What is the phase change from solid to gas?
O A. Condensation
O B. Sublimation
O C. Freezing
O D. Vaporization

Answer:

108.6 g

Explanation:

First we use the PV=nRT formula to calculate the number of nitrogen moles:

Inputting the data:

Then we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:

Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:

Answer:

Carboxyl, primary amine, amide, ester, and phenyl.

Explanation:

The functional groups present in the compound of aspartame are carboxyl, primary amine, amide, ester, and phenyl. Aspartame is an artificial non-saccharide sweetener which is 200 times sweeter than sucrose. This aspartame is commonly used as a sugar substitute in many foods and beverages. It has the trade names such as NutraSweet, Equal, and Canderel.

Solution :

a). The separation of 4-methylbenzoic acid from 1,4-dimethoxybenzene will work but it will result in lower recovery.

In the reaction of acid-base to form a sodium 4 - methoxy benzoate, that is soluble in the water, 4-methoxy benzoic acid reacts with the sodium bicarbonate to give sodium 4-methoxybenzoate as well as carbonic acid.

b). The pKa for the 4-methoxybenzoic acid is [tex]4.46[/tex], and that of carbonic acid is [tex]6.37[/tex]

c). The Keq for the reaction is [tex]10(6.37 - 4.46) = 101.91[/tex]

Therefore, the equilibrium lies to the right  and also the reaction favors the products and the separation works.

But the recovery will be low when compared to the extraction with Sodium hydroxide as the strong base will drive the equilibrium further to the right position, thus neutralizing all the acids virtually. And the weak base will partially neutralize the acid.

Answer:

Explanation:

From the given information:

Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.

Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.

If the molecular weight of camphor = 152.24 g/mol

and it mass = 200 mg

The its no of moles = 200 mg/ 152.24 g/mol

= 1.3137 mmol

Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol

= 6.831 mmol

since the molar mass of NaBH4 = 37.83 g/mol

Then, using the same formula:

No of moles = mass/molar mass

mass = No of moles × molar mass

mass = 6.831 mmol × 37.83 g/mol

mass of NaBH4 used = 258.42 mg  

Answer:

36.55kJ/mol

Explanation:

The heat of solution is the change in heat when the KNO3 dissolves in water:

KNO3(aq) → K+(aq) + NO3-(aq)

As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.

To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:

Moles KNO3 -Molar mass: 101.1032g/mol-

10.6g * (1mol/101.1032g) = 0.1048 moles KNO3

Change in heat:

q = m*S*ΔT

Where q is heat in J,

m is the mass of the solution: 10.6g + 251.0g = 261.6g

S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-

And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C

q = 261.6g*4.184J/g°C*3.5°C

q = 3830.87J

Molar heat of solution:

3830.87J/0.1048 moles KNO3 =

36554J/mol =

Answer:

Arrange the following compounds in order of acidity (highest to lowest): H2O, CH3COOH , HCl

A. CH3COOH > HCl > H2O

B. H2O > CH3COOH > HCl

C. HCl > H2O > CH3COOH

D. HCl > CH3COOH > H2O

Explanation:

The given substances are acetic acid, hydrochloric acid, and water.

Since HCl is a strong acid and it undergoes complete ionization.

CH3COOH acetic acid is a weak acid and it undergoes partial dissociation in water.

Pure water is a neutral substance.

Hence, the order of acidity is shown below:

HCl > CH3COOH > H2O.

Among the given options, option D is the correct answer.

Answer:

antimony-121 has the highest percent natural abundance

Explanation:

percent natural abundance;

121.76 = 120.90 x + 122.90 (1 - x)

121.76 = 120.90 x + 122.90 - 122.90x

121.76 = -2x + 122.90

121.76 - 122.90 = -2x

x= 121.76 - 122.90/ -2

x= 0.57

Where x and 1 - x refers to the relative abundance of each of the isotopes

Percent natural abundance of antimony-121 = 57 %

Percent natural abundance of antimony-123 = (1 - 0.57) = 43%

Let us remember that isotopy refers to a phenomenon in which atoms of the same element have the same atomic number but different mass numbers. This results from differences in the number of neutrons in atoms of the same element.

We can clearly see that antimony-121 has the highest percent natural abundance.

Answer:

Oxygen, O₂ is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2H₂ + O₂ —> 2H₂O

Next, we shall determine the masses of H₂ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 2 × 2 = 4 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mass of O₂O from the balanced equation = 1 × 32 = 32 g

SUMMARY:

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂.

Therefore, 10 g of H₂ will react with

= (10 × 32)/4 = 80 g of O₂.

