Answer:
b. reactant molecules collide more frequently and with greater energy per collision
Explanation:
As the temperature of a reaction is increased, the rate of the reaction increases because the reactant molecules collide more frequently and with greater energy per collision. When temperature is increased there is an increase in the kinetic energy of the molecules. The more the molecules move the more they collide. According to the collision theory there should be enough energy to allow bonds to be formed during a chemical reaction hence the need for greater energy per collision. Also the rate of reaction is directly proportional to the number of collisions that occur.
What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.163 M in sodium chloride and 0.019 M in magnesium sulfate? Assume complete dissociation for electrolytes.
Answer:
The minimum pressure should be 901.79 kPa
Explanation:
Step 1: Data given
Temperature = 25°C
Molarity of sodium chloride = 0.163 M
Molarity of magnesium sulfate = 0.019 M
Step 2: Calculate osmotic pressure
The formula for the osmotic pressure =
Π=MRT.
⇒ with M = the total molarity of all of the particles in the solution.
⇒ R = gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 25 °C = 298 K
NaCl→ Na+ + Cl-
MgSO4 → Mg^2+ + SO4^2-
M = 2(0.163) + 2(0.019 M)
M = 0.364 M
Π = (0.364 M)(0.08206 atm-L/mol-K)(25 + 273 K)
Π = 8.90 atm
(8.90 atm)(101.325 kPa/atm) = 901.79 kPa
The minimum pressure should be 901.79 kPa
Oxygen gas is compressed in a piston–cylinder device from an initial state of 0.8 m3 /kg and 25°C to a final state of 0.1 m3 /kg and 287°C. Determine the entropy change of the oxygen during this process. Assume constant specific heats.
Answer:
ΔS = -0.1076 kJ /kg*K
Explanation:
Step 1: Data given
Initial state = 0.8 m³/kg and 25 °C = 298.15 K
Final state = 0. 3³/kg and 287 °C = 560.15 K
Cv = 0.686 kJ/kg*K
Step 2: Calculate the average temperature
The average temperature = (25°C + 287 °C)/2 =156 °C ( = 429 K)
Step 3: Calculate the ΔS
ΔS =(Cv, average) * ln(T2/T1) + R*ln(V2/V1)
ΔS = 0.686 * ln(560.15/298.15) + 0.2598*ln( 0.1/0.8)
ΔS = -0.1076 kJ /kg*K
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.3 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN−. Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution?
Answer:
The concentration of Cd2+ is 0.0175 M
The concentration of Mn2+ is 0.0305 M
Explanation:
Step 1: Data given
A 50.0 mL sample contains Cd2+ and Mn2+
volume of 0.05 M EDTA = 56.3 mL
Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.
Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+
Step 2: Calculate mole ratio
The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+ in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed in the back titration with Ca2+:
Step 3: Calculate total mol of EDTA
Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA
Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA
Step 4: Calculate total moles of CD2+ and Mn2+
So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol
Step 5: Calculate remaining moles of Cd2+
The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction with cyanide:
Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.
Step 6: Calculate remaining moles of Mn2+
The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+
Step 7: Calculate initial concentrations
The initial concentrations must have been:
(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+
(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+
The concentration of Cd2+ is 0.0175 M
The concentration of Mn2+ is 0.0305 M
Automobile airbags contain solid sodium azide, NaN 3 , that reacts to produce nitrogen gas when heated, thus inflating the bag. 2 NaN 3 ( s ) ⟶ 2 Na ( s ) + 3 N 2 ( g ) Calculate the value of work, w , for the system if 16.5 g NaN 3 reacts completely at 1.00 atm and 22 ∘ C.
