Answer:
mol LiCl = 4.83 m
Explanation:
GIven:
Solution of LiCl in water XLiCl = 0.0800
Mol of water in kg = 55.55 mole
Find:
Molality
Computation:
mole fraction = mol LiCl / (mol water + mol LiCl)
0.0800 = mol LiCl / (55.55 mol + mol LiCl)
0.0800 mol LiCl + 4.444 mol = mol LiCl
mol LiCl - 0.0800 mol LiCl = 4.444 mol
0.92 mol LiCl = 4.444 mol
mol LiCl = 4.83 m
What is the balanced equation for the reaction of aqueous cesium sulfate and aqueous barium perchlorate?
Answer:
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
Explanation:
When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.
Two elements represents by the letter Q and R atomic number 9 and 12 respectively. Write the electronic configuration of R
Answer:
Atomic no = 12 = Mg
Explanation:
It is given that,
The atomic number of two elements that are represented by letter Q and R are 9 and 12.
We need to write the electronic configuration of R. Atomic number shows the number of protons in atom.
For R, atomic number = 12
Its electronic configuration is : 2,8,2
It has two valance electrons in its outermost shell. The element is Magnesium (Mg).
What would happen to the measured cell potentials if 30 mL solution was used in each half-cell instead of 25 mL
Answer:
The answer is "[tex]\bold{\log \frac{[0] mole}{[R]mole}}[/tex]"
Explanation:
[tex]E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\[/tex]
In the above-given equation, we can see from [tex]E_{ceu}[/tex], of both oxidant [tex]conc^n[/tex]as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor and each other suspend
[tex]\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\[/tex]
[tex]\to {\log \frac{[0] mole}{[R]mole}}[/tex]
The cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.
The cell potential has been given as the difference in the potential of the two half cells in the electrochemical reaction.
The two cells has been set with the concentration of solutions in the oxidation and reduction half cells.
Cell potential changeThe cell potential has been changed when there has been a change in the potential of the half cells.
The volume of 30 mL to the solution has been, resulting in the cell potential difference of x.
With the volume of 25 mL, there has been the difference in the potential being similar to the 30 mL solution, i.e. x.
Thus, the cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.
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Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: specific heat (ice) = 2.1 J/g·°C; specific heat (water) = 4.18 J/g·°C; ΔH fus = 6.0 kJ/mol.
Answer:
The amount of heat to absorb is 6,261 J
Explanation:
Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.
The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.
So:
Heat required to raise the temperature of ice from -20 °C to 0 °CBeing the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).
In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C
Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J
Heat required to convert 0 °C ice to 0 °C waterThe heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)
Heat required to raise the temperature of water from 0 °C to 60 °CIn this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C
Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J
Finally, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
The amount of heat to absorb is 6,261 J.
Calculation for heat:Heat required to raise the temperature of ice from -20 °C to 0 °C.
The formula for specific heat is used to calculate the amount of heat
Q = c * m * ΔT
Where,
Q =heat exchanged by a body,
m= mass of the body
c= specific heat
ΔT= change in temperature
Given:
m= 10 g,
specific heat of the ice= 2.1
ΔT=0 C - (-20 C)= 20 C
On substituting the values:
Q= 10 g*2.1 *20 C
Q=420 J
Heat required to convert 0 °C ice to 0 °C water.
The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
Heat required to raise the temperature of water from 0 °C to 60 °C
m= 10 g,
Specific heat of the water= 4.18
ΔT=60 C - (0 C)= 60 C
On substituting:
Q= 10 g*4.18 *60 C
Q=2,508 J
Thus, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
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In which of the following compounds does the carbonyl stretch in the IR spectrum occur at the lowest wavenumber?
a. Cyclohexanone
b. Ethyl Acetate
c. λ- butyrolactone
d. Pentanamide
e. Propanoyl Chloride
Answer:
a. Cyclohexanone
Explanation:
The principle of IR technique is based on the vibration of the bonds by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is a specific energy that generates a specific vibration. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.
Now, we must remember that the lower the wavenumber we will have less energy. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.