From the calculations made above, we can see that a higher mass (i.e 80 g) of O₂ than what was given (i.e 5 g) is required to react completely with 10 g of H₂. Therefore, O₂ is the limiting reactant.

Oxygen has been the limiting reactant in the reaction.

A limiting reactant can be defined as the reactant in the reaction in which the product concentration has been dependent.

The balanced equation for the formation of water has been:

[tex]\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O[/tex]

For the formation of reaction to form 2 moles of water, 2 moles of hydrogen reacts with 1 mole of oxygen.

The moles can be calculated as:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

The moles of Hydrogen in 10 g [tex]\rm H_2[/tex]:

Moles = [tex]\rm \dfrac{10}{2}[/tex]

Moles of hydrogen = 5 mol.

Moles of Oxygen in 5 grams Oxygen:

Moles = [tex]\rm \dfrac{5}{32}[/tex]

Moles of oxygen = 0.156 mol.

For the reaction with 2 moles of Hydrogen 1 mole of Oxygen has been required.

For reacting with 5 mol of Hydrogen, moles of oxygen required are:

Moles of oxygen = [tex]\rm \dfrac{1}{2}\;\times\;5[/tex]

Moles of oxygen required = 2.5 moles.

The available oxygen = 0.156 moles.

Since the moles of oxygen available is lesser than required, the formation of the product has been dependent on the concentration of the oxygen.

Thus, oxygen has been the limiting reactant in the reaction.

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Answer: There are 0.779 moles of gas were added to the bottle.

Explanation:

Given: [tex]n_{1}[/tex] = 0.650 mol,     [tex]P_{1}[/tex] = 730 mm Hg (1 mm Hg = 0.00131579 atm) = 0.96 atm

[tex]n_{2}[/tex] = ?,           [tex]P_{2}[/tex] = 1.15 atm

Formula used is as follows.

[tex]\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}\\\frac{0.96 atm}{0.650 mol} = \frac{1.15 atm}{n_{2}}\\n_{2} = 0.779 mol[/tex]

Thus, we can conclude that there are 0.779 moles of gas were added to the bottle.

Answer:

Explanation:

This is a halogenation reaction i.e substitution or replacement of a single or more than a single hydrogen atom in the organic alkane compound with the halogen(here it is chlorine).

The chlorination of 1-chloro-2,2,3,3-tetramethylpentane under UV light resulted in the formation of five (5) dichloro derivatives which are shown in the image attached below.

Also, the compounds containing a stereocenter (i.e a location within the compound composing of various substituents in which the interchangeability of these substituents has the tendency of resulting into a stereoisomer) are indicated with an asterisk in the image below.

From the image below:

compound 1 ⇒  1,1-dichloro-2,2,3,3-tetramethylpentane = 2° C

∴

The given relative reactivity rate for 2°  = 3.6x

For compound 2 ⇒  1,4-dichloro-2,2,3,3-tetramethylpentane = 2°  = 3.6x

For compound 3 ⇒ 1,5-dichloro-2,2,3,3-tetramethylpentane = 1°  = 1x

For compound 4 ⇒ 1-chloro-2-chloromethyl-2,3,3-trimethylpentane

= 1°  = 1x

For compound 5 ⇒ 1-chloro-3-chloromethyl-2,2,3-trimethylpentane

= 1°  = 1x

As such, we have:

2(3.6x) + 3(1x) = 100

7.2x + 3x = 100

10.2x = 100

x = 100/10.2

x = 9.803°

∴

For compound (1) = 3.6(9.803) = 35.3%

For compound (2) = 3.6(9.803) = 35.3%

For compound (3) = 1(9.803) = 9.803°%

For compound (4) = 1(9.803) = 9.803°%

For compound (5) = 1(9.803) = 9.803°%

Answer:

a. minus 10

Explanation:

An element in group 13 = Boron ,valence electrons = 3 , therefore, valence electrons in group 13 = group no. -10

An element in group 18 = Neon, valence electrons = 8 , therefore, valence electrons in group 18 = group no. - 10

For groups 13 through 18, the number of valence electrons in an atom is equal to the group number minus 10. Therefore, option (A) is correct.

Valence electrons in an atom can be described as the electrons occupying the outer most electron shell of an atom while the electrons in the inner shell are core electrons. Lewis structures can be helpful to calculate the number of valence electrons.

Valence electrons can be filled in several electron shells as they are caused interaction between atoms and responsible for the formation of chemical bonds. Only valence electrons can contribute to the formation of a chemical bond and decide the reactivity of the element.

The general electronic configuration of group 13 is ns²np¹ has three valence electrons. It can be determined as group number - 10 = 13 - 10 = 3.

