molar mass of water = 18 grams per mol

Number of moles = Given mass / molar mass

Moles of water = 5 / 18

Moles of water = 0.28 moles (approx)

Mole Fractions:

Mole fraction of a specific compound is the number of moles of that compound divided by the total number of moles in the solution

Mole fraction of Ethanol:

Moles of ethanol / (moles of ethanol + moles of water)

2 / (2 + 0.28)

2 / (2.28) = 0.9 (approx)

Mole fraction of Water:

Moles of water / (Moles of ethanol + moles of water)

0.28 / (2 + 0.28)

0.28 / (2.28) = 0.1 (approx)

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What does it mean when say that a 200-ml sample and a 400-ml sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different?

Answers

Answer:

Read explanation.

Explanation:

Molarity is a unit used to measure the $(solute)/(solution)$ ratio per unit volume of the solution. So, in other words, and in our case, we call it the $(moles~of~salt)/(per~ml~of~solution)$ ratio.

So, the amount of solute (moles of salt) per ml of solution will be the same, but the number of moles in each solution will be obviously different.

Hope it helped,

BiologiaMagister

(1) The samples of a salt solution are identical because they have the same amount of solute dissolved in them.

(2) The samples are different because the volume of their solution is different.

What is molarity?

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute dissolved in one liter of solution.

That is, molarity quantifies the amount of the solute in a given volume of the solution.

Molarity (M) = Moles of Solute / Volume of Solution (in liters)

Thus, when it's stated that a 200-ml sample and a 400-ml sample of a solution of salt have the same molarity, it means that both samples have an equal concentration of salt molecules per unit volume, while molecules each sample have different volume.

Learn more about molarity here: brainly.com/question/30404105

#SPJ3

1. What is a wave?a.) a vibrating disturbance that transfers energy from place to place
b.) a material which a wave travels through
c.) an area where coils spread out
d.) the maximum distance a medium moves from its rest position

Answers

Answer:

Explanation:

A wave is a disturbance that moves energy from one place to another. Only energy — not matter — is transferred as a wave moves. The substance that a wave moves through is called the medium. That medium moves back and forth repeatedly, returning to its original position. But the wave travels along the medium

Which statement is true about the cell membrane? It allows an unlimited amount of water to enter. It stops waste substances from passing through. It is not selective about which gases enter or leave the cell. It is selective about which substances enter or leave the cell.

Answers

Answer:

It is selective about which substances enter or leave the cell.

Explanation:

Also srry for spamming u all the time.

UwU hope this helps tho

Answer:

It is selective about which substances enter or leave the cell

Explanation:

I have learned this already

Hydrazine, N2H4, is a weak base and is used as fuel in the space shuttle.N2H4(aq)+H2O(l)âN2H5+(aq)+OHâ(aq)

Part A

If the pH of a 0.133 M solution is 10.66, what is the ionization constant of the base?

Express your answer using two significant figures.

Answers

Answer:

Kb = 1.6*10^-6

Explanation:

The given reaction is:

$N2H4(aq)+H2O(l)\rightarrow N2H5+(aq)+OH-(aq)$

The ionization constant of the base Kb is given as:

$Kb = ([N2H5+][OH-])/([N2H4])------(1)$

The pH = 10.66

therefore, pOH = 14-pH = 14-10.66 =3.34

$[OH-] = 10^(-pOH) =10^(-3.34) =4.57*10^(-4) M$

[N2H5+] = [OH-] = 4.57*10^-4M

[N2H4] = 0.133 M

Based on eq(1)

$Kb = ([4.57*10^(-4)]^(2))/([0.133])=1.6*10^(-6)$

Consider the combination reaction of samarium metal and oxygen gas. If you start with 33.7 moles of samarium metal, how many moles of oxygen gas would be required to react completely with all of the samarium metal? For this reaction, samarium has a +3 oxidation state within the samarium/oxygen compound.

Answers

Answer:

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Explanation:

$4Sm+3O_2\rightarrow 2Sm_2O_3$

Number of moles samarium metal = 33.7 moles

According to reaction, 4 moles of samarium reacts with 3 moles of oxygen gas.

Then 33.7 moles of samarium will react with:

$(3)/(4)* 33.7 mol=25.275 mol$ of oxygen gas.

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Answer:

Moles of oxygen gas required to react completely with 33.7 moles of samarium metal is $\fbox{25.3 \text{ mol}}$ .

Explanation:

A chemical equation in which the number of atoms of each element is the same on the reactant and product side is called a balanced chemical equation.

The balanced chemical equation can be used to determine the stoichiometric ratio between the reactant and the product. The stoichiometric ratio thus enables us to calculate:

1. Amount of one reactant required to react completely with the other reactant.

2. Amount of the product that can be produced from the given amount of the reactant.

Step 1: Write the chemical equation for the reaction between samarium metal and oxygen gas.

The chemical formula for oxygen gas is $\text{O}_(2)$ .

Samarium has +3 oxidation state within the samarium/oxygen compound. So, the chemical formula of the samarium oxygen compound is $\text{Sm}_(2)\text{O}_(3)$ .

The chemical equation is as follows:

$\fbox{\text{Sm}+\text{O}_(2) \rightarrow \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

Step 2: Balance the chemical equation for the reaction between oxygen and samarium metal.

The number of oxygen atoms in the product side is 3 and in the reactant side is 2. Put coefficient 2 in front of $\text{Sm}_(2)\text{O}_(3)$ and 3 in front of $\text{O}_(2)$ to balance the oxygen atoms.

$\fbox{\text{Sm}+3\text{O}_(2) \rightarrow 2 \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

The number of samarium atoms in the product side is 4 and in the reactant side is 1. Put coefficient 4 in front of Sm in the reactant side.

$\fbox{\n4\text{Sm}+3\text{O}_(2) \rightarrow 2 \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

Step 3: Determine the stoichiometric ratio between samarium and oxygen from the above balanced chemical equation.

According to the balanced chemical equation, we can see that the stoichiometric ratio between samarium and oxygen is 4:3.

Step 4: Use unitary method and calculate the moles of oxygen required to completely react with the given moles of samarium metal as follows:

$\text{moles of O}_(2) = \left( \text{moles of Sm} \right)\left( \frac{3 \text{ mol of O}_(2)}{4 \text{ mol of Sm}} \right)$ ...... (1)

Step 5: Substitute 33.7 mol for moles of Sm in equation (1).

$\text{Moles of O}_(2) = \left( \text{33.7 mol} \right) \left( \frac{3 \text{ mol of O}_(2)}{4 \text{ mol of Sm}} \right)\n\text{Moles of O}_(2)= 25.275 \text{ mol}\n\text{Moles of O}_(2)= 25.3 \text{ mol}$

Note:

Do not forgot to balance the reaction. The reaction must be balanced in order to calculate the amount (mol) of oxygen required to completely react with the given amount of samarium.

Learn more:

1. Balanced chemical equation brainly.com/question/1405182

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Some basic concept of chemistry

Keywords: samarium, oxygen gas, samarium/oxygen compound, 33.7 moles, 25.3 mol, balanced equation, stoichiometric ratio, coefficient, balance, moles, completely react.

Which isomer would you expect to undergo E2 elimination faster, trans-1-bromo-4-tert-butylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane? Draw each molecule in its more stable chair conformation, and explain your answer.

Answers

Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.

Explanation:

The two primary requirements for an E-2 elimination reaction are:

1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.

2.The hydrogen and leaving group must have a anti-periplanar position .

Any substrate which would follow the above two requirements can give elimination reactions.

For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane to be stable it must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.

Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.

The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.

The trans-1-bromo-4-tert-butylcyclohexane does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.

so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.

Final answer:

Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.

Explanation:

In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.

In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.

In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.