Answer: The value of work for the system is -935.23 J
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of sodium azide = 16.5 g
Molar mass of sodium azide = 65 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of sodium azide}=\frac{16.5g}{65g/mol}=0.254mol[/tex]
The given chemical equation follows:
[tex]2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)[/tex]
By Stoichiometry of the reaction:
2 moles of sodium azide produces 3 moles of nitrogen gas
So, 0.254 moles of sodium azide will produce = [tex]\frac{3}{2}\times 0.254=0.381mol[/tex] of nitrogen gas
To calculate volume of the gas given, we use the equation given by ideal gas, which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 1.00 atm
V = Volume of the gas = ?
T = Temperature of the gas = [tex]22^oC=[22+273]K=295K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of nitrogen gas = 0.381 moles
Putting values in above equation, we get:
[tex]1.00atm\times V=0.381mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 295K\\\\V=\frac{0.381\times 0.0821\times 295}{1.00}=9.23L[/tex]
To calculate the work done for expansion, we use the equation:
[tex]W=-P\Delta V[/tex]
We are given:
P = pressure of the system = [tex]1atm=1.01325\times 10^5Pa[/tex] (Conversion factor: 1 atm = 101325 Pa)
[tex]\Delta V[/tex] = change in volume = [tex]9.23L=9.23\times 10^{-3}m^3[/tex] (Conversion factor: [tex]1m^3=1000L[/tex] )
Putting values in above equation, we get:
[tex]W=-1.01325\times 10^5Pa\times 9.23\times 10^{-3}m^3\\\\W=-935.23J[/tex]
Hence, the value of work for the system is -935.23 J
The work done is -919 J.
The equation of the reaction is;
2 NaN 3 ( s ) ⟶ 2 Na ( s ) + 3 N 2 ( g )
Number of moles of NaN3 = 16.5 g/65 g/mol = 0.25 moles
If 2 moles of NaNa3 yields 3 moles of N2
0.25 moles of NaN3 yields 0.25 moles × 3 moles/2 moles
= 0.375 moles of N2
We need to find the volume change using;
PV = nRT
P = 1.00 atm
V = ?
n = 0.375 moles of N2
R = 0.082 atmLK-1mol-1
T = 22 ∘ C + 273 = 295 K
V = nRT/P
V = 0.375 × 0.082 × 295/ 1.00
V = 9.07 L
Recall that during expansion the gas does work. Work done by the gas is;
W = -PΔV
W =-( 1 atm × 9.07 L)
W = -9 atmL
Again;
1 L atm = 101.325 J
So,
-9 atmL = -9 atmL × 101.325 J/1 L atm
= -919 J
The work done is -919 J.
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Using the following data, determine the standard cell potential E^o cell for the electrochemical cell constructed using the following reaction, where zinc is the anode and lead is the cathode.
Zn(s) + Pb2+(aq) -> Zn2+(aq) + Pb(s)
Half-reaction: Standard Reduction Potential:
Zn2+(aq) + 2e- -> Zn(s)= -0.763
Pb2+(aq) + 2e- -> Pb(s)= -0.126
a. -0.889 V
b. +0.889 V
c. +0.637 V
d. +1.274 V
e. -0.637 V
The standard cell potential is determined by subtracting the anode potential from the cathode potential, resulting in a standard cell potential of +0.637 V.
The standard cell potential Eo cell can be determined by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode.
Given:
Zn: -0.763 V
Pb: -0.126 V
Calculation: (-0.126) - (-0.763) = +0.637 V
Therefore, the correct answer is c. +0.637 V.
Hemiacetals are _____. A. There is more than one correct answer choice listed here. B. partial acetals that can be converted by a reduction reaction C. semiacetals produced by alcohol subtraction from a ketone D. unstable compounds that convert to acetals in the presence of an alcohol and an acid.
Hemiacetals are unstable compounds that can convert to acetals when subjected to an alcohol and acid. They form from reactions involving aldehydes, ketones, and alcohols. Their transformation to full acetals can also be manipulated under acidic conditions.
Explanation:Hemiacetals are unstable compounds that can convert to acetals in the presence of an alcohol and an acid. They are also considered as partial acetals and mainly form through the reaction of aldehydes or ketones with an alcohol. The formation of a hemiacetal can be represented by the equation: R2C=O + R'OH → R2C(OH)OR'.
