Answer:
D) Oxygen is oxidized and hydrogen is reduced.
Explanation:
The electrolysis of water is the decomposition of water ( as the name suggests) of water into hydrogen and oxygen. The general equation is
2 H₂O(l) ⇒ 2 H₂ (g) + O₂ (g)
The hydrogen atom in the water molecule has been reduced since its oxidation snumber goes from 1⁺ to 0, and the oxygen has oxidized from 2⁻ to 0 , and in the balanced equation. The overall exchange of electron is 4.
Answer: D
Explanation:
Electrolysis of water which is also called water splitting, is the decomposition of water into oxygen and hydrogen gas when an electric current is passed through it, from a platinum electrode.
Pure water (H2O) is used to produce hydrogen gas (H2), which is fuel, and breathable oxygen gas(O2).
Hydrogen is collected at cathode and oxygen is collected at anode.
Chemical reaction
2 H2O(l) → 2 H2(g) + O2(g)
In the reactant H20, the oxidation state of Hydrogen is (+1) the oxidation state of Oxygen is (-2)
In the products, the oxidation number of elements in the uncombine state is zero so, the oxidation state of Hydrogen is 0 and Oxygen is 0.
Therefore Hydrogen is reduced (from +1 to 0)and oxygen is oxidized (from -2 to 0)
The half reaction are:
Reduction: 2 H+(aq) + 2e− → H2(g)
Oxidation: 2 H2O(l) → O2(g) + 4 H+(aq) + 4e−
Consider the sulfur difluoridemolecule.What is the central atom? Enter its chemical symbol.How many lone pairs are around the central atom?What is the ideal angle between the sulfur-fluorine bonds?Compared to the ideal angle, you would expect the actual angle between the sulfur-fluorine bonds to be ...
Answer:
a. The central atom is sulfur
b. SF2
c. The central atom has two lone pairs
d. The ideal angle between the sulfur-fluorine bonds is 109.5°
e. I expect the actual angle between the sulfur-fluorine bonds to be less than 109.5° because unbonded pairs repel bonded pairs more than bonded pairs repel other bonded pairs. So the bonds here will be pushed closer than normal
Explanation:
Question #1: What is the central atom?
The central atom of this molecule is Sulfur, S.
Question #2: Enter its chemical symbol.
The chemical symbol of the molecule is SF2 but the chemical symbol of the central atom is S.
Question #3: How many lone pairs are around the central atom?
There are two lone pairs around the central atom of Sulfur.
Question #4: What is the ideal angle between the sulfur-fluorine bonds?
The ideal angle between the Sulfur-Fluorine bonds is 109.5 degrees.
Question #5: Compared to the ideal angle, you would expect the actual angle between the sulfur-fluorine bonds to be.
I would expect the actual angle between the Sulfur-Fluorine bonds to be less than 109.5 degrees since the unbonded pairs have a greater repulsion with bonded pairs than the repulsion that happens between two bonded pairs. Therefore, the bonds would be closer to each other causing a smaller angle.
The normal freezing point of a certain liquid
X
is
0.4°C
, but when
5.90g
of urea
NH22CO
are dissolved in
450.g
of
X
, it is found that the solution freezes at
−0.5°C
instead. Use this information to calculate the molal freezing point depression constant
Kf
of
X
the molal freezing point depression constant [tex](\(K_f\))[/tex] of liquid X is approximately [tex]\(4.13 \, \text{°C/molal}\)[/tex].
To calculate the molal freezing point depression constant (\(K_f\)) of liquid X, we can use the formula:
[tex]\[ \Delta T_f = K_f \times m \][/tex]
Where:
- [tex]\( \Delta T_f \)[/tex] is the freezing point depression (given as [tex]\(0.4^\circ \text{C} - (-0.5^\circ \text{C}) = 0.9^\circ \text{C}\)[/tex]),
- [tex]\( m \)[/tex] is the molality of the solution,
- [tex]\( K_f \)[/tex] is the molal freezing point depression constant.
First, we need to calculate the molality of the solution, which is defined as the number of moles of solute per kilogram of solvent.
Given:
- Mass of urea [tex](\( \text{NH}_2\text{CO} \))[/tex] = 5.90 g
- Mass of liquid X = 450.0 g
We need to find the moles of urea first:
[tex]\[ \text{moles of urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} \][/tex]
The molar mass of urea [tex](\( \text{NH}_2\text{CO} \))[/tex] is the sum of the molar masses of nitrogen, hydrogen, carbon, and oxygen:
[tex]\[ \text{molar mass of urea} = 14.01 + 2(1.01) + 12.01 + 16.00 = 60.03 \, \text{g/mol} \][/tex]
[tex]\[ \text{moles of urea} = \frac{5.90 \, \text{g}}{60.03 \, \text{g/mol}} = 0.0983 \, \text{mol} \][/tex]
Now, we can calculate the molality of the solution:
[tex]\[ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \][/tex]
[tex]\[ \text{molality} = \frac{0.0983 \, \text{mol}}{0.450 \, \text{kg}} = 0.218 \, \text{mol/kg} \][/tex]
Now, we can rearrange the formula for [tex]\(K_f\)[/tex] and solve for it:
[tex]\[ K_f = \frac{\Delta T_f}{m} \][/tex]
[tex]\[ K_f = \frac{0.9^\circ \text{C}}{0.218 \, \text{mol/kg}} \][/tex]
[tex]\[ K_f \approx 4.13 \, \text{°C/molal} \][/tex]
Therefore, the molal freezing point depression constant [tex](\(K_f\))[/tex] of liquid X is approximately [tex]\(4.13 \, \text{°C/molal}\)[/tex].
The molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X is 4.124 °C/(mol/kg).
To calculate the molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X, we can use the formula for freezing point depression:
[tex]\[ \Delta T_f = i \cdot K_f \cdot m \][/tex]
where:
[tex]\( \Delta T_f \)[/tex] is the freezing point depression, which is the difference between the normal freezing point of the solvent and the freezing point of the solution.
i is the van 't Hoff factor, which is the number of moles of solute particles per mole of solute dissolved. For urea, i is typically 1 because urea does not dissociate in solution.
