Answer:
B. CH3Br
Explanation:
Dipole -Dipole interactions take place in polar molecules.
CH3Br exhibits dipole -dipole forces as its strongest attraction between molecules because it is a polar molecule due to the slightly negative dipole present on the Br molecule.
While O2 is a nonpolar molecule due to its linear structure, CCl4 has zero resultant dipole moment, Helium is non-polar and BrCH2CH2OH is a non polar compound having net dipole moment is zero.
Hence, the correct option is B. CH3Br.
The compound that exhibits dipole -dipole forces is CH3Br
Dipole -Dipole interactions:It should be taken place in polar molecules. Also, CH3Br should be the strongest attraction that lies between the molecules since it is treated as the polar molecule because of the slightly negative. While on the other hand, O2 should be non-polar molecule because of the linear structure.
Therefore, The compound that exhibits dipole -dipole forces is CH3Br
Learn more about force here: https://brainly.com/question/14379797
Which is a nonpolar molecule?
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.
The following initial rate data apply to the raction
F2(g) + 2Cl2O(g) ---> 2FClO2(g) +Cl2(g)
Expt. [F2] (M) [Cl2O] (M) Intitial rate (M/s)
1 0.05 0.010 5 x 10^-4
2 0.05 0.040 2.0 x 10^-3
3 0.10 0.010 1.0 x 10^-3
Which of the following is the rate law (rate equation) for this reaction?
A. rate= k[F2]^2 [Cl2O]^4
B. rate= k[F2]^2 [Cl2O]
C. rate= k[F2] [Cl2O]
D. rate= k[F2] [Cl2O]^2
E. rate= k[F2]^2 [Cl2O]^2
Answer:
C. rate = k[F₂] [Cl₂O]
Explanation:
Based on the reaction, rate law can be obtained from the initial concentration of reactants thus:
rate = k[F₂]ᵃ [Cl₂O]ᵇ
Where the exponents a and b can be finded doing a experiment changing initial concentrations and seeing how a variation contribute in rate law.
If you analize experiments 1 and 2, the only change is [Cl₂O] (From 0.010 to 0.040, four times more) that changes its concentration in four times. This change produce rate law change from 5x10⁻⁴ to 2.0x10⁻³, also four times. That means the exponent b of [Cl₂O] is 1.
rate = k[F₂]ᵃ [Cl₂O]ᵇ
rate = k[F₂]ᵃ [Cl₂O]¹
Now, comparing experiments 1 and 3, the [F₂] change from 0.05 to 0.10, (Twice), and initial rate change from 5x10⁻⁴ to 1x10⁻³ (Also, twice). That means a = 1 and rate law is:
rate = k[F₂]¹ [Cl₂O]
rate = k[F₂] [Cl₂O]
Thus, right answer is:
C. rate = k[F₂] [Cl₂O]Draw the Lewis structure of N₂O₄ and then choose the appropriate pair of hybridization states for the two central atoms. Your answer choice is independent of the orientation of your drawn structure.
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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State five difference between ionic compound and covalent compound
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.
After heating a sample of hydrated CuSO4, the mass of released H2O was found to be 2.0 g. How many moles of H2O were released if the molar mass of H2O is 18.016 g/mol
Answer:
0.1110 mol
Explanation:
Mass = 2g
Molar mass = 18.016 g/mol
moles = ?
These quantities are realted by the following equation;
Moles = Mass / Molar mass
Substituting the values of the quantities and solving for moles, we have;
Moles = 2 / 18.016 = 0.1110 mol
Draw the Lewis structure for methane (CH4) and ethane (C2H6) in the box below. Then predict which would have the higher boiling point. Finally, explain how you came to that conclusion.
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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Where are the lanthanides and actinides found on the periodic table?
A. Columns 7 and 8
B. Columns 3 - 12, in the center of the table
C. Rows 6 and 7, separated from the rest of the table
D. Columns 1 and 2
Answer:
C. Rows 6 and 7, separated from the rest of the table
Explanation:
The lanthanides and actinides are groups of elements in the periodic table, that are thirty (30) in number. They are separated from the rest of the periodic table, usually appearing as separate rows at the bottom. They are often called the inner transition metals, because they all fill the f-block.
Therefore, the correct option is C
" They are found in Rows 6 and 7, separated from the rest of the table"
For the reaction X + Y → Z, the reaction rate is found to depend only upon the concentration of X. A plot of 1/X verses time gives a straight line. What is the rate law for this reaction?
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
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 1.5 L flask with 0.59 atm of sulfur dioxide gas and 2.9 atm of oxygen gas at 35.0 °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.53 atm.
Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.
Kp=_______.