Essentially, hemiacetals are semiacetals produced by alcohol subtraction from a ketone or the addition of an alcohol to an aldehyde. This process often takes place under acidic conditions. Furthermore, they can be converted into a full acetal by additional alcohol in the presence of an acid.
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Hemiacetals are unstable compounds that convert to acetals in the presence of an alcohol and an acid. They are formed when an alcohol reacts with an aldehyde or ketone, resulting in the addition of an -OH group to the carbon atom attached to the carbonyl group.
Explanation:Hemiacetals are unstable compounds that convert to acetals in the presence of an alcohol and an acid. They are formed when an alcohol reacts with an aldehyde or ketone, resulting in the addition of an -OH group to the carbon atom attached to the carbonyl group. The remaining oxygen atom in the hemiacetal is attached to a hydrogen atom. Hemiacetals can further react to form full acetals by adding an alcohol molecule to replace the hydrogen atom, resulting in the formation of a new carbon-oxygen bond.
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The boiling points for a set of compounds in a homologous series can be qualitatively predicted using intermolecular force strengths. Using their condensed structural formulas, rank the homologous series for a set of alkanes by their boiling point.Rank from highest to lowest boiling point. To rank items as equivalent, overlap them.a)CH3CH2CH2CH3 b) CH3CH2CH2CH2CH2CH3CH3c) CH3CH2CCH2CH3CH3d)CH3CH2CH2CH2CH2CH2CH3answer is D, B , C, A
Answer:
Heptane (d) > Hexane (b) 3,3 dimethyl pentane (c) > butane (a)
Explanation:
The boiling points of the alkanes are depends on the following factors:
Hydrocarbons with strong intermolecular forces have high boiling points.Longer carbon chain molecules have high boiling points.Straight chain hydrocarbon molecules have high boiling points than the branched chain hydrocarbons molecules because straight chain molecule have large surface area.The given molecules are as follows.
a) Butane
b) Hexane
c) 3,3 dimethyl pentane
d) Heptane.
The Highest to lowest boiling point order is as follows.
Heptane (d) > Hexane (b) 3,3 dimethyl pentane (c) > butane (a)
Given only the following data, what can be said about the following reaction?3H2(g) + N2(g)---> 2NH3(g) ΔH=-92kJA.) The enthalpy of products is greater than the enthalpy of reactantsB.) The total bond energies of products are greater than the total bond energies of reactantsC.) The reaction is very fastD.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia
Answer:
D.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.
Explanation:
For the reaction:
3H₂(g) + N₂(g) → 2NH₃(g)
The enthalpy change is ΔH = -92kJ
This enthalpy change is defined as the enthalpy of products - the enthalpy of reactants. As the enthalpy is <0, The enthalpy of products is lower than the enthalpy of reactants.
Also, it is possible to obtain the enthalpy change from the bond energies of products - bond energies of reactants, thus, The total bond energies of products are lower than the total bond energies of reactants.
The rate of the reaction couldn't be determined using ΔH.
As the bond energy of ammonia is lower than bonds of nitrogen and hydrogen, D. Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.
I hope it helps!
Final answer:
The correct statement is that the enthalpy of products is less than the enthalpy of reactants, indicating that heat is released and bond energies are lower in the products than in the reactants. Reaction speed and relative bond stability cannot be inferred from the given data.
Explanation:
From the data provided, several conclusions about the chemical reaction forming ammonia can be drawn. It is evident that the reaction is exothermic since the enthalpy change (ΔH) is negative, meaning that heat is released when hydrogen and nitrogen react to form ammonia. Specifically, ΔH is -92kJ, which indicates that the enthalpy of the reactants is higher than the enthalpy of the products, and thus, the correct statement is:
A.) The enthalpy of products is less than the enthalpy of reactants because energy is released in the formation of ammonia from nitrogen and hydrogen.