[tex]\( K_f \)[/tex] is the molal freezing point depression constant for the solvent.
m is the molality of the solution, which is the number of moles of solute per kilogram of solvent.
First, we calculate the freezing point depression [tex]\( \Delta T_f \)[/tex]:
[tex]\[ \Delta T_f = T_f^0 - T_f = 0.4C - (-0.5C) = 0.9C \][/tex]
Next, we need to calculate the molality m of the solution. The molar mass of urea [tex]\( NH_2CONH_2 \)[/tex] is 60.06 g/mol. The number of moles of urea is:
[tex]\[ n_{urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} = \frac{5.90 \text{ g}}{60.06 \text{ g/mol}} \] \[ n_{urea} = 0.0982 \text{ mol} \][/tex]
The molality m is:
[tex]\[ m = \frac{n_{urea}}{\text{mass of solvent in kg}} = \frac{0.0982 \text{ mol}}{0.450 \text{ kg}} \] \[ m = 0.2182 \text{ mol/kg} \][/tex]
Now we can rearrange the freezing point depression formula to solve for [tex]\( K_f \)[/tex]:
[tex]\[ K_f = \frac{\Delta T_f}{i \cdot m} \] Since \( i = 1 \) for urea, we have: \[ K_f = \frac{0.9C}{1 \cdot 0.2182 \text{ mol/kg}} \] \[ K_f = 4.124 \text{ C/(mol/kg)} \][/tex]
Therefore, the molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X is 4.124 °C/(mol/kg).
Ag+ (aq) + Cl- (aq) → AgCl (s) Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid NaCl must be added to 25.0 mL of 0.251 M AgNO3 solution to completely precipitate the silver?
Answer:
0.367 g
Explanation:
Precipitation of silver and Cl requires 1:1 mol ratio of silver and Cl each, so the amount of mol NaCl (provides 1 mol of Cl⁻) and AgNO₃ (privides 1 mol of Ag⁺) must be equal for complete precipitation.
mol of Ag⁺ we have is
[tex]mol_{Ag^+}=[AgNO_3]*V_{solution}\\ =0.251*0.025=0.006275\ mol[/tex]
Thus we need 0.00625 mol of NaCl as well
The molar mass of NaCl is 58.44 g/mol
Therefore the total grams of NaCl needed is
[tex]m_{NaCl}=58.44*0.006275\\ =0.367\ g[/tex]
To completely precipitate the silver, approximately 0.900 grams of solid NaCl must be added to the 25.0 mL of 0.251 M AgNO3 solution.
Explanation:The question asks how many grams of solid NaCl must be added to 25.0 mL of 0.251 M AgNO3 solution to completely precipitate the silver. To find the number of grams of NaCl needed, we need to calculate the number of moles of AgNO3 in the 25.0 mL solution. Then, using the stoichiometry of the reaction, we can determine the number of moles of AgCl that will precipitate. Finally, we can convert the moles of AgCl to grams of NaCl using the molar mass of NaCl.
First, we calculate the number of moles of AgNO3:
Volume of solution = 25.0 mL = 0.0250 LMolarity of solution = 0.251 MNumber of moles of AgNO3 = volume × molarity = 0.0250 L × 0.251 mol/L = 0.00628 molAccording to the balanced equation, 1 mole of AgNO3 produces 1 mole of AgCl. Therefore, 0.00628 mol of AgNO3 will produce 0.00628 mol of AgCl.
Next, we convert moles of AgCl to grams of NaCl:
Molar mass of AgCl = 143.32 g/molGrams of NaCl = moles of AgCl × molar mass of AgCl = 0.00628 mol × 143.32 g/mol = 0.900 g of NaClTherefore, approximately 0.900 grams of solid NaCl must be added to the 25.0 mL of 0.251 M AgNO3 solution to completely precipitate the silver.
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Tarnish on copper is the compound CuO. A tarnished copper plate is placed in an aluminum pan of boiling water. When enough salt is added so that the solution conducts electricity, the tarnish disappears. Imagine that the two halves of this redox reaction were separated and connected with a wire and a salt bridge.
Calculate the standard cell potential given the following standard reduction potentials:
Al3++3e-?Al; E?=?1.66 V
Cu2++2e-?Cu; E?=0.340 V
Answer: 2.00 V
Explanation:
The balanced redox reaction is:
[tex]2Al+3Cu^{2+}\rightarrow 2Al^{3+}+3Cu[/tex]
Here Al undergoes oxidation by loss of electrons, thus act as anode. Copper undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]Al^{3+}+3e^-\rightarrow Al[/tex] [tex]E^0_{[Al^{3+}/Al]}=-1.66V[/tex]
[tex]Cu^{2+}+2e^-\rightarrow Cu[/tex] [tex]E^0_{[Cu^{2+}/Cu]}=0.340V[/tex]
[tex]E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Al^{3+}/Al]}[/tex]
[tex]E^0=+0.34- (-1.66V)=2.00V[/tex]
Thus the standard cell potential is 2.00 V
In the water molecule, the H-O-H bond angle is 105º. Which distribution of electrons around the central atom provides the best explanation for this bond angle?
4 shared pairs
2 shared pairs, 2 lone pairs
1 shared pair, 3 lone pairs
3 shared pairs, 1 lone pair
The water molecule has an H-O-H bond angle of 104.5° which is due to the central oxygen atom having 2 shared pairs and 2 lone pairs of electrons, resulting in a bent molecular geometry due to the repulsion of lone pairs and sp³ hybridization.
Explanation:The bond angle in a water molecule is closer to 104.5° rather than the stated 105°, and the distribution of electrons around the central oxygen atom that best explains this bond angle is 2 shared pairs, 2 lone pairs.
In the water molecule, the central oxygen atom is sp³ hybridized, with its four hybrid orbitals occupied by two lone pairs of electrons and two bonding pairs that form covalent bonds with hydrogen atoms.