The general electronic configuration of group 18 is ns²np⁶ has eight valence electrons. It can be determined as group number - 10 = 18 - 10 = 8.

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Answer:

2

Explanation:

The number of carbon atoms that are sp²-hybridized in this alkene is 2

Because all the single bonded carbon atoms in the alkene are  sp²-hybridized

There are three(3) single formed via sp² orbitals and one ( 1 ) PI bond formed via Pure-P-orbital

attached below is the some part of the solution

Answer:

A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine was formed

Explanation:

The balanced chemical equation for the decomposition of calcium fluoride is shown below:

[tex]CaF_2(s)->Ca(s)+F_2(g)[/tex]

The sample produced 294 g of calcium then, how many grams of fluorine is formed?

From the balanced chemical equation,

1 mol of CaF2 forms 1mol of calcium and 1 mol of fluorine.

That is:

40g of calcium and 38.0 g of fluorine are formed.

then,

If 294 g of calcium is formed then how many grams of fluorine is formed?

[tex]294g Ca * 38g F2 / 40g Ca\\=279.3 g F_2[/tex]

Hence, 279.3 g of fluorine will be formed.

Answer:

azertyuiopazertyuiiop

Explanation:

High carbon concentration in the deep ocean means increased absorption of carbon to the atmosphere resulting to even greater and harmful amounts of carbon in the atmosphere. Therefore we need to keep a close eye of the deep ocean in the quest to monitor and pump out excess carbon from this part of marine life.

Answer:

Calcium

Explanation:

Zinc cannot displace Ca because calcium is above it in the reactivity series

Answer:

Explanation:

Atomicity is defined as the total number of atoms present in a molecule.

The question is incomplete, the complete question is shown in the image attached

Answer:

A and B

Explanation:

The electrophilic substitution of arenes yields a cation intermediate. The positive charge of the cation is delocalized over the entire ring.

The -CN group directs incoming electrophiles to the ortho/para position. The resonance structures for the chlorination of benzonitrile are shown in the question.

Recall that -CN is an electron withdrawing group. The resonance forms that destablize the carbocation intermediate are those in which the -CN group is directly attached to the carbon atom bearing the positive charge as in structures A and B.

Answer:

The appropriate answer is "9.225 g".

Explanation:

Given:

Required level,

= 63 ppm

Initial concentration,

= 22 ppm

Now,

The amount of free SO₂ will be:

= [tex]Required \ level -Initial \ concentration[/tex]

= [tex]63-22[/tex]

= [tex]41 \ ppm[/tex]

The amount of free SO₂ to be added will be:

= [tex]41\times 225[/tex]

= [tex]9225 \ mg[/tex]

∵ 1000 mg = 1 g

So,

= [tex]9225\times \frac{1}{1000}[/tex]

= [tex]9.225[/tex]

Thus,

"9.225 g" should be added.

Answer:

0.1 L

Explanation:

From the question given above, the following data were obtained:

Concentration of stock solution (C₁) = 1.25 M

Volume of diluted solution (V₂) = 0.5 L

Concentration of diluted solution (C₂) = 0.25 M

Volume of stock solution needed solution (V₁) =?

The volume of the stock solution needed can be obtained as follow:

C₁V₁ = C₂V₂

1.25 × V₁ = 0.25 × 0.5

1.25 × V₁ = 0.125

Divide both side by 1.25

V₁ = 0.125 / 1.25

V₁ = 0.1 L

Therefore, the volume of the stock solution needed is 0.1 L

Answer:

0.1775 M

Explanation:

The reaction that takes place is:

Where HA is the unknown weak acid.

At the equivalence point all HA moles are converted by NaOH. First we calculate how many NaOH moles reacted, using the given concentration and volume:

That means that in 4.00 mL of the weak acid solution, there were 0.71 weak acid mmoles. With that in mind we can now calculate the concentration:

Answer:

I don't know what to do

Explanation:

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Answer:

a. Single DNA strand used to produce mRNA ⇒ Transcription

b. Requires tRNA ⇒ Translation

c. Ribosome ⇒ Translation

d. DNA polymerase ⇒ Replication

e. Both DNA strands are duplicate  ⇒ Replication

f. Described as semi-conservative  ⇒ Replication

g. Amino acids added to peptide chain ⇒ Translation

Explanation:

Replication: the double-strand DNA is separated into two strands. Each strand is used as a template by DNA polymerase to produce the other strand. The leading strand is read by DNA polymerase in a continuous form (3' - 5') and the lagging strand is read in a discontinuous form (5'-3'). In this way, both strands are duplicated. The process is semi-conservative because the DNA molecule produced conserves 1 original strand and the other strand is the new synthesized one. The corresponding options are: d, e, f.