If we look at the structure of all the molecules we will find that in the last three we have heteroatoms (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of resonance structures which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.
The molecule that fulfills this condition is the cyclohexanone.
See figure 1
I hope it helps!
What's the mass in grams of 0.442 moles of calcium bromide, CaBr2? The atomic
weight of Ca is 40.1 and the atomic weight of Br is 79.9.
A) 452.3 g
B) 53.04 g
C) 44.2 g
D) 88.4 g
Answer:
Below
Explanation:
Let n be the quantity of matter in the Calcium Bromide
● n = m/ M
M is the atomic weight and m is the mass
M of CaBr2 is the sum of the atomic wieght of its components (2 Bromes atoms and 1 calcium atom)
M = 40.1 + 2×79.9
● 0.422 = m/ (40.1+2×79.9)
●0.422 = m/ 199.9
● m = 0.422 × 199.9
● m = 84.35 g wich is 88.4 g approximatively
88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 ,therefore option (d) is correct.
What do you mean by mass ?Mass is the amount of matter that a body possesses. Mass is usually measured in grams (g) or kilograms (kg) .
To calculate mass in grams of 0.442 moles of calcium bromide, CaBr2,
Let n be the quantity of matter in the Calcium Bromide
M is the atomic weight and m is the mass
n = m/ MM of CaBr2 is the sum of the atomic weight of its components
Mass of Ca = 40.1 , Mass of Br = 79.9
M = 40.1 + 2×79.9
0.422 = m/ (40.1+2×79.9)
0.422 = m/ 199.9
m = 0.422 × 199.9
m = 84.35 g which is 88.4 g approximatively .
Thus ,88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 , hence option (d) is correct .
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Im really confused and select all that apply questions scare me.
Answer:
The 3rd one
Explanation:
A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 198 picometers and its molar mass is 195.08 g/mol. Calculate the density of the metal in g/cm3.
Answer:
7.38 g/cm³ is the density of the metal
Explanation:
In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).
To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.
Volume of the unit cellVolume = a³
a = √8×r
(r = 198x10⁻¹²m)
a = 5.6x10⁻¹⁰ m
Volume = 1.756x10⁻²⁸ m³
1m = 100cm → 1m³ = (100cm)³:
1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =
1.756x10⁻²² cm³ → Volume of the unit cell in cm³Mass of the unit cell:There are 4 atoms of gold:
4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold
As 1 mole weighs 195.08g:
6.64x10⁻²⁴ moles of gold × (195.08g / mol) =
1.296x10⁻²¹g is the mass of the unit cellDensity of the metal:1.296x10⁻²¹g / 1.756x10⁻²² cm³ =
7.38 g/cm³ is the density of the metalThe density of the metal is 7.40 g/cm³
In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.
It is expressed by using the formula:
[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]
where;
[tex]\rho[/tex] = density of the metalZ = atoms coordination no = 4 (for FCC)Molar mass (M) = 195.8 g/molAvogadro's constant (NA) = 6.022 × 10²³ /mola = edge lengthFor face-centered cubic FFC;
The edge length [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]
[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]
[tex]\mathbf{a =560.0285 \ pm }[/tex]
a = 5.60 × 10⁻⁸ cm
Replacing it into the previous equation, we have:
[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]
[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]
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Assume that you are provided with the following materials:
• Strips of metallic zinc, metallic copper, metallic iron
• 1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine(I2)
• Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos,identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.For each cell created, include the following details.
A) Which electrode was the anode,and which was the Cathode?
B) The anode and cathode half reactions.
C) Balanced equation for each cell you propose to construct.
D) Calculated Eocelle Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed).