Lone pairs require more space than bonding pairs, which leads to a repulsion that pushes the hydrogen atoms closer together, resulting in the H-O-H bond angle being slightly less than the ideal 109.5° angle of a tetrahedron.
Consider a galvanic cell composed of the SHE and a half-cell using the following reaction: Ag+(aq) + e− → Ag(s)
(a) Calculate the standard cell potential. E o cell = V
(b) What is the spontaneous cell reaction under standard-state conditions?
(c) Calculate the cell potential when [H+] in the hydrogen electrode is changed to the following concentrations, while all other reagents are held at standard-state conditions:
(i) 4.2 × 10−2 M E = V
(ii) 9.6 × 10−5 M E = V
(d) Based on this cell arrangement, suggest a design for a pH meter.
Answer:
a. 0.80V b. 2Ag⁺(aq) + H2(g) ⇄ 2Ag(s) +2H⁺(aq) c. i) 0.88 ii) 1.03 d. Cell is a ph meter with the potential being a function of hydrogen ion concentration
Explanation:
a. The two half cell reactions are
1. 2H⁺(aq) +2e⁻ → H₂(g) Eanode = 0.00V
2. Ag⁺(aq) + e⁻ → Ag(s) Ecathode = 0.80V
The balanced cell reaction is
2Ag(aq)⁺ + H₂(g) ⇄ 2Ag(s) + 2H⁺(aq)
therefore Ecell = Ecathode - Eanode = 0.80 - 0.00 = +0.80V
b. Since the Ecell is positive, the spontaneous cell reaction under standar state conditions is
2Ag(aq)⁺ + H₂(g) ⇄ 2Ag(s) + 2H⁺(aq)
c. Use Nernst Equation
E = Ecell - (0.0592/n)log([H⁺]/[Ag⁺]²[P H₂]), where n is the number of moles of Ag and P H₂= 1.0 atm
i) E = 0.80 - (0.0592/2)log(4.2x10^-2)/(1.0)²(1.0) = 0.88V
ii) E = 0.80 - (0.0592/2)log(9.6x10^-5)/(1.0)²(1.0) = 1.03V
d . From the above calculation we can conclude that the cell acts as a pH meter as a change in hydrogen ion concentration results in a change in the potential of the cell. A change of ph of 2.64 changes the E of cell by 0.15 V.
What change occurs during the reaction:MnO4- --->Mn2+?Five electrons are lost.Three electrons are lost.Five electrons are gained.Three electrons are gained.
Answer:
Five electrons are gained.
Explanation:
Oxidation number or oxidation state of an atom in a chemical compound is the number of electrons lost or gained. It is also defined as the degree of oxidation of the atom in the compound.
This is a theoretical number which can help to decipher the oxidation and reduction in a redox reaction.
Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.
The given reaction is shown below as:
[tex]MnO_4^-\rightarrow Mn^{2+}[/tex]
Manganese in [tex]MnO_4^-[/tex] has oxidation state of +7
Manganese in [tex]Mn^{2+}[/tex] has an oxidation state of +2
It reduces from +7 to +2 . It means that 5 electrons are gained.
If you mix different amounts of two ideal gases that are originally at different temperatures, what must be true of the final state after the temperature stabilizes? (There may be more than one correct choice.)
a) Both gases will reach the same final temperature.
b) The final rms molecular speed will be the same for both gases.
c) The final average kinetic energy of a molecule will be the same for both gases.
Answer:
a,c are correct
Explanation:
a) On mixing two gases the final temperature of both the gases becomes the same. The heat will flow from high temp. gas to lower temp gas till the temp of both gases become equal (Thermal equilibrium). This is correct.
b) The rms speed of the molecule is inversely proportional to its molar mass so the final rsm will not be the same. This is incorrect.
c) The average kinetic energy of the system will remain the same. Hence this is also correct.
A two-phase system of liquid water and water vapor in equilibrium at 8000 kPa con sists of equal volumes of liquid and vapox, If the total volume V = 0.15 m, what is the total enthalpy H' and what is the total entropy S?
The total enthalpy H' is 2758.611 kJ/kg and the the total entropy S is 5.745 kJ/ kg . K.
What is entropy?Entropy is defined as the amount of thermal energy per unit temperature in a system that cannot be used for productive work. Entropy, which measures disorder, has an impact on every element of our daily life. In actuality, you could consider it to be nature's tax. Disorder, if unchecked, gets worse over time.
Water vapor will be in its saturated form because the question states that water and water vapor are in equilibrium.
According to the steam table, the total enthalpy Ht = 2758.611 kJ/kg for saturated steam at 8000 kPa (8MPa).
Entropy in its entirety St = 5.745 kJ/ kilogram. K
The saturated steam table at that particular pressure is seen in this figure.
Thus, the total enthalpy H' is 2758.611 kJ/kg and the the total entropy S is 5.745 kJ/ kg . K.
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Given the conditions stated, the water is in a supercritical fluid state where distinct liquid and gas phases don't exist. However, we can't determine the total enthalpy and entropy with the provided information, as we require additional data and the use of thermodynamic equations.
Explanation:The question pertains to a two-phase system of liquid water and water vapor at a specific pressure and volume. The system exists at 8000 kPa, which is beyond the critical point of water (22.064 Mpa or 218 atm). This signifies that under such conditions, the sample of water is in the state of a supercritical fluid, where distinct liquid and gas phases cease to exist and instead a single phase with properties intermediate of both liquid and gas forms.
Unfortunately, based on the available data, determining the total enthalpy (H) and the total entropy (S) isn't feasible. For that, we need extra data such as the specific heat capacity, the exact temperatures of the different phases of water, and/or the number of moles involved. Calculations for H and S would also involve complex thermodynamic equations and experimentally determined constants for water.
Supercritical fluids are noteworthy substances in which the line separating liquid and gas phases disappears at a critical temperature and pressure, leading to the creation of a new phase with intermediate characteristics between a liquid and a gas. The vapor pressure, the liquid-gas equilibrium, and the thermodynamic properties of substances are crucial concepts in this context.
