Answer:
[tex]Sc_2O_3[/tex] reacts with an acidic solution
Explanation:
Scandium Oxide [tex]Sc_2O_3[/tex] is a basic metal oxide which therefore reacts with acidic solution. An oxide is a compound that contains only two elements, one of which is oxygen .
The objective of this question is to Write a balanced chemical equation to support your answer.
The chemical equation to support the reaction of [tex]Sc_2O_3[/tex] with acidic solution is as follows:
Assuming the acidic solution to be HCl
[tex]\mathbf{Sc_2O_3_{(s)} + 6 HCl_{(aq)} ----> 2 ScCl_{3(aq)} + 3H_2O_{(l)}}[/tex]
The ionic equation :
[tex]\mathbf{Sc_2O_{3(s)} + 6H^+_{(aq)} ---> 2Sc^{3+}_{(aq)} + 3H_2O_{(l)}}[/tex]
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
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)
Give the formulas for all of the elements that exist as diatomic molecules under normal conditions. See if you can do this without looking anything up.
Answer:
They are:
H2, N2, O2, F2, Cl2, Br2, and I2.
Note: whether the element At molecule is monoatomic or diatomic is incredibly arguable. While some say it exists as diatomic because it is a halogen like bromine, iodine etc, At is in fact extremely unstable and no one has ever really studied the molecules on it, so, when others say it is monoatomic, this is also based on calculations. But the other 7 elements listen above is for sure diatomic.
Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2).
Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2) are the formulas of the elements that is present as diatomic molecules under normal environmental conditions. Diatomic molecules refers to those molecules that is composed of only two atoms of the same or different elements. There are large number of diatomic molecules which is made up of two similar elements or different elements.
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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.Two 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:
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.
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.
A student obtained a clean flask. She weighed the flask and stopper on an analytical balance and found the total mass to be 34.232 g. She then filled the flask with water and found the new mass to be 60.167 g. The temperature of the water was measured to be
Answer:
25.99mL is the volume internal volume of the flask
Explanation:
To complete the question:
The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask
The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.
To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:
Mass water = Mass filled flask - Mass of clean flask
Mass water = 60.167g - 34.232g
Mass water = 25.935g of water.
To convert this mass to volume:
25.935g × (1mL / 0.997992g) =
25.99mL is the volume internal volume of the flaskWhat is the balanced form of the chemical equation shown below?
Ca(OH)2(aq) + Na2CO3(aq) → CaCO3(s) + NaOH(aq)
Answer:
D
Explanation:
Double Displacement reaction
Both sides are balanced with option D
The balanced form of the chemical equation shown below is [tex]\rm Ca(OH)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaOH(aq).[/tex] The correct option is D.
What is a balanced equation?A balanced equation is where the reactant and the product have the number of moles of elements. According to the law, the reaction, and the product have the same number of moles after the reaction, so balancing an equation is important.
To balance an equation, it is significant to see the number of moles of reactant and the same number of moles is in the product side. Here the moles of sodium has to be balanced.
Thus, the correct option is D, [tex]\rm Ca(OH)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaOH(aq).[/tex]
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interpret the electron configuration
Answer:
Ca for calcium
20 electrons
2-2s electron
Which of the following statements about metal elements is correct?
A. Metals tend to easily gain more valence electrons.
B. Metal elements are always heavier than non-metal elements.
C. Metals tend to easily lose their valence electrons.
D. A metal atom can take an electron from a non-metal atom.
Answer: C. Metals tend to easily lose their valence electrons.
Explanation:
Metals are those substances which have tendency to loose their valence electrons to attain noble gas configuration and forms positive ions called as cations.
Example: Gold, potassium etc
[tex]M\rightarrow M^++e^-[/tex]
Non metals are those substances which have tendency to gain valence electrons to attain noble gas configuration and form negative ions called as anions.
Example: Sulphur, Chlorine
[tex]N+e^-\rightarrow N^-[/tex]
1. In the simple cubic unit cell, the centers of ____________ identical particles define the ____________ of a cube. The particles do touch along the cube's ____________ but do not touch along the cube's ____________ or through the center. There is/are ____________ particle per unit cell and the coordination number is ____________ .
2. In the body-centered cubic unit cell, the centers of ____________ identical particles define the ____________ of the cube plus ____________ particle at the ____________ of ____________ . The particles do not touch along the cube's ____________ or faces but do touch along the cube's ____________ . There is/are ____________ particles per unit cell and the coordination number is ____________ .
3. In the face-centered cubic cell, the centers of ____________ identical particles define the ____________ of the cube plus ____________ particle in the ____________ of ____________ . The particles on the ____________ do not touch each other but do touch those on the ____________ . There is/are ____________ particles per unit cell and the coordination number is ____________ .
Answer:please see below for answers in the spaces given.
Explanation:
There are three types of cubic-unit cells of a cubic system which include Simple cubic unit cell, body-centered cubic unit cell and face-centered cubic-unit cell and Thier characteristics are completed below.
1) In the simple cubic unit cell, the centers of _______eight _____ identical particles define the _________corners___ of a cube. The particles do touch along the cube's _______edges_____ but do not touch along the cube's ____diagonal_______ or through the center. There is/are _______one_____ particle per unit cell and the coordination number is
__six______ .
2. In the body-centered cubic unit cell, the centers of _______eight _____ identical particles define the _______corners_____ of the cube plus ______one______ particle at the _______center_____ of ______the cube______ . The particles do not touch along the cube's _______edges_____ or faces but do touch along the cube's ____diagonal________ . There is/are _____two_______ particles per unit cell and the coordination number is _____eight_______ .
3. In the face-centered cubic cell, the centers of ______eight______ identical particles define the _______corner____ of the cube plus ________one____ particle in the _____center_______ of ______each face______ . The particles on the _____corners_______ do not touch each other but do touch those on the ______faces____ . There is/are ________four___ particles per unit cell and the coordination number is _____twelve_______ .
Hydrogen iodide decomposes according to the equation: 2HI(g) H 2(g) + I 2(g), K c = 0.0156 at 400ºC A 0.660 mol sample of HI was injected into a 2.00 L reaction vessel held at 400ºC. Calculate the concentration of HI at equilibrium.
Answer:
[HI] = 0.264M
Explanation:
Based on the equilibrium:
2HI(g) ⇄ H₂(g) + I₂(g)
It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:
Kc = 0.0156 = [H₂] [I₂] / [HI]²
As initial concentration of HI is 0.660mol / 2.00L = 0.330M, the equlibrium concentrations will be:
[HI] = 0.330M - 2X
[H₂] = X
[I₂] = X
Where X is reaction coefficient.
Replacing in Kc:
0.0156 = [X] [X] / [0.330M - 2X]²
0.0156 = X² / [0.1089 - 1.32X + 4X² ]
0.00169884 - 0.020592 X + 0.0624 X² = X²
0.00169884 - 0.020592 X - 0.9376 X² = 0
Solving for X:
X = - 0.055 → False solution, there is no negative concentrations
X = 0.0330 → Right solution.
Replacing in HI formula:
[HI] = 0.330M - 2×0.033M
[HI] = 0.264MCalculate the pH of a buffer solution that contains 0.25 M benzoic acid (C 6H 5CO 2H) and 0.15M sodium benzoate (C 6H 5COONa). [K a = 6.5 × 10 –5 for benzoic acid]
Answer:
3.97
Explanation:
pH of buffer solution = pKa+Log(Cb/Ca)
pH of buffer solution = -log(Ka)+log(Cb/Ca)............... Equation 1
Where Ca = concentration of acid, Cb = concentration of base.
Given: Ka = 6.5×10⁻⁵, Ca = 0.25 M, Cb = 0.15 M
Substitute into equation 1
pH of buffer solution = -log(6.5×10⁻⁵)+log(0.15/0.25)
pH of buffer solution = 4.19+(0.22)
pH of buffer solution = 3.97.