Answer:
See explanation
Explanation:
First voltaic cell;
Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)
Anode;
Zinc
Cathode;
Copper
Oxidation half equation;
Zn(s)------> Zn^2+(aq) + 2e
Reduction half equation;
Cu^2+(aq) +2e -----> Cu(s)
Overall; Zn(s) + Cu^2+(aq) -----> Zn^2+(aq) + Cu(s)
E°cell = 0.34 -(-0.76) =1.1 V
Second voltaic cell;
Zn(s)|Zn^2+(aq)||Fe^2+(aq)|Fe(s)
Anode;
Zinc
Cathode;
Iron
Oxidation half equation;
Zn(s)------> Zn^2+(aq) + 2e
Reduction half equation;
Fe^2+(aq) +2e -----> Fe(s)
Overall; Zn(s) + Fe^2+(aq) -----> Zn^2+(aq) + Fe(s)
E°cell = (-0.44) -(-0.76) = 0.32 V
Third voltaic cell;
Fe(s)|Fe^2+(aq)||Cu^2+(aq)|Cu(s)
Anode;
Iron
Cathode;
Copper
Oxidation half equation;
Fe(s)------> Fe^2+(aq) + 2e
Reduction half equation;
Cu^2+(aq) +2e -----> Cu(s)
Overall; Fe(s) + Cu^2+(aq) -----> Fe^2+(aq) + Cu(s)
E°cell = 0.34 -(-0.44) = 0.78 V
Fourth voltaic cell
Cu(s)|Cu^2+(aq)||I2(aq)|C(s)|I^-(aq)
Anode;
Copper
Cathode;
Graphite rod
Oxidation half equation;
Cu(s)------> Cu^2+(aq) + 2e
Reduction half equation;
I2(aq) +2e -----> 2I^-(aq)
Overall; Cu(s) + I2(aq) -----> Cu^2+(aq) + 2I^-(aq)
E°cell = 0.54 -0.34 = 0.20 V
How many atoms are in 65.0g of zinc?
from
1moles=iatom
Mole=mass÷avogardos
Where
Avogadro's= 6.02×10²³
So moles = 65.0÷6.02×10²³
Atoms of zinc = 391.6 ×10²³
The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].
Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.
Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately [tex]6.022 \times10^{23}[/tex]atoms per mole.
First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.
Number of moles = Mass / Molar mass
Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol
Next, multiply the number of moles by Avogadro's number to find the number of atoms.
Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]
Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol
Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.
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An electrolysis cell has two electrodes. Which statement is correct? A. Reduction takes place at the anode, which is positively charged. B. Reduction takes place at the cathode, which is positively charged. C. Reduction takes place at the dynode, which is uncharged. D. Reduction takes place at the cathode, which is negatively charged. E. Reduction takes place at the anode, which is negatively charged.
Answer:
D. Reduction takes place at the cathode, which is negatively charged.
Explanation:
In an electrolytic cell there are two electrodes; the cathode and the anode. The anode is the positive electrode while the cathode is the negative electrode. Oxidation occurs at the anode while reduction occurs at the cathode.
At the anode, species give up electrons and become positively charged ions while at the cathode species accept electrons and become reduced.
Come up with a definition for density
Density measures how tightly packed particles are.
If particles are tightly packed together, they will be more dense.
If they are loosely together, they will be less dense.
However, a common mistake is thinking that if something
is more dense it means that it's heavier.
However, that's not the case.
It has to do with how particles are packed in an object.
0.25 L of aqueous solution contains 0.025g of HCLO4 (strong acid) what will be the Ph of the solution g
Answer:
The pH of the solution will be 3
Explanation:
The strength of acids is determined by their ability to dissociate into ions in aqueous solution. A strong acid is any compound capable of completely and irreversibly releasing protons or hydrogen ions, H⁺. That is, an acid is said to be strong if it is fully dissociated into hydrogen ions and anions in solution.
Being pH=- log [H⁺] or pH= - log [H₃O⁺] and being a strong acid, all the HClO₃ dissociates:
HClO₄ + H₂O → H₃O⁺ + ClO₄-
So: [HCLO₄]= [H₃O⁺]
The molar concentration is:
[tex]molar concentration=\frac{number of moles of solute}{volume solution}[/tex]
The molar mass of HClO₄ being 100 g / mole, then if 100 grams of the compound are present in 1 mole, 0.025 grams in how many moles are present?