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The lines on Absorption Atomic Spectra correspond to energies needed for electrons to be excited from a lower energy level to a higher energy level. Assume that the energy needed for an electron in 2p orbital in an O atom to jump to 3s orbital is 3.6*10^-19 J, what is its wavelength of the line atomic spectra in nanometer (nm)?
Note: please use whole numbers and 3 sig figs, or no decimal place.
Answer:
553 nm
Explanation:
When an electron from O absorbs radiation with an energy (E) of 3.6 × 10⁻¹⁹ J, it is excited from orbital 2p to orbital 3s. The wavelength (λ) associated with that radiation can be calculated using the Planck-Einstein equation.
E = h. ν = h . c . λ⁻¹
where,
h is the Planck's constant
c is the speed of light
ν is the frequency
[tex]\lambda = \frac{h.c}{E} =\frac{6.63 \times 10^{-34}J.s \times 3.00 \times 10^{8} m/s}{3.6 \times 10^{-19}J } .\frac{10^{9}nm }{1m} =553nm[/tex]
In calculating the concentration of [Cu(NH3)4]2+ from [Cu(H2O)4]2+, the stepwise formation constants are as follows: K1=1.90×104 K2=3.90×103 K3=1.00×103 K4=1.50×102 Calculate the overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+.
Answer:
Kf = 1.11x10¹³
Explanation:
The value of Kf for a multistep process that involves an equilibrium at each step, is the multiplication of the constant of the equilibrium of each step.
Kf = K1xK2xK3xK4
Kf = 1.90x10⁴ x 3.90x10³ x 1.00x10³ x 1.50x10²
Kf = 1.11x10¹³
Final answer:
The overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+ is 2.1 x 10^13.
Explanation:
The overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+ can be calculated by multiplying the stepwise formation constants (K1, K2, K3, K4). In this case, the value of Kf is calculated as follows:
Kf = K1 * K2 * K3 * K4 = 1.90×10^4 * 3.90×10^3 * 1.00×10^3 * 1.50×10^2 = 2.1 x 10^13
Therefore, the overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+ is 2.1 x 10^13.
For an aqueous solution of sodium chloride (NaCl), determine the molarity of 4.15 L of a solution that contains 173 g of sodium chloride. Determine the volume of this solution that would contain 3.93 moles of sodium chloride. Determine the number of moles of sodium chloride in 22.45 L of this solution.
Answer:
a) Molarity = 0.713 M
b) volume = 5.51 L
c) Number of moles = 16.01 moles
Explanation:
A) What is the molarity of a 4.15L solution that contains 173 g of sodium chloride?
Step 1: Data given
Volume = 4.15 L
Mass of NaCl = 173 grams
Molar mass of NaCl = 58.44 g/mol
Step 2: Calculate moles of NaCl
Number of moles NaCl = Mass NaCl / molar mass
Moles NaCl = 173 grams / 58.44 g/mol
Moles NaCl = 2.96 moles
Step 3: Calculate molarity of solution
Molarity = moles NacL/ volume
Molarity = 2.96 moles / 4.15L
Molarity = 0.713 M
b) Determine the volume of this solution that would contain 3.93 moles of sodium chloride.
Step 1: Data given
Number of moles = 3.93 moles
Molarity = 0.713 M
Step 2: Calculate volume
Volume = Moles/ Molarity
Volume = 3.93 mol/0.713 M
volume = 5.51 L
c) Determine the number of moles of sodium chloride in 22.45 L of this solution.
Step 1: Data given
Volume = 22.45 L
Molarity = 0.713 M
Step 2: Calculate number of moles
Moles NaCl = Molarity * volume
Moles NaCl = 0.713 * 22.45 L
Moles NaCl = 16.01 moles
The molarity of a solution containing 173 g of sodium chloride (NaCl) in 4.15 L of solution is 0.71 M. The volume of the solution that would contain 3.93 moles of NaCl is 5.52 L. There are 15.95 moles of NaCl in 22.45 L of the solution.
Explanation:The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. To determine the molarity of a solution containing 173 g of sodium chloride (NaCl) in 4.15 L of solution, we first need to calculate the number of moles of NaCl. The molar mass of NaCl is 58.44 g/mol, so 173 g of NaCl is equal to 2.96 moles. Dividing this by the volume of the solution (4.15 L), we find that the molarity is 0.71 M.
To determine the volume of the solution that would contain 3.93 moles of NaCl, we can rearrange the equation for molarity: Moles = Molarity x Volume. Solving for Volume, we find that Volume = Moles / Molarity. Plugging in the given values, we have Volume = 3.93 moles / 0.71 M = 5.52 L.
Finally, to determine the number of moles of NaCl in 22.45 L of the solution, we can use the molarity formula again: Moles = Molarity x Volume. Plugging in the given values, we have Moles = 0.71 M x 22.45 L = 15.95 moles. Therefore, there are 15.95 moles of NaCl in 22.45 L of the solution.
Assume 0.18 L of a 5.0 M solution of lead (II) nitrate, Pb(NO3)2, reacts with a 2.6 M solution of sodium phosphate, Na3PO4, to produce lead (II) phosphate, Pb3(PO4)2, and sodium nitrate, NaNO3. The problem requires that you determine the volume of sodium phosphate, Na3PO4, needed for the reaction to occur.
Answer:
0.23 L
Explanation:
Let's consider the following balanced equation.