Answer:
P SO₂ = 0.06atm
P O₂ = 2.635atm
P SO₃ = 0.53atm
Kp = 29.6
Explanation:
The reaction of Sulfur dioxide and oxygen react to form sulfur trioxide is as follows:
2SO₂(g) + O₂(g) ⇄ 2SO₃(g)
And Kp is defined as:
[tex]Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}[/tex]
Where P represents the pressure at equilibrium of each reactant.
If you add, in the first, 0.59atm of SO₂ and 2.9atm of O₂, the equilibrium pressures will be:
P SO₂ = 0.59atm - 2X
P O₂ = 2.9atm - X
P SO₃ = 2X
Where X represents the reaction coordiante.
As equilibrium pressure of SO₃ is 0.53atm:
0.53atm = 2X
0.265atm = X
Replacing, equilibrium pressures of each species will be:
P SO₂ = 0.59atm - 2×0.265atm
P O₂ = 2.9atm - 0.265atm
P SO₃ = 2×0.265atm
P SO₂ = 0.06atmP O₂ = 2.635atmP SO₃ = 0.53atmAnd Kp will be:
[tex]Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}[/tex]
[tex]Kp = \frac{0.53^2}{{0.06}^2*{2.635}}[/tex]
Kp = 29.6Two football players are running toward each other. One football player has a mass of 105 kg and is running at 8.6 m/s. The other player has a mass of 90 kg and is running at -9.0 m/s. What is the momentum of the system after the football players collide? 93 kg · m/s 1,713 kg · m/s. 810 kg · m/s. 903 kg · m/s.
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:
What compound is formed when 2,2-dimethyloxirane (2-methyl-1,2-epoxypropane) is treated with ethanol containing sulfuric acid
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!
* Question Completion Status:
QUESTION 1
'What compound represents conjugate base in the following chemical reaction? H2SO4 + H2O HSO4 + H30+
O a. H2SO4
O b. H2O
O c. HSO4
O d. H30+
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.
A sample of N2(g) was collected over water at 25 oC and 730 torr in a container with a volume of 340 mL. The vapor pressure of water at 25 oC is 23.76 torr. What mass of N2 was collected
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.
Permanganate ion reacts in basic solution with oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skeleton ionic equation for the reaction between NaMnO4 and Na2C2O4 in basic solution: Fill in all blanks with numbers so if the term is not in the equation make it 0.
Mno4^- (aq)+ C204^2- (aq)+
H^+(aq) + OH^-(aq)
H2O(l) MnO2(s)+
CO3^2 (aq)+ H^+(aq)+
OH^- (aq) + H2O(l)
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)
hen adding a solute to water, the vapor pressure will __________ and the boiling point will __________.
Answer: When a solute is added to water, the vapor pressure will decrease and the boiling point will increase.
Explanation:
When a solute is added to water, a solvent's vapor pressure will decrease because of the displacement of solvent molecules by the solute. i.e. some of the solvent molecules at the surface of the water are replaced by the solute.When a solute is added to water, a solvent's boiling point will increase because water molecules need more energy to produce required pressure to escape the boundary of the liquid , so as the number of particles increase in the liquid it increase the boiling point.How many milliliters of a 0.250 MNaOHMNaOH solution are needed to completely react with 500. gg of glyceryl tripalmitoleate (tripalmitolein)
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!
Green plants use light from the Sun to drive photosynthesis. Photosynthesis is a chemical reaction in which water and carbon dioxide chemically react to form the simple sugar glucose and oxygen gas . What mass of simple sugar glucose is produced by the reaction of 4.9 of carbon dioxide?
Answer:
3.3 g of glucose, C6H12O6.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
6CO2 + 6H2O —> C6H12O6 + 6O2
Next, we shall determine the mass of CO2 that reacted and the mass of C6H12O6 produced from the balanced equation.
This is illustrated below:
Molar mass of CO2 = 12 + (2x16) = 44 g/mol
Mass of CO2 from the balanced equation = 6 x 44 = 264 g
Molar mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 180 g/mol
Mass of C6H12O6 from the balanced equation = 1 x 180 = 180 g
From the balanced equation above,
264 g of CO2 reacted to produce 180 g of C6H12O6.
Finally, we shall determine the mass of C6H12O6 produced by reacting 4.9 g of CO2 as follow:
From the balanced equation above,
264 g of CO2 reacted to produce 180 g of C6H12O6.
Therefore, 4.9 g of CO2 will react to produce = (4.9 x 180)/264 = 3.3 g of C6H12O6.
Therefore, 3.3 g of glucose, C6H12O6 were obtained from the reaction.
What is the molar concentration of H atoms at equilibrium if the equilibrium concentration of H2 is 0.28 M? Express your answer to two significant figures and include the appropriate units.
Answer:
0.56M
Explanation:
Molar concentration is defined as the ratio between moles of solute and volume in liters of solution.
In a 0.28M H₂ there are 0.28moles of H₂ per liter of solution.