A 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.
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.
A student is given an antacid tablet that weighs 5.8400 g. The tablet is crushed and 4.2800 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 11.6 mL of a NaOH solution to titrate it to a methyl red end point. It takes 29.0 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the original 200. mL of stomach acid (in mL) is neutralized by the 4.2800 g crushed sample of the tablet
Answer:
Explanation:
Given that:
mass of the antacid tablet = 5.8400 g
required mass of the antacid tablet = 4.2800 g was added to 200. mL of simulated stomach acid.
The amount of the original 200. mL of stomach acid (in mL) needed to neutralize the 4.2800 g crushed sample of the tablet can be calculated as:
= 11.6 mL of NaOH × 25.00 mL /29.0 mL NaOH
= 10.00 mL original stomach acid
Now; since it requires 11.6 mL of NaOH o neutralize 10.00 mL of original acid , then:
the antacid neutralized = 200 mL - 10.00 mL
the antacid neutralized = 190.00 mL
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.
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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 = 29.6Calculate 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
2) 2.5 mol of an ideal gas at 20 oC under 20 atm pressure, was expanded up to 5 atm pressure via; (a) adiabatic reversible and (b) adiabatic irreversible process. Calculate the values of w, q, ΔU, ΔH for each process. (Cv = 5 cal / mol.K ≈ 5/2 R; R ≈ 2 cal / mol.K) (Please find the desired values by making the corresponding derivations
Answer:
a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)
workdone (w) = -8442.6 J ≈ -8.443 KJ
heat transferred (q) of the ideal gas = - w
q = 8.443 KJ
b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0
the workdone(w) in the ideal gas= - 4567.5 J ≈ - 4.57 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Explanation:
given
mole of an ideal gas(n) = 2.5 mol
Temperature (T) = 20°C
= (20°C + 273) K = 293 K
Initial pressure of the ideal gas(P₁) = 20 atm
Final pressure of the ideal gas(P₂) = 5 atm.
2) (a)for adiabatic reversible process,
note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.
Work done (w) = nRT ln[tex]\frac{P_{1} }{P_{2} }[/tex]
= 2.5 mol × 8.314 J/mol K × 293 K × ln[tex]\frac{5atm}{20atm}[/tex]
= 6090.01 J × [-1.3863]
= -8442.6 J ≈ -8.443 KJ
So, the work done (w) of ideal gas = -8.443 KJ
For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-8.443 KJ)
q = 8.443 KJ
heat transfer (q) of the ideal gas = 8.443 KJ
(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.
Work done (w) = -nRT(1 - ln[tex]\frac{P_{1} }{P_{2} }[/tex] )
= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]
= - 6090.01 J × 0.75
= - 4567.5 J ≈ - 4.57 KJ
∴work done(w) of an ideal gas = - 4.57 KJ
For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-4.5675 KJ)
q = 4.5675 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
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]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
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
Why was it important to establish the Clean Air Act?
Answer: The Clean Air Act was important because it emphasized cost-effective methods to protect the air; encouraged people to study the effects of dirty air on human health; and created a regulation that makes any activities that pollute the air illegal.
Explanation:
Answer:
Clean Air Act (CAA), U.S. federal law, passed in 1970 and later amended, to prevent air pollution and thereby protect the ozone layer and promote public health. The Clean Air Act (CAA) gave the federal Environmental Protection Agency (EPA) the power it needed to take effective action to fight environmental pollution.
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.
Learn more:
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.
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.
Which element would have the most valence electrons and also be able to react with hydrogen?
Answer:
Fluorine, Chlorine, Bromine, or Iodine
Explanation:
These all have an ALMOST full valence shell. And they need one more electron so they'd react with hydrogen
Answer:
its chlorine
Explanation:
just trust me do i look like i would lie too you ;-)
btw i just took the test :-)
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