3 Pb(NO₃)₂(aq) + 2 Na₃PO₄(aq) ⇄ Pb₃(PO₄)₂(s) + 6 NaNO₃(aq)
The moles of Pb(NO₃)₂ are:
[tex]0.18L\times \frac{5.0mol}{L} =0.90mol[/tex]
The molar ratio of Pb(NO₃)₂ to Na₃PO₄ is 3:2. The moles of Na₃PO₄ are:
[tex]0.90molPb(NO_{3})_{2}.\frac{2molNa_{3}PO_{4}}{3molPb(NO_{3})_{2}} =0.60molNa_{3}PO_{4}[/tex]
The volume of Na₃PO₄ required is:
[tex]\frac{0.60mol}{2.6mol/L} =0.23L[/tex]
A student performs an experiment to determine the volume of hydrogen gas produced when a given mass of magnesium reacts with excess HCl(aq) , as represented by the net ionic equation above. The student begins with a 0.0360 g sample of pure magnesium and a solution of 2.0 M HCl(aq).
(a) Calculate the number of moles of magnesium in the 0.0360 g sample
Answer:
1.48x10⁻³ mol
Explanation:
The balanced equation is
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
The aqueous solution will disociate, so:
Mg(s) + 2H⁺(aq) + 2Cl⁻(aq) → Mg²⁺(aq) + 2Cl⁻(aq) + H₂(g)
Simplifying by eliminating the bolded substances that have the same amount in both sides, we have the net ionic equation:
Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).
a) The number of moles of Magnesium is the mass divided by the molar mass (24.305 g/mol):
n = 0.0360/24.305
n = 1.48x10⁻³ mol
Describe the relationship between predator and prey in a balanced ecosystem.
Predators grow along with their victims. Over time, prey animals are now developing and avoiding themselves to get eaten by their predators. Such tactics and modifications can take many forms that make their work easier, including disguise, mimicry, defense mechanisms, flexibility, distance, habits and even tool use.
Though fact an equilibrium appears to occur within an ecosystem between predators and prey, there are several factors which affect it, including the birth and death rates of predators and preys.
Select the following solutions that would form a buffer. Select all that apply.
A. 0.1M HNO3 and 0.15M NH4Cl
B. 0.5M NaNO2 and 0.5M HNO2
C. 0.1M KC6H5COO and 0.05M C6H5COOH
D. 0.1M NH3 and 0.1M KClO
E. 0.05M NaOH and 0.1M HCHO2
F. 0.1M KF and 0.15M HCl
G. 0.1M HBr and 0.1M NaOH
Answer:
B. 0.5M NaNO₂ and 0.5M HNO₂ .
C. 0.1M KC₆H₅COO and 0.05M C₆H₅COOH .
E. 0.05M NaOH and 0.1M HCHO₂.
Explanation:
A buffer is made of 2 components:
A weak acid and its conjugate base orA weak base and its conjugate acidSelect the following solutions that would form a buffer. Select all that apply.
A. 0.1M HNO₃ and 0.15M NH₄Cl. NO. HNO₃ is a strong acid
B. 0.5M NaNO₂ and 0.5M HNO₂. YES. HNO₂ is a weak acid and NO₂⁻ (coming from NaNO₂) its conjugate base.
C. 0.1M KC₆H₅COO and 0.05M C₆H₅COOH. YES. C₆H₅COOH is a weak acid and C₆H₅COO⁻ (coming from KC₆H₅COO) its conjugate base.
D. 0.1M NH₃ and 0.1M KClO. NO. It does not have the components of a buffer system.
E. 0.05M NaOH and 0.1M HCHO₂. YES. HCHO₂ is a weak acid that can react with NaOH to produce CHO₂⁻, its conjugate base.
F. 0.1M KF and 0.15M HCl. NO. HCl is a strong acid.
G. 0.1M HBr and 0.1M NaOH. NO. HBr is a strong acid.
The options that would form a buffer are B, C, and E.
Explanation:A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. Using this information, we can analyze the given options:
0.1M HNO3 and 0.15M NH4Cl: HNO3 is a strong acid, while NH4Cl is a salt. Therefore, this combination does not form a buffer.0.5M NaNO2 and 0.5M HNO2: NaNO2 is a salt, while HNO2 is a weak acid. Thus, this combination can form a buffer.0.1M KC6H5COO and 0.05M C6H5COOH: KC6H5COO is a salt, and C6H5COOH is a weak acid. Hence, this combination can form a buffer.0.1M NH3 and 0.1M KClO: NH3 is a weak base, while KClO is a salt. Therefore, this combination can form a buffer.0.05M NaOH and 0.1M HCHO2: NaOH is a strong base, while HCHO2 is a weak acid. Thus, they can form a buffer.0.1M KF and 0.15M HCl: KF is a salt, and HCl is a strong acid. Therefore, this combination does not form a buffer.0.1M HBr and 0.1M NaOH: HBr is a strong acid, and NaOH is a strong base. Hence, this combination does not form a buffer.Based on this analysis, the options that would form a buffer are B, C, and E.
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Standard Reduction Potentials in Aqueous Solution at 25 °C
Acidic Solution (V) 1.77 1.68 1.50 1.08 0.96 0.799 0.789 0.337 0.27
–0.14 –0.25 –0.28 –0.41 –0.44 –0.763 –0.91 –1.66 –2.37
Which metal ion from the selection is the weakest oxidizing agent? Which metal ion from the selection is the strongest oxidizing agent? Which metal from the selection is the strongest reducing agent? Which metal from the selection is the weakest reducing agent? Will reduce to ? yes no Will reduce to ? yes no Which metal from the selection can be oxidized by ?
Answer:
Hope this helps:)
Explanation:
The values for the table entries are reduction potentials, so lithium at the top of the list has the most negative number, indicating that it is the strongest reducing agent. The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential.
Elemental fluorine, for example, is the strongest common oxidizing agent.
Lithium metal is therefore the strongest reductant (most easily oxidized) of the alkali metals in aqueous solution. The standard reduction potentials can be interpreted as a ranking of substances according to their oxidizing and reducing power
The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and 269.2 J/K*mol, respectively. Calculate ΔH° , ΔS° , and ΔG° for the following process at 25.00°C.
C6H6(l) ------> C6H6(g)
Final answer:
The standard enthalpy change (ΔH°) for the process C6H6(l) → C6H6(g) is 82.93 kJ/mol. The standard entropy change (ΔS°) for the process is 269.2 J/K*mol. The standard free energy change (ΔG°) for the process is 7361.033 kJ/mol.
