Answer:
1. 36
2. Two
Explanation:
The Lewis structure shows the valence electrons present in a compound. Usually the valence electrons are shown as dot structures around the symbol of the elements involved in the compound.
For a compound AB4 where B is more electronegative than A and A has 8 electrons in its valence shell, there will be thirty six valence electrons on the outermost shell of the molecule.
There are six electron pair domains present in the molecule, four bond pairs and two lone pairs. The molecule is in a square planar geometry.
Answer: a- 36 valence electrons
b- 14 lone pairs
Explanation:
Valence is equal to A + 4B = 8 + 4(7)
With 4 bonds between A and the 4 B, that is 36 valence minus 8 electrons in those pairs = 28. 28 is 14 lone pairs.
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.
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]
Learn more about face-centered cubic arrangement here:
https://brainly.com/question/14786352?referrer=searchResults
"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
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.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.
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
9
What might happen if acidic chemicals were emitted into
the air by factories? Choose the best answer.
A
The acid would destroy metallic elements in the air
B
The acid would be neutralized by bases within clouds
C
Acid rain might destroy ecosystems and farmland
D
Violent chemical reactions would take place within the
atmosphere
co search
O
BI
How many grams of H2O will be formed when 32.0 g H2 is mixed with 73.0 g of O2 and allowed to react to form water
hope this helps u
pls mark as brainliest .-.
Calculate the molarity of a solution containing 29g of glucose (C 6 H 12 O 6 ) dissolved in 24.0g of water. Assume the density of water is 1.00g/mL.
Answer:
whats the ph ofpoh=9.78
Explanation:
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.
Question 8 of 30
When is a redox reaction spontaneous?
A. When a metal electrode is in contact with an electrolyte
B. When a power source supplies an electrical current
C. When the cell potential is positive
D. When the cell potential is negative
According to the concept of redox reactions, the answer to this question is option C.
When the cell potential is positive. A redox reaction is spontaneous when the cell potential is positive.The cell potential is the measure of the driving force of the chemical reaction occurring in the electrochemical cell. In an electrochemical cell, a redox reaction occurs, which leads to the production of an electric potential.
If this potential is positive, then the redox reaction is considered spontaneous. However, if the potential is negative, then the reaction is non-spontaneous.In general, a redox reaction is spontaneous if the potential difference between the two electrodes of the cell is positive. This means that the reaction will occur spontaneously without any external energy input.
Thus, the correct option is C.
Learn more about redox reaction,here:
https://brainly.com/question/28300253
#SPJ4
Write the equations that represent the first and second ionization steps for sulfuric acid (H2SO4) in water.
Answer:
[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}rightarrow[/tex]
Explanation:
Hello,
In this case, given that the sulfuric acid is a diprotic acid (two hydrogen ions) we can identify two ionization reactions, the first one, showing up the dissociation of the first hydrogen to yield hydrogen sulfate ions and the second one, showing up the dissociation of the hydrogen sulfate ions to hydrogen ions and sulfate ions by separated as shown below:
[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}[/tex]
They are have one-sensed arrow, since sulfuric acid is a strong acid.
Regards.
The equations that represent the first and second ionization steps for sulfuric acid in water are H₂SO₄→HSO₄+H⁺ & HSO₄⁻→SO₄⁻+H⁺ respectively.
What is ionization reaction?Ionization reactions are those reactions in which atom or molecule will convert into ion by bearing a positive or negative charge on itself.
In water in the following way ionization of sulphuric acid takes place:
In the first ionization step one hydrogen atom (H⁺) will loose from the sulphuric acid molecule as:H₂SO₄ → HSO₄⁻ + H⁺
In the second ionization step another hydrogen atom will also loose and we get the sulphate ion (SO₄⁻) and one proton (H⁺) as:HSO₄⁻ → SO₄⁻ + H⁺
Hence, two steps are shown above.
To know more about ionization reaction, visit the below link:
https://brainly.com/question/1445179
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.
Im really confused and select all that apply questions scare me.
Answer:
The 3rd one
Explanation:
A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate. Fill in the missing part of this equation. (0.030 cm^3) x ? =m^3
Answer:
\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}
Explanation:
0.030 cm³ × ? = x m³
You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.
For example, you know that centi means "× 10⁻²", so
1 cm = 10⁻² m
If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).
If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.
So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.
We choose the former because it has the desired units on top.
The "cm" is cubed, so we must cube the conversion factor.
The calculation becomes
[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]
Which of the following is required for the flow of current in all systems?
a) the presence of ions
b) an electrical potential ofo
c) a closed circuit
d) a short circuit
Answer:
I would say c) a closed circuit.
