230g sample of a compound contains 136.6g carbon, 26.4g hydrogen, and 31.8g nitrogen. What is masspercentif oxygen

Answer:

Molecule C

Explanation:

In this case, on the first reaction, we will have the production of a Grignard reagent. This molecule will react with [tex]D_2O[/tex] and a deuterium atom will be transferrred to the benzene ring. Then at the top of the molecule, we will have an acetal structure. This acetal can be broken by the action of the acid [tex]DCl[/tex], In the mechanism at the end, we will obtain a carbonyl group bonded to a hydrogen atom. Therefore we will have in the final product the aldehyde group. See figure 1 to further explanations.

I hope it helps!

Answer:

35%

Explanation:

Percentage yield = actual yield / theoretical yield × 100.

Given:

Actual yield = 63.07g

Theoretical yield = ?

Mole ratio of C7H6O3 to C4H6O3 = 1 : 1

1 mole of C7H6O3 - 138.1g

Which implies that only 1 mole s[tex]\frac{63.07}{180.2} * 100[/tex]hould be used up in the reaction, yielding 180.2 g of C9H8O4. ⇒ Theoretical yield = 180.2g

∴ % Yield = [tex]\frac{63.07}{180.2} * 100[/tex]

= 35% yield.

Let me know if you found this easy to understand.

Answer:

20.0

Explanation:

NaOH = (25.0) (0.100m) \ 0.125M = 20.0mL

Answer:

r = k [X]²

Explanation:

X + Y → Z

Generally, the rate of reaction depends on the concentration of reactants. However, the question stated that the rate depends only on reactant X.

The plot of 1/X versus time giving a straight line signifies that this is a second order reaction.

For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.

From this, our rate law is r = k [X]²

Answer:

How the chemical reacts

Explanation:

Answer:

[tex]7.48X10^3~mL[/tex]

Explanation:

For this question we have:

-) A solution NaOH 0.25 M

-) 500 g of glyceryl tripalmitoleate (tripalmitolein)

We can start with the reaction between NaOH and tripalmitolein. NaOH is a base and tripalmitolein is a triglyceride, therefore we will have a saponification reaction. The products of this reaction are glycerol and (E)-hexadec-9-enoate.

Now, with the reaction in mind, we can calculate the moles of NaOH that we need if we use the molar ratio between NaOH and tripalmitolein (3:1) and the molar mass of tripalmitolein (801.3 g/mol). So:

[tex]500~g~tripalmitolein\frac{1~mol~tripalmitolein}{801.3~g~tripalmitolein}\frac{3~mol~NaOH}{1~mol~tripalmitolein}=1.87~mol~NaOH[/tex]

With the moles of NaOH we can calculate the volume (in litters) if we use the molarity equation and the Molarity value:

[tex]M=\frac{mol}{L}[/tex]

[tex]0.25~M=\frac{1.87~mol~NaOH}{L}[/tex]

[tex]L=\frac{1.87~mol~NaOH}{0.25~M}[/tex]

[tex]L=7.48[/tex]

Now we can do the conversion to mL:

[tex]7.48~L~\frac{1000~mL}{1~L}=~7.48X10^3~mL[/tex]

I hope it helps!

Answer:

Compound are defined as the containing two or more different element .

(1) Ionic compound and (2) Covalent compound.

Explanation:

Covalent compound :- covalent compound are the sharing of electrons two or more atom.

Covalent compound are physical that lower points and compared to ionic .

Covalent compound that contain bond are carbon monoxide (co), and methane .

Covalent compound are share the pair of electrons.

Covalent compound are bonding a hydrogen atoms electron.

Ionic compound a large electrostatic actions between atoms.

Ionic compound are higher melting points and covalent compound.

Ionic compound are bonding a nonmetal electron.

Ionic electron can be donate and received ionic bond.

Ionic compound bonding kl.

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

[tex]w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln} *100\%\\\\w/w\%=3.15\%[/tex]

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

For the w/w% of the solution, information about the molecular mass of the solute, and density of the solution has been required.

Molarity can be defined as the moles of the solute per liter of the solution. The molarity can be used for the determination of the weight of the solute, by the information about the molecular weight of the compound.

Thus, for the w/w% of the solution, the weight of the solute has been determined with information about the molecular mass of the solute.

The weight of the solvent has been determined with the density of the solution. The density has been defined as the mass per unit volume.

Thus, for the w/w% of the solution, the weight of the solvent has been determined by the density of the solution.

For more information about the w/w% of the solution, refer to the link:

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Answer:

The mass of the element is 141.03701 amu

Explanation:

The catch here is that it notes a " newly found element. " Otherwise you could just refer to the average atomic mass of the element in the periodic table, and receive your solution in a much faster way.