Explanation:
The standard enthalpy change, ΔH°, for the process C6H6(l) → C6H6(g) can be calculated using the standard enthalpy of formation of benzene in the gaseous state and the liquid state. The equation for ΔH° is ΔH° = ΣΔH°(products) - ΣΔH°(reactants). In this case, since benzene is the only product and there are no reactants, the equation simplifies to ΔH° = ΔH°(C6H6(g)). Therefore, ΔH° = 82.93 kJ/mol.
The standard entropy change, ΔS°, for the process can be calculated using the standard entropy of benzene in the gaseous state and the liquid state. The equation for ΔS° is ΔS° = ΣΔS°(products) - ΣΔS°(reactants). Similar to the calculation for ΔH°, since benzene is the only product and there are no reactants, the equation simplifies to ΔS° = ΔS°(C6H6(g)). Therefore, ΔS° = 269.2 J/K*mol.
The standard free energy change, ΔG°, for the process can be calculated using the equation ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin. Since the temperature is given as 25.00°C, we need to convert it to Kelvin by adding 273.15. Therefore, T = 25.00°C + 273.15 = 298.15 K. Substituting the values into the equation, we get ΔG° = 82.93 kJ/mol - (298.15 K)(269.2 J/K*mol) = 7361.033 kJ/mol.
The vapor pressure of the liquid SO2 is measure at different temperatures. the following vapor pressure data are obtained.Temp (K) Pressure mmHg241.3 271.2259.3 641.8Calculate the enthalpy of vaporization(delta H vap) in KJ/mol for this liquid.
Answer:
24.895 kJ/mol
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
[tex]\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c [/tex]
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314 J /mol K)
c is the constant.
For two situations and phases, the equation becomes:
[tex]\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)[/tex]
Given:
[tex]P_1[/tex] = 271.2 mmHg
[tex]P_2[/tex] = 641.8 mmHg
[tex]T_1[/tex] = 241.3 K
[tex]T_2[/tex] = 259.3 K
So,
[tex]\ln \:\left(\:\frac{271.2}{641.8}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{259.3}-\:\frac{1}{241.3}\:\right)[/tex]
[tex]\Delta \:H_{vap}=\ln \left(\frac{271.2}{641.8}\right)\frac{8.314}{\left(\frac{1}{259.3}-\:\frac{1}{241.3}\right)}\ J/mol[/tex]
[tex]\Delta \:H_{vap}=\left(-\frac{520199.41426}{18}\right)\left(\ln \left(271.2\right)-\ln \left(641.8\right)\right)\ J/mol[/tex]
ΔHvap = 24895.015 J/mol = 24.895 kJ/mol ( 1 J = 0.001 kJ )
Final answer:
This detailed answer explains how to calculate the enthalpy of vaporization for liquid SO2 using the Clausius-Clapeyron equation and vapor pressure data at different temperatures.
Explanation:
The enthalpy of vaporization (ΔHvap) can be calculated using the Clausius-Clapeyron equation:
In this case, the enthalpy of vaporization for SO2 can be calculated using the given vapor pressure data at different temperatures:
Convert temperatures to Kelvin.
Use the Clausius-Clapeyron equation to find ΔHvap.
By applying the equation with the provided data, the calculated enthalpy of vaporization for liquid SO2 is 24.895 kJ/mol.
Lauryl alcohol is a nonelectrolyte obtained from coconut oil and is used to make detergents. A solution of 6.80 g of lauryl alcohol in 0.200 kg of benzene freezes at 4.5 ∘C. What is the approximate molar mass of lauryl alcohol?
What is the approximate molar mass of lauryl alcohol?
Answer:
The approximate molar mass of lauryl alcohol is 174.08 g/m
Explanation:
An excersise to apply the colligative property of Freezing-point depression.
This is the formula: ΔT = Kf . m
First of all, think the T° of fusion of benzene → 5.5°C
ΔT = T° pure solvent - T° fusion solution
Kf for benzene: 5.12 °C/m
5.5°C - 4.5°C = 5.12 °C /m . m
1°C / 5.12 m /°C = m
0.195 m = molality
This moles of lauryl alcohol, solute, are in 1 kg of benzene, solvent.
I have to find out in 0.2 kg.
1 kg sv ____ 0.195 moles solute
0.2 kg sv ____ (0.195 . 0.2)/1 = 0.039 moles solute
The mass for these moles is 6.80 g, so if I want to know the molar mass, I have to divide mass / moles
6.80 g/ 0.039 moles = 174.08 g/m
For which of the following equilibria does Kc correspond to the acid-dissociation constant, Ka, of H2PO4-?
A) H2PO4-(aq) + H3O+(aq)--> H3PO4(aq) + H2O(l)
B) H2PO4-(aq) + H2O(l)--> H3PO4(aq) + OH-(aq)
C)H2PO4-(aq) + H2O(l)--> H3O+(aq) + HPO42(aq)
D) H3PO4(aq) + H2O(l)--> H3O+(aq) + H2PO4-(aq)
E) HPO42-(aq) + H2O(l)--> H2PO4-(aq) + OH-(aq)
Answer:
The correct answer is C)H2PO4-(aq) + H2O(l)--> H3O+(aq) + HPO42-(aq)
Explanation:
The acid dissociation equilibrium involves the loss of a proton of the acid to give the conjugated acid. In this case, the acid is H₂PO₄⁻ and it losses a proton (H⁺) to give the conjugated acid HPO₄²⁻ (without a proton and with 1 more negative charge). In the aqueous equilibrium, the proton is taken by H₂O molecule to give the hydronium ion H₃O⁺.
H₂PO₄⁻(aq) + H₂O(l)--> H₃O⁺(aq) + HPO₄²⁻(aq)
Answer:
C) H₂PO₄⁻(aq) + H₂O(l) → H₃O⁺(aq) + HPO₄²⁻(aq)
Explanation:
For which of the following equilibria does Kc correspond to the acid-dissociation constant, Ka, of H₂PO₄⁻?