Now, in 1 molecule of H₂ there are 2 atoms of H. Following this idea, in 0.28 moles of H₂ there are 0.28*2 = 0.56 moles of H atoms.
Thus, molar concentration of H atoms in a 0.28M H₂ is 0.56M
If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)?
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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A 25.0-mL sample of 0.100M Ba(OH)2(aq) is titrated with 0.125 M HCl(aq).
How many milliliters of the titrant will be needed to reach the equivalence point?
Answer:
20.0
Explanation:
NaOH = (25.0) (0.100m) \ 0.125M = 20.0mL
AB2AB2 has a molar solubility of 3.72×10−4 M3.72×10−4 M. What is the value of the solubility product constant for AB2AB2? Express your answer numerically.
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
Ksp = 4X³As molar solubility of the salt is 3.72x10⁻⁴M:
Ksp = 4X³
Ksp = 4(3.72x10⁻⁴)³
Ksp = 2.06x10⁻¹⁰A major component of gasoline is octane, C8H18. When octane is burned in air, it chemically reacts with oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O) . What mass of carbon dioxide is produced by the reaction of 3.2g of oxygen gas? Round your answer to 2 significant digits.
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.
Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter? 
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]
A newly found element with the symbol J has two naturally occurring isotopes. Isotope one has an atomic mass of 139.905 amu and an abundance of 37.25%. Isotope two has an atomic mass of 141.709 amu and an abundance of 62.75%. Calculate the mass of the element.
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
interpret the electron configuration
Answer:
Ca for calcium
20 electrons
2-2s electron
Calculate the entropy change for the reaction: HCl(g) + NH3(g) -> NH4Cl(s) Entropy data: HCl: 187 J/K mol NH3: 193 J/K mol NH4Cl: 94.6 J/K mol
Answer:
-285.4 J/K
Explanation:
Let's consider the following balanced equation.
HCl(g) + NH₃(g) ⇒ NH₄Cl(s)
We can calculate the standard entropy change for the reaction (ΔS°r) using the following expression.
ΔS°r = 1 mol × S°(NH₄Cl(s)) - 1 mol × S°(HCl(g)) - 1 mol × S°(NH₃(g))
ΔS°r = 1 mol × 94.6 J/K.mol - 1 mol × 187 J/K.mol - 1 mol × 193 J/K.mol
ΔS°r = -285.4 J/K
Answer:
-198.3 J/K mol
Explanation:
I got it correct on founders edtell
Identify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. do you think it is reasonable to exclude these types of substances as contaminants that would give a false positive when you tested your reaction residue to verify that it is KCl?
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.
Calculate the moles of acetic acid used in each trial and record in Data Table 3.Volume acetic acid in Liters = (Mass/Density)/1000Moles = Volume ∗ Concentration
Calculate the moles of magnesium hydroxide (Mg(OH)2) used in each trial and record in Data Table 3.Moles Mg(OH2) = Moles acetic acid
Calculate the neutralization capacity of each trial and record in Data Table. Neutralization Capacity = Moles Mg(OH)2 / Mass of Milk of Magnesium
Data Table Trial 1 Trial 2
Mass of milk of magnesia 2.5g 2.5g
Density of milk of magnesia 1.14 g/mL 1.14 g/mL
Volume of acetic acid, initial 10mL 10mL
Volume of acetic acid, final 2.2mL 1.8mL
Volume of acetic acid, total 7.8mL 8.2mL
Concentration of acetic acid 0.88 M 0.88 M
Moles of acetic acid
Moles of Mg(OH)2
Moles Mg(OH)2 / g milk of magnesia
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
Hydrazine, N2H4 , reacts with oxygen to form nitrogen gas and water. N2H4(aq)+O2(g)⟶N2(g)+2H2O(l) If 2.45 g of N2H4 reacts with excess oxygen and produces 0.450 L of N2 , at 295 K and 1.00 atm, what is the percent yield of the reaction?
Answer:
24.15%
Explanation:
According to the given situation the computation of the percent yield of the reaction is shown below:-
PV = NRT = N = [tex]\frac{PV}{RT}[/tex]
Mole of [tex]N_2[/tex] = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{1\times 0.450}{0.0821\times 295}[/tex]
= [tex]\frac{0.450}{24.2195}[/tex]
= 0.0186
Mole of [tex]N_2H_4 = \frac{2.45}{32}[/tex]
= 0.077
Now, the percentage of yield is
= [tex]\frac{Practical\ yield}{Theoretical\ yield}\times 100[/tex]
= [tex]\frac{0.0186}{0.077}\times 100[/tex]
= 24.15%
Therefore for computing the percentage of yield we simply divide the practical yield by theoretical yield and multiply with 100 so that we can get the result into the percentage form.
What is the mass of a sample of water that takes 2000 kJ of energy to boil into steam at 373 K. The latent heat of vaporization of water is 2.25 x 10^6 J kg-1
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