[tex]moles of HClO_{4} =\frac{0.025 grams*1 mole}{100 grams}[/tex]
moles of HClO₄= 0.00025
Then:
[tex][HClO_{4}]=\frac{0.00025 moles}{0.25 L}[/tex]
[tex][HClO_{4}]=0.001 \frac{ moles}{ L}[/tex]
Being [HCLO₄]= [H₃O⁺]:
pH= - log 0.001
pH= 3
The pH of the solution will be 3
An atom of 120In has a mass of 119.907890 amu. Calculate the mass defect (deficit) in amu/atom. Use the masses: mass of 1H atom
Answer:
a
Explanation:
answer is a on edg
What is the ph of 0.36M HNO3 ?
Answer:
0.44
Explanation:
We know that the pH of any acid solution is given by the negative logarithm of its hydrogen ion concentration. Hence, if I can obtain the hydrogen ion concentration of any acid, I can obtain its pH.
For the acid, HNO3, [H^+] = [NO3^-]= 0.36 M
pH= -log [H^+]
pH= - log[0.36]
pH= 0.44
What is the final volume V2 in milliliters when 0.551 L of a 50.0 % (m/v) solution is diluted to 23.5 % (m/v)?
Answer:
[tex]V_2=1.17L[/tex]
Explanation:
Hello,
In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:
[tex]V_1C_1=V_2C_2[/tex]
Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:
[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]
Best regards.
Refer to the figure.
30. How many planes are shown in the figure?
31. How many planes contain points B, C, and E?
32. Name three collinear points.
3. Where could you add point G on plane N
so that A, B, and G would be collinear?
4. Name a point that is not coplanar with
A, B, and C.
5. Name four points that are coplanar.
BN
Answer:
30. 5 planes are shown
31. 1 plane
32. CEF
33. on line AB
34. E or F
35. ABCD or BCEF or CDEF or ACEF
Explanation:
30. Each of the surfaces of the rectangular pyramid is a plane. There are 5 planes.
__
31. 3 points define one plane only.
__
32. The only points shown on the same line segment are points E, F, and C.
__
33. If G is to be collinear with A and B, it must lie on line AB.
__
34. The only points shown that are not on plane N are points E and F. Either of those will do.
__
35. There are three planes that have 4 points shown on them. The four points that are on the same plane are any of ...
ABCDBCEFCDEFPlane ACEF is not shown on the diagram, but we know that those 4 points are also coplanar. (Any point not on line CE, together with the three points on that line, will define a plane with 4 coplanar points.)
place the following substances in Order of decreasing boiling point H20 N2 CO
Answer:
-195.8º < -191.5º < 100º
Explanation:
Water, or H20, starts boiling at 100ºC.
Nitrogen, or N2, starts boiling at -195.8ºC.
Carbon monoxide, or C0, starts boiling at -191.5ºC.
When we place these in order from decreasing boiling point:
-195.8º goes first, then -191.5º, and 100º goes last.
Answer:
therefore, N2, CO, H20
Decreasing boiling point
Explanation:
the bond existing in H2O is hydrogen bond
bond existing in N2 is covalent bond, force existing is dipole-dipole-interaction
bond existing in CO is covalent bond , force existing between is induced -dipole- induced dipole-interaction
hydrogen bond is the strongest , followed by dipole-dipole-interaction and induced -dipole- induced dipole-interaction
the stronger the bond , the higher the boiling point
therefore, N2, CO, H20
-------------------------------------->
Decreasing boiling point
please help guys the question is
give reasons
a. we have to separate the mixture
b. All impure substances are not harmful.
c. A mixture of iron fillings and sand can be separated by using a magnet
d. A sentences "shake before well use" is written on the bottle of the medicine.
Answer:
(a )people separate mixtures in order to ger a specific substance that they need.