Transcription: is a process in which the genetic code of DNA is copied into a molecule called messenger RNA (mRNA). The double-strand DNA is opened and one strand is read. The enzyme involved is RNA polymerase, which binds the DNA (in a sequence called promoter) and uses the nucleotide code of DNA as a template to produce a molecule of RNA (the mRNA). Therefore, the correct option for this process is a.

Translation: is the process in which the mRNA sequence (copied from a DNA molecule) is translated into an amino acid sequence to produce a protein. This process is carried out within the cell ribosome. The mRNA is read in groups of three nucleotides (a codon) that codifies amino acids. The translation between codon and amino acid is assisted by molecules called transference RNA (tRNA). As each codon is decoded, an amino acid is added and the new polypeptide sequence is synthesized. Therefore, the correct options are: b, c, g.

Answer: The density of the given vapor is 0.939 g/L.

Explanation:

Given: Pressure = 233 mm Hg (1 mm Hg = 0.00131579 atm) = 0.31 atm

Temperature = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K

According to the ideal gas equation,

[tex]PV = \frac{m}{M}RT[/tex]

where,

P = pressure

V = volume

m = mass

M = molar mass

R = gas constant = 0.0821 L atm/mol K

T = temperature

This formula can be re-written as follows.

[tex]PM = \frac{m}{V}RT[/tex]    (where, [tex]Density = \frac{mass (m)}{Volume (V)}[/tex] )

Hence, formula used to calculate density of diethy ether (molar mass = 74.12 g/mol) vapor is as follows.

[tex]d = \frac{PM}{RT}[/tex]

Substitute values into the above formula as follows.

[tex]d = \frac{PM}{RT}\\= \frac{0.31 atm \times 74.12 g/mol}{0.0821 L atm/mol K \times 298 K}\\= \frac{22.9772}{24.4658}\\= 0.939 g/L[/tex]

Thus, we can conclude that the density of the given vapor is 0.939 g/L.

Explanation:

This is a decomposition reaction. Firstly, you will want to write the chemical equation out and balance it.

[tex]2Hg_2O->4Hg+O_2[/tex] (The -> is supposed to be an arrow, sorry!)

We see that there's only 1mol of Oxygen made in the products, we can do some simple math to solve for the amount of grams of Oxygen produced according to the amount of the reactant (Hg2O).

[tex]32.2gHg_2O*\frac{1molHg_2O}{417.18gHg_2O}*\frac{1molO_2}{2molHg_2O}*\frac{32gO_2}{1molO_2}[/tex]

I want to break this down, just in case:

The 417.18gHg2O is the molecular mass of the molecule (so I doubled Hg and added 16 to it to get this number).

As we can see in the chemical equation, 1mol Hg2O produces 2mol O because Oxygen is a diatomic molecule (so there will always be two of it when it's by itself).

And finally, in 1mol O2 there are 32g of O2.

** When you do math like this, always make sure that all of your units cancel out except for the units you're looking for. For example, here we're looking for the grams of Oxygen, so after everything else cancels out, we should only have grams O2.

So, 1.23gO2 should be your answer.

Answer:

21.63 %

Explanation:

The molar mass of Mn₃[Mn(CO)₄]₃ is 665.64 g/mol.

Let's assume we have 1 mol of Mn₃[Mn(CO)₄]₃, if that were the case then we would have 665.64 grams.

There are 12 C moles per Mn₃[Mn(CO)₄]₃, with that in mind we calculate the weight of 12 C moles:

Finally we calculate the mass percent of carbon:

The question is incomplete, the complete question is:

What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.

Answer: 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

Explanation:

We first calculate the number of moles of soft drink in a volume of 10 mL

The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(1)

Taking the concentration of soft drink from the example be = 0.375 M

Volume of solution = 10 mL

Putting values in equation 1, we get:

[tex]0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol[/tex]

Calculating volume of sweetened tea:

Moles of sugar = 0.00375 mol

Molarity of sweetened tea = 0.05 M

Putting values in equation 1, we get:

[tex]0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL[/tex]

Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

Water is in the solid phase at 0.25 atm and 0°C.

A pressure of 50 kPa and a temperature of 50 °C correspond to the “water” region—here, water exists only as a liquid. At 25 kPa and 200 °C, water exists only in the gaseous state.

Under standard atmospheric conditions, water exists as a liquid. But if we lower the temperature below 0 degrees Celsius, or 32 degrees Fahrenheit, water changes its phase into a solid called ice.

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Answer:

d

Explanation: