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mol fraction Codein : 0.054

mol fraction Ethanol : 0.946

Given

46.85 g of codeine

125.5 g of ethanol

Required

mol fraction

Solution

The mole fraction : the mole ratio of a substance to the mole of solution /mixture  

mol Codeine, C₁₈H₂₁NO₃(MW=299.4 g/mol) :

[tex]\tt mol=\dfrac{46.85}{299.4 g/mol}\\\\mol=0.1565[/tex]

mol Ethanol, C₂H₅OH(MW=46.07 g/mol)

[tex]\tt mol=\dfrac{125.5}{46.07}=2.7241[/tex]

mol of solution :

[tex]\tt mol~codein+mol~ethanol=0.1565+2.7241=2.8806[/tex]

[tex]\tt \dfrac{0.1565}{2.8806}=0.054[/tex]

mol fraction Ethanol :

[tex]\tt \dfrac{2.7241}{2.8806}=0.946[/tex]

Answer:

Because of time zones?

Explanation: The only thing that would make sense to me is the time zones.

Answer:

I believe the answer is Reflecting Telescope

Explanation:

-leeknowishawt

Answer:

Li to Zn2+

Explanation:

Because oxidation number of Li changes from 0 to 2+ it means that Li has lost two electrons and oxidation number of Zn2+ changes to Zn which means Zn received two electrons.

Answer:

4

Explanation:

Because of single replacement the ideal shape comes in form

Answer:

H2CO

Explanation:

a)CH3-CH2-CH2-CH=CH2

b)CH3-CH2-CH2-CH2-CH2-CH2-CH3

c)CH≡C-CH2-CH2-CH2-CH3

d)CH3-CH2-CH2-CH2-CH2-CH2-CH=CH2

e)CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3

give me brain list

Answer:

D

Explanation:

The other three are related to

a. selling a home

b. buying a home

c. owning a home

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Answer:

4 Al(s) + 3 O₂(g) ⇒ 2 Al₂O₃(s)

Explanation:

Let's consider the unbalanced molecular equation when aluminum metal and oxygen gas combine to produce aluminum oxide. This is a combination reaction.

Al(s) + O₂(g) ⇒ Al₂O₃(s)

We can balance O  atoms by multiplying O₂ by 3 and Al₂O₃ by 2.

Al(s) + 3 O₂(g) ⇒ 2 Al₂O₃(s)

Finally, we multiply Al by 4 to get the balanced equation.

4 Al(s) + 3 O₂(g) ⇒ 2 Al₂O₃(s)

Answer:

erosion

Explanation:

erosion is the term that refers to the process by which land is worn away by natural forces or human activity.

Answer:

Erosion

Explanation:

Just took the test, hope it helps!

Answer:

M = 1.04 M

Explanation:

Given data:

Molarity of solution = ?

Mass of HCl = 6.27 g

Volume of solution = 163 mL (163 mL× 1L /1000 mL = 0.163 L)

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 6.27 g / 36.5 g/mol

Number of moles = 0.17 mol

Molarity:

M = 0.17 mol/ 0.163 L

M = 1.04 M

Answer:

4.6 M HCI

Explanation:

on edge

We can see here that the number of lone pairs surround the boron atom is A. 0.

A boron atom is an atom of the chemical element boron, which is represented by the symbol B in the periodic table. Boron is a metalloid with atomic number 5, meaning it has five protons in its nucleus. It belongs to Group 13 (formerly known as Group IIIA) of the periodic table and is located in period 2.

The Lewis structure of [tex]BF_{3}[/tex] indicates that boron (B) forms three single bonds with three fluorine (F) atoms. Each bond consists of a pair of electrons shared between boron and fluorine. Since there are no unshared electron pairs around the boron atom, the answer is option a. 0.

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Answer:

V₂ =  116126.75 cm³

Explanation:

Given data:

Radius of balloon = 15 cm

Initial pressure = 2 atm

Initial temperature = 35 °C (35 +273 = 308K)

Final temperature = -20°C (-20+273 = 253 K)

Final pressure = 0.3 atm

Final volume = ?

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

Initial volume of balloon:

V = 4/3πr³

V = 4/3×22/7×(15cm)³

V = 14137.17 cm³

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 2 atm × 14137.17 cm³ × 253 K / 308 K × 0.3 atm

V₂ = 7153408.02 atm .cm³. K / 61.6 K.atm

V₂ =  116126.75 cm³

Explanation:

A) Highest occupied energy level and Group Number.

Answer:

See explanation

Explanation:

A reducing agent in chemistry is any substance that accepts electrons and its oxidation number is decreased in a redox reaction.

Manganese occurs in a variety of oxidation states; +7, +6, +5, +4, +3, +2. This makes it an exceptionally good reducing agent since it can accept different number of electrons and change from one oxidation state to another.

The commonest compound of manganese used as a strong oxidizing agent is KMnO4 in which manganese has an oxidation number of +7.

Answer:

i dunno

Explanation:

Answer:

Explanation:

Evidence for early forms of life comes from fossils. By studying fossils, scientists can learn how much (or how little) organisms have changed as life developed on Earth. There are gaps in the fossil record because many early forms of life were soft-bodied, which means that they have left few traces behind.

Answer: Positive ions or cations

Explanation:

The ions are formed when an atom gains or lose electrons.

The ions are classified into two which are called the cations and anions. The cation is classified as the positive charge formed by the loss of electrons. Example: [tex]Na^+[/tex] is formed by the loss of one electron by [tex]Na[/tex]

The anion is classified as the negative charge ion formed by the gain of electrons. Example : [tex]F^-[/tex] is formed the gain of one electron by [tex]F[/tex]

Thus when an element loses electrons, it forms positive ions or cations.

Answer:

Height and mass

Explanation:

The potential energy of a body is the energy due to the position of a body.

It is mathematically expressed as:

 Potential energy  = m g h

m is the mass of the body

g is the acceleration due to gravity

h is the height of the body

Acceleration due to gravity on the earth surface is a constant.

As mass and height of a body increases, the acceleration due to gravity will also increase.

Answer:

it is 7

Explanation:

dogs are 7 HOPE THIS HELPED kidding i don get the question

Explanation:

DON"T EAT MY DOG OR I"LL EAT YOU

Answer:

d

Explanation:

According to law of conservation of mass, mass  is conserved during burning of paper although the mass of ash is less than paper as some of the products are  given out as gas.

According to law of conservation of mass, it is evident that mass is neither created nor destroyed rather it is restored at the end of a chemical reaction .

Law of conservation of mass and energy are related as mass and energy are directly proportional which is indicated by the equation E=mc².Concept of conservation of mass is widely used in field of chemistry, fluid dynamics.

Law needs to be modified in accordance with laws of quantum mechanics under the principle of mass and energy equivalence.This law was proposed by Antoine Lavoisier in the year 1789.

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Answer:

4.289 mol

Explanation:

Given that the molar mass of HBr (known as Hydrobromic Acid or Hydrogen Bromide) has a molar mass of 80.91 g/mol.

To find the number of moles, we have the formula

(Grams of the substance ÷ moles of the substance) * (molar mass of the substance in grams ÷ one mol)

Here the Grams of the substance = 347grams,

Moles of the substance =?

Molar mass of the substance = 80.91 g/mol

By cross multiplying, we have

347 grams * 1 mol = Z * (80.91)

Z = 347 grams ÷ 80.91g/mol

Z = 4.289 mol

Answer:

Explanation:

The [H3O+] of HNO2 can be found by using the formula

[tex][H_3O^{ + } ] = \sqrt{Kac} [/tex]

Ka is the acid dissociation constant

c is the concentration of the acid

From the question we have

[tex][H3O^{ + } ] = \sqrt{4.46 \times {10}^{ - 4} \times 0.1} \\ = 0.006678...[/tex]

We have the final answer as

Hope this helps you

Answer:

8H2SO4 + 2KMnO4 + 5Na2O2 => 8H2O + 2MnSO4 + 5Na2SO4 + 5O2 + K2SO4

have:

Explanation:

Answer:

Mass of hydrogen needed = 10 g

29.22 gram of sodium are required

Explanation:

1 )Given data:

Mass of oxygen = 80.0 g

Mass of hydrogen needed = ?

Solution:

Chemical equation:

2H₂ +  O₂       →        2H₂O

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles =  80 g/ 32 g/mol

Number of moles = 2.5 mol

now we will compare the moles of oxygen and hydrogen.

               O₂          :         H₂

                 1           :            2

               2.5        :          2× 2.5 = 5 mol

Mass of hydrogen:

Mass = number of moles × molar mass

Mass = 5 mol × 2 g/mol

Mass = 10 g

2 : Given data:

initial mass ration of chlorine and sodium react with each other  = 1.54 g : 1.0 g

Mass of sodium required for 45 g of chlorine = ?

Solution:

                 Cl₂             :          Na

                   1.54          :           1.0

                     45           :          1.0/1.54× 45 = 29.22 g

Thus, 29.22 gram of sodium are required.

Answer:

CH3CH2CH2CH3 <CH3CH2OCH3<CH3CH2OCH2CH3< CH3CH2CH2CH2NH2

Explanation:

Some important factors affect the boiling point of aliphatic compounds. Prominent among them are;

1) relative molecular mass

2) nature of intermolecular

3) surface area of compound.

In this case, the nature of intermolecular forces and the molecular mass of the compounds are worthy of mention.

The degree of dipole interaction in CH3CH2CH2CH2NH2 is far greater than that in the ethers hence the amine has the highest boiling point followed by the ether of higher molecular mass then the ether of lower molecular mass and finally the alkane has the lowest boiling point because only dispersion forces are involved.