"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Answer:
245.66g of NH₄Cl is the mass we need to add to obtain the desire pH
Explanation:
The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical
First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.
pKa NH₃/NH₄⁺pKb = - log Kb
pKb = -log 1.8x10⁻⁵ = 4.74
pKa = 14 - pKb
pKa = 14 - 4.74
pKa = 9.26
Moles NH₃2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃
H-H equation:pH = pKa + log [NH₃] / [NH₄Cl]
8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]
-1.06 = log [0.400 moles] / [NH₄Cl]
0.0087 = [0.400 moles] / [NH₄Cl]
[NH₄Cl] = 0.400 moles / 0.0087
[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):
4.59 moles NH₄Cl ₓ (53.491g / mol) =
245.66g of NH₄Cl is the mass we need to add to obtain the desire pH
NEED HELP ASAP
In 1988, three gray whales were trapped in Arctic ice. Television crews captured the frantic
attempts of hundreds of people to save the whales. Eventually, a Soviet icebreaker and U.S.
National Guard helicopters arrived to help free the whales. The cost of the rescue mission
exceeded $5 million.
i. Write a scientific question related to the whale story. (1 point)
A piece of plastic sinks in oil but floats in water. Place these three substances in order from lowest density to greatest density.
Answer:
[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]
Explanation:
Hello,
In this case, since water and oil are immiscible due to the oil's nonpolarity and water's polarity, when mixed, the oil remains on the water since it is less dense than water. In such a way, for a plastic sunk in the oil and floating on the water (in middle of them) we can conclude that the plastic have a mid density, therefore, the required organization is:
[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]
Best regards.
For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.
Answer:
-138.9 kJ/mol
Explanation:
Step 1: Convert 235.8°C to the Kelvin scale
We will use the following expression.
K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K
Step 2: Calculate the standard enthalpy of reaction (ΔH°)
We will use the following expression.
ΔG° = ΔH° - T.ΔS°
ΔH° = ΔG° / T.ΔS°
ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K
ΔH° = -3.583 kJ (for 1 mole of balanced reaction)
Step 3: Convert -9.9°C to the Kelvin scale
K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K
Step 4: Calculate ΔG° at 263.3 K
ΔG° = ΔH° - T.ΔS°
ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K
ΔG° = -138.9 kJ/mol
The electrolysis of molten AlCl 3 for 2.50 hr with an electrical current of 15.0 A produces ________ g of aluminum metal.
is the general formula of a certain hydrate. When 256.3 g of the compound is heated to drive off the water, 214.2 g of anhydrous compound is left. Further analysis shows that the percentage composition of the anhydrate is 21.90% Ca, 43.14% Se, and 34.97% O.. (Hint: Treat the anhydrous compound and water just as you have treated elements in calculating in the formula of the hydrate.) (Use an asterisk to enter the dot in the formula. If a subscript is 1, omit it.) Find the empirical formula of the anhydrous compound. Find the empirical formula of the hydrate.
Answer:
The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.
Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams
The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,
Moles of water = weight of water/molecular weight
= 42.1 grams / 18 = 2.3
The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,
= 214.2 * 0.2190 = 46.91 grams
The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,
= 214.2 * 0.4314 = 92.40 grams
The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,
= 214.2 * 0.3497 = 74.91 grams
The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17
The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,
92.40/78.96 = 1.17
The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,
74.91/15.999 = 4.68.
Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O
A saturated sodium carbonate solution at 0°C contains 7.1 g of dissolved sodium carbonate per 100. mL of solution. The solubility product constant for sodium carbonate at this temperature is
Answer:
[tex]Ksp=1.2[/tex]
Explanation:
Hello,
In this case, as the saturated solution has 7.1 grams of sodium carbonate, the solubility product is computed by firstly computing the molar solubility by using its molar mass (106 g/mol):
[tex]Molar \ solubility=\frac{7.1gNa_2CO_3}{0.1L}*\frac{1molNa_2CO_3}{106gNa_2CO_3}=0.67M[/tex]
Next, as its dissociation reaction is:
[tex]Na_2CO_3(s)\rightleftharpoons 2Na^+(aq)+CO_3^{2-}(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Na^+]^2[CO_3^{2-}][/tex]
And the concentrations are related with the molar solubility (2:1 mole ratio between ionic species):
[tex]Ksp=(2*0.67)^2*(0.67)\\\\Ksp=1.2[/tex]
Best regards.