The first isotope has an atomic mass of 139.905 amu, and a respective percent abundance of 37.25%. The second isotope has an atomic mass of 141.709 amu, and the remaining percent abundance, 100% - 37.25% = 62.75% ( given ). We can calculate the mass of the unknown element by associating each percentage with the mass of their respective isotope, over 100%.

Mass = ( ( 139.905 amu )( 37.25% ) + ( 141.709 amu )( 62.75% ) )/ 100,

Mass = ( ( 5211.46125 ) + ( 8892.23975 ) ) / 100,

Mass = ( 14103.701 ) / 100 = 141.03701 amu

Answer:

0.89kg

Explanation:

Q=mL L=specific latent heat

Q=energy required in J

m=mass in Kg

Q=mL

m=Q/L

m=2000000J/2.25 x 10^6 J kg-1

m=0.89kg

Answer:

ΔH = [tex]q_{p}[/tex]

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

[tex]q_{com}[/tex] = [tex]q_{p}[/tex]

where

[tex]q_{p}[/tex] = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.

[tex]q_{com}[/tex]  = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

ΔH = the total heat absorbed by the system

ΔQ = the internal heat absorbed by the system which in this case is [tex]q_{p}[/tex]

ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0

substituting into the heat change equation

ΔH = [tex]q_{p}[/tex] + 0

==> ΔH = [tex]q_{p}[/tex]

Answer:

The balanced equation is: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

Explanation:

Zn(s) is a simple substance (its oxidation number is zero) and it is oxidized to Zn²⁺. It loses two electrons, so the half reaction is the following:

Zn(s) → Zn²⁺(aq) + 2 e-   (oxidation reaction)

Hydrogen ion (H⁺) is reduced to hydrogen gas (H₂). The oxidation number is decreased from +1 to 0 (because H₂ is a simple substance). H⁺ gains 1 electron per H atom, so the half reaction is the following:

2H⁺(aq) + 2 e- → H₂(g) (reduction reaction)

We obtain the overall reaction from the addition of the two half reactions. We write the reduction reaction first and then the oxidation reaction, as follows:

2H⁺(aq) + 2 e- → H₂(g)

+

Zn(s) → Zn²⁺(aq) + 2 e-

---------------------------------

Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

The two electrons at both sides of the equation (2 e-) are canceled. The overall reaction is in acidic solution due to the presence of H⁺ ions. The net charge at both sides is the same : +2, so the mass and the charge are balanced.

Answer:

Ethane would have a higher boiling point.

Explanation:

In this case, for the lewis structures, we have to keep in mind that all atoms must have 8 electrons (except hydrogen). Additionally, each carbon would have 4 valence electrons, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.

Now, the main difference between methane and ethane is an additional carbon. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have more area of interaction for ethane. If we have more area of interaction we have to give more energy to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.

I hope it helps!

The Lewis structure shows the valence electrons in a molecule. Ethane will have a higher boiling point than methane.

We can deduce the number of valence electrons in a molecule by drawing the Lewis structure of the molecule. The Lewis structure consists of the symbols of elements in the compound and the valence electrons in the compound.

We know that the higher the molar mass of a compound the greater its boiling point. Looking at the Lewis structures of methane and ethane, we cam see that ethane has a higher molecular mass (more atoms) and consequently a higher boiling point than methane.

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Answer:

2-ethoxy-2-methylpropan-1-ol

Explanation:

On this reaction, we have an "epoxide" (2-methyl-1,2-epoxypropane). Additionally, we have acid medium (due to the sulfuric acid [tex]H_2SO_4[/tex]). The acid medium will produce the hydronium ion ([tex]H^+[/tex]). This ion would be attacked by the oxygen of the epoxide. Then a carbocation would be produced, in this case, the most stable carbocation is the tertiary one. Then an ethanol molecule acts as a nucleophile and will attack the carbocation. Finally, a deprotonation step takes place to produce 2-ethoxy-2-methylpropan-1-ol.

See figure 1

I hope it helps!

Answer:

I think its B

Explanation:

Precipitation reactions leave a solid behind. The solid is called a precipitate.

Answer:

B

Explanation:

An insoluble solid will form when ions in dissolved compounds switch places.

Answer:

Explanation:

A nonpolar molecule has no separation of charge, so no positive or negative poles are formed. In other words, the electrical charges of nonpolar molecules are evenly distributed across the molecule. Nonpolar molecules tend to dissolve well in nonpolar solvents, which are frequently organic solvents. The answer is hydrogen cyanide.

Answer:

0.36 g of N2.

Explanation:

The following data were obtained from the question:

Temperature (T) = 25 °C

Volume (V) = 340 mL

Measured pressure = 730 torr

Vapour pressure = 23.76 torr

Mass of N2 =..?

First, we shall determine the true pressure of N2. This can be obtained as follow:

Measured pressure = 730 torr

Vapor pressure = 23.76 torr

True pressure =..?

True pressure = measured pressure – vapor pressure

True pressure = 730 – 23.76

True pressure = 706.24 torr.

Converting 706.24 torr to atm, we have:

760 torr = 1 atm

Therefore,

706.24 torr = 706.24 / 760 = 0.929 atm

Next, we shall convert 340 mL to L. This is illustrated below:

1000 mL = 1 L

Therefore,

340 mL = 340/1000 = 0.34 L

Next, we shall convert 25 °C to Kelvin temperature. This is illustrated below:

Temperature (K) = Temperature (°C) + 273

T(K) = T (°C) + 273

T (°C) = 25 °C

T(K) = 25 °C + 273

T (K) = 298 K

Next, we shall determine the number of mole of N2. This can be obtained as follow:

Pressure (P) = 0.929 atm

Volume (V) = 0.34 L

Temperature (T) = 298 K

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =...?

PV = nRT

0.929 x 0.34 = n x 0.0821 x 298

Divide both side by 0.0821 x 298

n = (0.929 x 0.34 ) /(0.0821 x 298)

n = 0.0129 mole

Finally, we shall determine the mass of N2 as shown below:

Mole of N2 = 0.0129 mole

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 =.?

Mole = mass /Molar mass

0.0129 = mass of N2/ 28

Cross multiply

Mass of N2 = 0.0129 x 28

Mass of N2 = 0.36 g

Therefore, 0.36 g of N2 was collected.

Answer:

The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

Explanation:

Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.

C = k*P

where C is the solubility, P the partial pressure and k is the Henry constant.

So, being the concentration [tex]C=\frac{ngas}{V}[/tex]  

where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So [tex]n=\frac{P*V}{R*T}[/tex]

In this case:

Then:

[tex]n=\frac{1 atm*5.16*10^{-2} L}{0.082 \frac{atm*L}{mol*K} *298K}[/tex]

Solving:

n= 2.11 *10⁻³ moles

So: [tex]C=\frac{ngas}{V}=\frac{2.11*10^{-3} moles}{1 L} =2.11*10^{-3} \frac{moles}{L}= 2.11*10^{-3} M[/tex]

Using Henry's Law and being C=CAr and P =PAr:

2.11*10⁻³ M= k* 1 atm

Solving:

[tex]k=\frac{2.11*10^{-3} M}{1 atm}[/tex]

You get:

[tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

The Henry's law constant for argon gas in 1 litre of water is 2.1 × 10⁻³M/atm.

Henry's law of gas states that solubility of a gas in any liquid at particular temperature is directly proportional to the partial pressure of the gas.

C∝P

C = kP, where

k = Henry's constant

P = partial pressure of gas

C is the solubility and it is present in the form of concentration and will be calculated as:
C = n/V

n = no. of moles

V = volume

And moles of the gas will be calculated by using the ideal gas equation as:

PV = nRT

n = (1)(5.16×10⁻²) / (0.082)(298) = 2.1 × 10⁻³ moles

And Concentration in liquid will be:

C = 2.1 × 10⁻³mol / 1L = 2.1 × 10⁻³ M

Now we put all these values in the first equation to calculate the value of k as:

k = (2.1 × 10⁻³M) / (1atm) = 2.1 × 10⁻³M/atm

Hence required value of k is 2.1 × 10⁻³M/atm.

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Answer:

Ca for calcium

20 electrons

2-2s electron

Answer:

See explanation

Explanation:

In this case, we have to keep in mind the valence electrons for each atom:

N => 5 electrons

O => 6 electrons

If the formula is [tex]N_2O_4[/tex], we will have in total:

[tex](5*2)+(6*4)=34~electrons[/tex]

Additionally, we have to remember that each atom must have 8 electrons. So, for oxygens 5 and 3 we will have 3 lone pairs and 1 bond (in total 8 electrons. For oxygens, 6 and 4 we will have 2 lone pairs and 2 bonds (in total 8 electrons) and for nitrogens 1 and 2 we will have 4 bonds (in total 8 electrons).

To find the hybridization, we have to count the atoms and the lone pairs around the nitrogen. We have 3 atoms and zero lone pairs. If we take into account the following rules:

[tex]Sp^3~=~4[/tex]

[tex]Sp^2~=~3[/tex]

[tex]Sp~=~2[/tex]

With this in mind, the hybridization of nitrogen is [tex]Sp^2[/tex].

See figure 1

I hope it helps!

The central nitrogen atoms in N2O4 are both sp2 hybridized.

The Lewis structure shows the number of electron pairs that surround the atoms in a molecule as dots. It is quite easy to determine the number of valence electrons in a molecule simply by observing its Lewis dot structure.