A) H₂PO₄⁻(aq) + H₃O⁺(aq) → H₃PO₄(aq) + H₂O(l)
NO. This is the inverse of the acid dissociation of H₃PO₄.
B) H₂PO₄⁻(aq) + H₂O(l) → H₃PO₄(aq) + OH⁻(aq)
NO. This is the basic dissociation of H₂PO₄⁻.
C) H₂PO₄⁻(aq) + H₂O(l) → H₃O⁺(aq) + HPO₄²⁻(aq)
YES. This is the acid dissociation of H₂PO₄⁻. The acid-dissociation constant is:
[tex]Ka=\frac{[H_{3}O^{+}].[HPO_{4}^{2-} ]}{[H_{2}PO_{4}^{-} ]}[/tex]
D) H₃PO₄(aq) + H₂O(l) → H₃O⁺(aq) + H₂PO₄⁻(aq)
No. This is the acid dissociation of H₃PO₄.
E) HPO₄²⁻(aq) + H₂O(l) → H₂PO₄⁻(aq) + OH⁻(aq)
NO. This is the basic dissociation of HPO₄²⁻.
A solenoid 2 cm long consists of 5653 loops of wire. If the magnetic field inside the solenoid is 4 T, what is the magnitude of the current that flows through it? A solenoid 2 cm long consists of 5653 loops of wire. If the magnetic field inside the solenoid is 4 T, what is the magnitude of the current that flows through it? 140 A 9.1×10−2 A 11 A 2.0 A
Answer:
11 A
Explanation:
The magnetic field inside a solenoid can be calculated by the equation:
B = μ*(N/L)*i
Where B is the magnetic field, μ is the magnetic permeability (which is 1.256x10⁻⁶ T/m.A at vacuum), N is the number os loops, L is the length of the solenoid (2 cm = 0.02 m), and i the current.
4 = 1.256x10⁻⁶ *(5653/0.02)*i
0.355i = 4
i ≅ 11 A
For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
The entropy change ΔSo for a reaction can be obtained by considering the sum of decompositions and formations of reactants and products. In this scenario, while we were only able to compute the enthalpy change ΔHxn of -136.80 kJ, the entropy change, ΔSo, cannot be calculated without additional data.
Explanation:To find the entropy change ΔSo for the reaction 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g), we start by writing this as the sum of decompositions of 3NO₂(g) and 1H₂O(1) into their constituent elements. Similarly, we identify the formation of 2HNO3(aq) and 1NO(g) from their constituent elements. By summing the enthalpy changes obtained from standard enthalpy changes of formation (ΔHf) for these compounds, we find the result for ΔHxn, which in this case equates to -136.80 kJ. However, the original question asked for ΔSo, not ΔHxn. Without knowing the ΔSo for the individual substances in this reaction, we cannot calculate ΔSo for the entire reaction.
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Consider the dissolution of AB(s):AB(s)⇌A+(aq)+B−(aq)Le Châtelier's principle tells us that an increase in either [A+] or [B−] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A+ or B− ions. This is an example of the common-ion effect.The generic metal hydroxide M(OH)2 has Ksp = 1.05×10−18. (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH− from water can be ignored. However, this may not always be the case.)What is the solubility of M(OH)2 in pure water?
Answer:
S = 6.40 × 10⁻⁷ M
Explanation:
In order to calculate the solubility (S) of M(OH)₂ in pure water we will use an ICE Chart. We recognize 3 stages: Initial, Change and Equilibrium, and we complete each row with the concentration or change in concentration.
M(OH)₂(s) ⇄ M²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product (Kps) is:
Kps = 1.05 × 10⁻¹⁸ = [M²⁺].[OH⁻]²=S.(2S)²
1.05 × 10⁻¹⁸ = 4S³
S = 6.40 × 10⁻⁷ M
Rank the following in order of decreasing acid strength in aqueous solution: HCl, HOCl, HOBr,HOI.
A) HCl > HClO > HBrO > HIO
B) HIO > HBrO > HClO > HCl
C) HCl > HIO > HBrO > HClO
D) HClO > HCl > HBrO > HIO
E) HClO > HBrO > HCl > HIO
The correct order of decreasing acid strength in an aqueous solution among HCl, HOCl, HOBr, and HOI is HCl > HIO > HBrO > HClO. This order is based on the strength of the H-X bond and the stability of the X- ion, where HCl is the strongest acid.
Explanation:The correct order of decreasing acid strength in aqueous solution: HCl, HOCl, HOBr, and HOI, is option C: HCl > HIO > HBrO > HClO. This order is determined by two main factors: the strength of the H-X bond and the stability of the X-ion. Hydrochloric acid (HCl) is the strongest because it can easily donate protons in solution. The other compounds are oxyacids, and their strength increases with increasing electronegativity and size of the halogen attached to oxygen. So, when comparing HOCl, HOBr, and HOI, the more polarizable (or larger in size) the halogen, the stronger the acid. Iodine is the largest here and thus HOI is more acidic than HOBr and HOCl.
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Which of the following best describe the particles present in a 2.0 M aqueous solution of Al(NO3)3?2 M Al3+(aq) and 18 M NO3-(aq)2 M Al(NO3)3(aq)2 M Al3+(aq) and 6 M NO3-(aq)2 M Al3+(aq) and 3 M NO3-(aq)
Answer:
The correct answer is: 2M Al3+(aq) and 6 M NO3-(aq)
Explanation:
Step 1: Data given
2.0 M Al(NO3)3
Step 2:
Al(NO3)3 in water will dissociate as following:
Al(NO3)3 → Al^3+ + 3NO3^-
For 1 mol of Al(NO3)3 we will have 1 mol of Al^3+ and 3 moles of NO3^-
We know that the molarity of Al(NO3)3 = 2.0 M, this means 2.0 mol/ L
The mol ratio Al(NO3)3 and Al^3+ is 1:1 so the molarity of Al^3+ is 2.0 M
The mol ratio Al(NO3)3 and NO3^- is 1:3 so the molarity of NO3^- is 6.0M
The correct answer is: 2M Al3+(aq) and 6 M NO3-(aq)
When solid Fe metal is put into an aqueous solution of Sn(NO3)2, solid Sn metal and a solution of Fe(NO3)2 result. Write the net ionic equation for the reaction. (Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)
Answer:
Fe(s) + Sn²⁺(aq) ⇒ Sn(s) + Fe²⁺(aq)
Explanation:
When solid Fe metal is put into an aqueous solution of Sn(NO₃)₂, solid Sn metal and a solution of Fe(NO₃)₂ result. The resulting molecular equation is:
Fe(s) + Sn(NO₃)₂(aq) ⇒ Sn(s) + Fe(NO₃)₂
The full ionic equation includes all the ions and the species that do not dissociate in water.