The correct ranking of the given organic compounds from highest to lowest boiling points is given by;

Option D; CH3CH2CH2CH2NH2 < CH3CH2OCH2CH3 < CH3CH2OCH3 < CH3CH2CH2CH3

We are given the organic compounds;

CH3CH2OCH2CH3

CH3CH2OCH3

CH3CH2CH2CH3

CH3CH2CH2CH2NH2

The family names of the given organic compounds are;

CH3CH2OCH2CH3 is an ether

CH3CH2OCH3 is an ether

CH3CH2CH2CH3 is an alkane

CH3CH2CH2CH2NH2 is an amine

The boiling points are given in the order from highest to lowest as;

Alkanes < Ethers < Amines

This is because Amines have higher possibility of hydrogen bonding than the others.

Lastly, a longer chain ether will have a higher boiling point than the shorter chain one because it can have more Vander walls force interactions.

Thus, the correct order of arrangement of the given organic compounds from highest to lowest boiling point is;

CH3CH2CH2CH2NH2 > CH3CH2OCH2CH3 > CH3CH2OCH3 > CH3CH2CH2CH3

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Answer:

3.20 moles of carbon dioxide

Explanation:

The reaction of C₂H₂ with O₂ is:

C₂H₂ + 5/2O₂ → 2CO₂ + H₂O

Where 1 mole of acetylene produce 2 moles of carbon dioxide.

Based on the chemical equation of this reaction, there are produced twice the moles of acetylene in moles of carbon dioxide assuming a compete reaction.

If 1.60 moles of acetylene reacts there are produced:

1.60 moles * 2 =

Answer:

18.6 ( had to be 20 letter :/)

Explanation:

Answer:

it is actually 17 but the closest would be 15.

B. 15

Answer:

Approximately [tex]0.224\;\rm L[/tex], assuming that this reaction took place under standard temperature and pressure, and that [tex]\rm CO_2[/tex] behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that [tex]\rm CaCO_3[/tex] is the limiting reactant. How many moles of [tex]\rm CO_2[/tex] would be produced?

Look up the relative atomic mass of [tex]\rm Ca[/tex], [tex]\rm C[/tex], and [tex]\rm O[/tex] on a modern periodic table:

Calculate the formula mass of [tex]\rm CaCO_3[/tex]:

[tex]\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}[/tex].

Calculate the number of moles of formula units in [tex]1\; \rm g[/tex] of [tex]\rm CaCO_3[/tex] using its formula mass:

[tex]\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}[/tex].

In the balanced chemical equation, the ratio between the coefficient of [tex]\rm CaCO_3[/tex] and that of [tex]\rm CO_2[/tex] is [tex]\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1[/tex].

In other words, for each mole of [tex]\rm CaCO_3[/tex] formula units consumed, one mole of [tex]\rm CO_2[/tex] would be produced.

If [tex]\rm CaCO_3[/tex] is indeed the limiting reactant, all that approximately [tex]1.00\times 10^{-2}\; \rm mol[/tex] of [tex]\rm CaCO_3\![/tex] formula would be consumed. That would produce approximately [tex]1.00\times 10^{-2}\; \rm mol\![/tex] of [tex]\rm CO_2[/tex].

On the other hand, assume that [tex]\rm HCl[/tex] is the limiting reactant.

Convert the volume of [tex]\rm HCl[/tex] to [tex]\rm dm^{3}[/tex] (so as to match the unit of concentration.)

[tex]\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}[/tex].

Calculate the number of moles of [tex]\rm HCl[/tex] molecules in that [tex]5.00\times 10^{-2}\; \rm dm^{3}[/tex] of this [tex]\rm 0.05\; \rm mol \cdot dm^{-3}[/tex]

[tex]\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}[/tex].

Notice that in the balanced chemical reaction, the ratio between the coefficient of [tex]\rm HCl[/tex] and that of [tex]\rm CO_2[/tex] is [tex]\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}[/tex].

In other words, each mole of [tex]\rm HCl[/tex] molecules consumed would produce only [tex]0.5\;\rm mol[/tex] of [tex]\rm CO_2[/tex] molecules.

Therefore, if [tex]\rm HCl[/tex] is the limiting reactant, that [tex]2.50 \times 10^{-3}\; \rm mol[/tex] of [tex]\rm HCl\![/tex] molecules would produce only one-half as many (that is, [tex]1.25\times 10^{-3}\; \rm mol[/tex]) of [tex]\rm CO_2[/tex] molecules.

If [tex]\rm CaCO_3[/tex] is the limiting reactant, [tex]\rm 1.00\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2[/tex] molecules would be produced. However, if [tex]\rm HCl[/tex] is the limiting reactant, [tex]1.25\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2\![/tex] molecules would be produced.

In reality, no more than [tex]\rm 1.00\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2[/tex] molecules would be produced. The reason is that all [tex]\rm CaCO_3[/tex] would have been consumed before [tex]\rm HCl[/tex] was.

After finding the limiting reactant, approximate the volume of the [tex]\rm CO_2\![/tex] produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately [tex]22.4\; \rm L[/tex].

If [tex]\rm CO_2[/tex] behaves like an ideal gas, the volume of that [tex]\rm 1.00\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2\![/tex] molecules would be approximately [tex]\rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L[/tex].