The molecule N2O4 has 34 valence electrons as shown in its dot electron structure. The central nitrogen atoms in N2O4 are both sp2 hybridized as shown. The formal charges on each atom in N2O4 are also shown.

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Answer:

1.55 × 10²⁵ atoms of H  

Explanation:

3.21mol C₃H₈ × 8mol H × (6.022×10²³)

Answer:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Explanation:

First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.

Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.

Reduction:

MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)

Oxidation:

C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-

Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.

2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)

3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-

Now combine both equations and eliminate repeating H+ and H2O.

2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)

turns into:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Answer: b. HSO4-

Explanation:

H2SO4 (sulfuric acid) will donate a hydrogen ion in solution to form H3O+(hydronium). The remaining HSO4- would be the conjugate base of this dissociation.

A conjugate base contains one less H atom and one more - charge than the acid that formed it.

A conjugate acid contains one more H atom and one more + charge than the base that formed it.

Answer:

[tex]m_{CO_2}=2.8gCO_2[/tex]

Explanation:

Hello,

In this case, the combustion of octane is chemically expressed by:

[tex]C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O[/tex]

In such a way, due to the 25/2:8 molar ratio between oxygen and carbon dioxide, we can compute the yielded grams of carbon dioxide (molar mass 44 g/mol) as shown below:

[tex]m_{CO_2}=3.2gO_2*\frac{1molO_2}{32gO_2} *\frac{8molCO_2}{\frac{25}{2}molO_2 } *\frac{44gCO_2}{1molCO_2}\\ \\m_{CO_2}=2.8gCO_2[/tex]

Best regards.

Answer:

Ksp = 2.06x10⁻¹⁰

Explanation:

For AB₂. solubility product constant, Ksp, is written as follows:

AB₂(s) ⇄ A²⁺ + 2Br⁻

Ksp = [A²⁺] [Br⁻]²

Molar solubility represents how many moles of AB₂ are soluble per liter of solution. and is obtained from Ksp:

AB₂(s) ⇄ A²⁺ + 2Br⁻

AB₂(s) ⇄ X + 2X

where X are moles that are soluble (Molar solubility)

Ksp = [X] [2X]²

As molar solubility of the salt is 3.72x10⁻⁴M:

Ksp = 4X³

Ksp = 4(3.72x10⁻⁴)³

Answer:

Trial 1: Moles acetic acid = 0.00686 moles;

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = 0.00137 mol/g

Trial 2: Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = 0.00144 mol/g

Explanation:

Equation of the reaction: 2CH₃COOH + Mg(OH)₂ ---> Mg(CH₃COO)₂ + 2H₂O

Trial 1:

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 7.8 mL = (7.8/1000) Litres = 0.0078 L

Moles acetic acid = 0.88 M * 0.0078 L

Moles acetic acid = 0.00686 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00686 moles of acetic acid will react with 0.00686/2 moles of Mg(OH)₂ = 0.00343 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00343 mole /2.5 g

Neutralization capacity = 0.00137 mol/g

Trial 2.

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 8.2 mL = (8.2/1000) Litres = 0.0082 L

Moles acetic acid = 0.88 * 0.0082

Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00722 moles of acetic acid will react with 0.00722/2 moles of Mg(OH)₂ = 0.00361 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00361 mole /2.5 g

Neutralization capacity = 0.00144 mol/g

Answer:

-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.

-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides

Explanation:

In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.

With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.

Answer:

1. Na + O2 → Na2O (Balanced)

2. 4Al + 3O2 → 2(Al2O3) (Balanced)

3. H2 + i2 → 2HI (Balanced)

4. Mg + 2H2O → Mg(OH)2+ H2 (Balanced)

5. 2Ca +O2 → 2CaO (Balanced)

Answer:

Total momentum of both player after collision =93  Kg m/s

Explanation:

According to law of conservation of momentum

For an isolated system of bodies , momentum of bodies before and after collision remains same.

momentum is given by mass* velocity

_________________________________________

Here the isolated system of bodies are

two football players.

Momentum of player before collision

Momentum of player 1 = 105*8.6 = 903 Kg m/s

Momentum of player 2 = 90*-9 = -810 Kg m/s

Total momentum of both player before collision = 903 + (-810) = 93 Kg m/s

as by conservation of

Total momentum of both player before collision = Total momentum of both player after collision

Total momentum of both player after collision =93  Kg m/s

Answer:A is the Answer

Explanation:

Answer:

The side chains of the amino acids alternate above and below the sheet

Explanation:

Hydrogen bonds are formed between the amine and carbonyl groups across strands. ... Antiparallel ß sheets are slightly more stable than parallel ß sheets because the hydrogen bonding pattern is more optimal.