Fe(s) + Sn²⁺(aq) + 2 NO₃⁻(aq) ⇒ Sn(s) + Fe²⁺(aq) + 2 NO₃⁻(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and species that do not dissociate in water.
Fe(s) + Sn²⁺(aq) ⇒ Sn(s) + Fe²⁺(aq)
Final answer:
The net ionic equation for the reaction of solid Fe metal with an aqueous solution of Sn(NO3)2 is Fe(s) + Sn2+(aq)
ightarrow Fe2+(aq) + Sn(s), a single displacement redox reaction.
Explanation:
When solid Fe metal is introduced to an aqueous solution of Sn(NO3)2, a single displacement reaction occurs where Fe displaces Sn from the stannous nitrate. The net ionic equation for this chemical reaction, considering that nitrates are soluble and metallic tin will precipitate out of the solution as a solid, can be written as follows:
Fe(s) + Sn2+(aq)
ightarrow Fe2+(aq) + Sn(s)
In this equation, solid iron (Fe) reacts with stannous ions (Sn2+) in solution to form ferrous ions (Fe2+) and solid tin (Sn). This type of reaction is also referred to as a redox reaction.
Why is a mechanism needed to remove carbon dioxide gas from your cells?
Answer:
Respiration is responsible for removing carbon dioxide and providing oxygen to our body cells. Carbondioxide is eliminated from the body as a byproduct.
Explanation:
The shape of the BF3 molecule is best described as
T-shaped
see-saw
tetrahedral
trigonal planar
Answer:
The shape of the BF3 molecule is best described as trigonal planar.
Explanation:
The Lewis Structure for BF3 is like this:
_ _
| F | | F |
\ /
B
|
| F |
---
It forms three angles of 120° each. The bonds are in the same planar that's why it is trigonal planar and they are exactly the same.
Boron and Fluorine have 3 covalent bonds, produced by electronic promotion that enables the 2py and 2pz orbitals, leaving an electron to pair in the 2px. So boron will have 3 possible electrons to pair in 2s1, 2px and 2py, remember that electronic configuration for B is 1s2, 2s2, 2p1
By hybridization between the orbitals 2s2 and 2p1, the electrons of F, can joined to make the covalent bond. The new B configuration is 1s2, 2s1, 2px1, 2py1 (these last three, hybrid orbitals)
The shape of the BF3 molecule is trigonal planar.
Explanation:The shape of the BF3 molecule is trigonal planar.
To determine the shape of a molecule, we need to look at its electron geometry and molecular geometry. The central atom in BF3 is Boron, with three Fluorine atoms surrounding it. Boron has three valence electrons, and each Fluorine atom contributes one electron. When we draw the Lewis structure for BF3, we can see that Boron forms three bonds with the Fluorine atoms, resulting in a trigonal planar electron geometry.
Since there are no lone pairs on the central atom, the molecular geometry of BF3 is also trigonal planar.
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The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following:
Sn2+(aq)+2e−→Sn(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.140 V E∘red=+0.337 V
What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.
Express your answer to three decimal places and include the appropriate units.
Answer:
0.477 V
Explanation:
When a substance is gaining electrons, it's reducing, and when the substance loses electrons, it's oxidizing. In a galvanic cell, one substance oxides giving electrons for the other, which reduces. Then, the substance with higher reduction potential must reduce and the other must oxide.
E°cell = E°red(red) - E°red(oxid)
Where, E°red(red) is the reduction potential of the substance that reduces, and E°red(oxid) is the reduction potential of the substance that oxides. For the value given, Cu⁺² reduces, so:
E°cell = +0.337 - (-0.140)
E°cell = 0.477 V
The standard cell potential for this galvanic cell is [tex]0.477 \ V[/tex].
To calculate the standard cell potential [tex](\( E^\circ_\text{cell} \))[/tex] for the galvanic cell using the given reduction potentials, we use the formula:
[tex]\[ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \][/tex]
Given reduction potentials:
[tex]\( E^\circ_\text{red}(\text{Sn}^{2+} \rightarrow \text{Sn}) = -0.140 \, \text{V} \)[/tex]
[tex]\( E^\circ_\text{red}(\text{Cu}^{2+} \rightarrow \text{Cu}) = +0.337 \, \text{V} \)[/tex]
Since the reduction potential for [tex]\( \text{Cu}^{2+} \)[/tex] is more positive, it acts as the cathode:
[tex]\[ E^\circ_\text{cathode} = +0.337 \, \text{V} \][/tex]
And the reduction potential for [tex]\( \text{Sn}^{2+} \)[/tex] is less positive (more negative), so it acts as the anode:
[tex]\[ E^\circ_\text{anode} = -0.140 \, \text{V} \][/tex]
Now, calculate the standard cell potential:
[tex]\[ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \][/tex]
[tex]\[ E^\circ_\text{cell} = (+0.337 \, \text{V}) - (-0.140 \, \text{V}) \][/tex]
[tex]\[ E^\circ_\text{cell} = +0.337 \, \text{V} + 0.140 \, \text{V} \][/tex]
[tex]\[ E^\circ_\text{cell} = +0.477 \, \text{V} \][/tex]