When we react a weak acid with a strong base of equal amounts and concentration, the component of the reaction that will have the greatest effect on the pH of the solution is:______.
a. the acid.
b. the base.
c. the conjugate acid.
d. the conjugate base.
Answer:
d. the conjugate base.
Explanation:
The general reaction of a weak acid, HA, with a strong base YOH, is:
HA + YOH → A⁻ + H₂O + Y⁻
Where A⁻ is the conjugate base of the weak acid and Y⁻ usually is a strong electrolyte.
That means after he complete reaction you don't have weak acid nor strong base, just conjugate base that will be in equilibrium with water, thus (Strong electrolyte doesn't change pH:
A⁻ + H₂O ⇄ HA + OH⁻
As the equilibrium is producing OH⁻, the pH of the solution is being affected for the conjugate base
Right option:
d. the conjugate base.The surface temperature on Venus may approach 753 K. What is this temperature in degrees Celsius?
Answer:
461.85 degrees Celsius
Calculate the [H+] and pH of a 0.0010 M acetic acid solution. The Ka of acetic acid is 1.76×10−5. Use the method of successive approximations in your calculations.
Answer:
[tex][H^+]=0.000123M[/tex]
[tex]pH=3.91[/tex]
Explanation:
Hello,
In this case, dissociation reaction for acetic acid is:
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
For which the equilibrium expression is:
[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
Which in terms of the reaction extent [tex]x[/tex] could be written as:
[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]
Thus, solving by using a solver or quadratic equation we obtain:
[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]
And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:
[tex][H^+]=0.000123M[/tex]
Now, the pH is computed as follows:
[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]
Best regards.
The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.350 atm? (∆Hvap = 28.5 kJ/mol)
Answer:
The temperature at which the vapor pressure would be 0.350 atm is 201.37°C
Explanation:
The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.
Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.
The equation for Clausius- Clapeyron Equation can be expressed as:
[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]
where ;
[tex]P_1[/tex] is the vapor pressure at temperature 1
[tex]P_ 2[/tex] is the vapor pressure at temperature 2
∆Hvap = enthalpy of vaporization
R = universal gas constant
Given that:
[tex]P_1[/tex] = 1 atm
[tex]P_ 2[/tex] = 0.350 atm
∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol
[tex]T_1[/tex] = 282 °C = (282 + 273) K = 555 K
R = 8.314 J/mol/k
Substituting the above values into the Clausius - Clapeyron equation, we have:
[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]
[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }[/tex]
[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }[/tex]
[tex]\dfrac{1}{T_2} = 0.002108048708[/tex]
[tex]T_2 = \dfrac{1}{0.002108048708}[/tex]
[tex]\mathbf{T_2 }[/tex] = 474.37 K
To °C ; we have [tex]\mathbf{T_2 }[/tex] = (474.37 - 273)°C
[tex]\mathbf{T_2 }[/tex] = 201.37 °C
Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C
The temperature of the liquid at the given vapor pressure is 201.5 ⁰C.
The given parameters;
boiling point temperature, = 282 ⁰Cvapor pressure, P₂ = 0.35 atmenthalpy of vaporization, ∆Hvap = 28.5 kJ/molThe temperature of the liquid will be determined by applying Clausius- Clapeyron Equation;
[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )[/tex]
where;
R is ideal gas constant = 8.314 J/mol.kT₁ is the initial temperature in Kelvin = 282 + 273 = 555 KP₁ is the initial pressure = 1 atm[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )\\\\ln(\frac{0.35}{1} ) = \frac{28.5 \times 10^3}{8.314} (\frac{T_2 - 555}{555T_2} )\\\\-1.049 = 6.176- \frac{3427.95}{T_2} \\\\\frac{3427.95}{T_2} = 6.176 + 1.049\\\\\frac{3427.95}{T_2} = 7.225\\\\T_2 = \frac{3427.95}{7.225} \\\\T_2 = 474.5 \ K\\\\T_2 = 474.5 - 273 = 201.5 \ ^0C[/tex]
Thus, the temperature of the liquid at the given vapor pressure is 201.5 ⁰C.
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