Hope I was right.
Which of the following contains a nonpolar covalent bond?
O A. Co
B. NaCl
O C. 02
O D. HE
Answer:
The answer is o2
Explanation:
I took the test
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
Find more information about Specific heat here:
brainly.com/question/13439286
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.
Learn more about atoms here:
https://brainly.com/question/1566330
#SPJ2
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.)
Heterocyclic aromatic compounds undergo electrophilic aromatic substitution in a similar fashion to that undergone by benzene with the formation of a resonance-stabilized intermediate. Draw all of the resonance contributors expected when the above compound undergoes bromination
Answer:
See explanation
Explanation:
When we talk about electrophilic substitution, we are talking about a substitution reaction in which the attacking agent is an electrophile. The electrophile attacks an electron rich area of a compound during the reaction.
The five membered furan ring is aromatic just as benzene. This aromatic structure is maintained during electrophilic substitution reaction. The attack of the electrophile generates a resonance stabilized intermediate whose canonical structures have been shown in the image attached.
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.
Learn more about cell potential, here:
https://brainly.com/question/1313684
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an elevation where the pressure is 0.73 atm, then the new volume is 1.8 L. What is the temperature (in °C) of the air at this elevation?
Answer:
The temperature of the air at this given elevation will be 53.32425°C
Explanation:
We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.
Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm ; Final Volume = 1.8 L ; Final pressure = 0.73 atm
We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T[tex]_2[/tex] ),
P[tex]_1[/tex]V[tex]_1[/tex] / T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / T[tex]_2[/tex],
T[tex]_2[/tex] = P[tex]_2[/tex]V[tex]_2[/tex]T[tex]_1[/tex] / P[tex]_1[/tex]V[tex]_1[/tex],
T[tex]_2[/tex] = 0.73 atm [tex]*[/tex] 1.8 L [tex]*[/tex] 298.15 K / 1 atm [tex]*[/tex] 1.2 L = ( 0.73 [tex]*[/tex] 1.8 [tex]*[/tex] 298.15 / 1 [tex]*[/tex] 1.2 ) K = 326.47425 K,
T[tex]_2[/tex] = 326.47425 K = 53.32425 C
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
Consider these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)|Cu(s) Ag+(aq)|Ag(s) -0.40 V -0.76 V ‑0.25 V +0.34 V +0.80 V Based on the data above, which species is the best reducing agent?
Answer:
The best reducing agent is Zn(s)
Explanation:
A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:
Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V
Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
Ni²⁺(aq)|Ni(s) ⇒‑0.25 V
Cu²⁺(aq)|Cu(s) ⇒ +0.34 V
Ag⁺(aq)|Ag(s) ⇒ +0.80 V
The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).
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.
3,3-dibromo-4-methylhex-1-yne
Explanation:
see the attachment. hope it will help you...What creation would you make if you had access to any of the chemical elements? can someone answer this for me please.
Answer:
Depends on the chemicals.
Explanation:
The creation of something depends on the chemical we have. If I have chemicals such as acid and base so I can produce salt and water by mixing acid and base. If I have ethanol, vegetable oil and sodium hydroxide which is a catalyst, I can produce biodiesel which can be used in vehicles instead of fossil fuel. If I have sodium element and chlorine gas, I can produce sodium chloride also known as table salt.
Which of the following is a covalent bond? A NaCl B K2O C H2O D MgO
Answer:
[tex]H_2O[/tex]
Explanation: Research has proven that ;
Water is a Polar Covalent Molecule
It consists of 2 Hydrogen molecules bonded to one Oxygen molecule and the two hydrogen atoms are not evenly distributed around the oxygen atom.
g Which ONE of the following pairs of organic compounds are NOT pairs of isomers? A) butanol ( CH3-CH2-CH2-CH2-OH ) and diethyl ether ( CH3–CH2–O–CH2–CH3 ) B) isopentane ( (CH3)2-CH-CH2-CH3 ) and neopentane ( (CH3)4C ) C) ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ) D) acrylic acid ( CH2=CH-COOH ) and propanedial ( OHC–CH2–CHO ) E) trimethylamine ( (CH3)3N ) and propylamine ( CH3-CH2-CH2-NH2 )
Answer:
ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 )
Explanation:
Isomers are compounds that have the same molecular formula but different structural formulas. Hence any pair of compounds that can be represented by exactly the same molecular formula are isomers of each other.
If we look at the pair of compounds; ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ), one compound has molecular formula, C2H7ON while the other has a molecular formula, C2H5ON, hence they are not isomers of each other.