Moreover, a negative ΔH suggests that the total bond energies of products are less than that of reactants because forming stronger bonds in the products releases energy. This also aligns with the principle that an exothermic reaction strengthens product bonds compared to reactant bonds, as seen in the enthalpy change provided.
However, the claim that the reaction is very fast (C.) cannot be substantiated by the given data. In fact, the synthesis of ammonia from nitrogen and hydrogen is known to be slow under ambient conditions and requires specific industrial processes, such as the Haber process, to enhance its rate. As for the last option (D.), while nitrogen and hydrogen do form very stable bonds, the statement that they are more stable compared to the bonds in ammonia is not specified by the given ΔH value and requires additional bond energy data for a direct comparison.
For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [Co3+]= 0.765 M , [Co2+]= 0.369 M , and [Cl−]= 0.486 M , and the pressure of Cl2 is PCl2= 9.30 atm ? Express your answer with the appropriate units.
Answer:
Ecell = 0.500 V
Explanation:
For a chemical cell, the cell potential (Ecell) is:
Ecell = E° - (0.0592/n)*logQ (Nernst equation)
Where n is the number of electrons involved in the redox reaction, and Q is the reaction coefficient.
Let's see a half-reaction:
2Co³⁺(aq) → 2Co²⁺(aq)
The charge goes from +6 (2*3) to +4 (2*2), so n = 6 - 4 = 2 electrons.
The reaction coefficient is the multiplication of the concentration of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients:
Q = ([Cl₂]*[Co⁺²]²)/([Co⁺³]²*[Cl⁻]²)
The concentration of Cl₂ is the number of moles (n) divided by the volume(V), and can be calculated by the ideal gas law:
PV = nRT
n/V = P/RT (P is the pressure, R is the gas constant, and T is the temperature)
P = 9.30 atm, R = 0.082 atm.L/mol.K, T = 25°C + 273 = 298 K
n/V = [Cl₂] = 9.30/(0.082*298)
[Cl₂] = 0.381 M
Q = [0.381*(0.369)²]/[(0.765)²*(0.486)²]
Q = 0.051877/0.138228
Q = 0.3753
Ecell = 0.483 - (0.0592/2)*log(0.3753)
Ecell = 0.483 - 0.0296*(-0.4256)
Ecell = 0.500 V
The cell potential of the electrochemical cell with the cobalt and chlorine electrodes has been 0.5 V.
What is an electrochemical cell?The electrochemical cell has been the cell in which the chemical and electrical energy is converted by the charge transfer.
The cell potential ([tex]E_{cell}[/tex]) can be given as:
[tex]E^\cell=E^\circ-\dfrac{0.059}{n}\;\times\;log\; Q[/tex]
The standard cell potential ([tex]E^\circ[/tex]) of the cell has been:
[tex]E^\circ=0.483\;\text V[/tex]
The charge transferred in the reaction ([tex]n[/tex]) can be given as 2.
The concentration of chlorine has been:
[tex]\rm M=\dfrac{P}{RT} \\\\Cl_2=\dfrac{9.30}{0.0821\;\times\;298}\\\\ Cl_2=0.981\;M[/tex]
The value of Q is given as:
[tex]\rm Q=\dfrac{[Cl_2][CO_2]^2}{[CO_3]^2[Cl]^2}\\\\ Q=\dfrac{(0.981)(0.369)^2}{(0.765)^2(0.486)^2} \\\\Q=0.3753[/tex]
Substituting the values for the cell potential:
[tex]E_{cell}=0.483-\dfrac{0.059}{2}\;\times\;\text {log}\;0.3753 \\E_{cell}=0.5\;\rm V[/tex]
The cell potential is 0.5 V.
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According to the bottle, a serving size of Gatorade is 240 mL. The molar mass of Red Dye #40 is 496.43 g/mol. Use this information along with the concentration of Red Dye #40 in Gatorade to calculate the mass (in mg) of Red Dye #40 ingested per serving of Gatorade.
Answer:
mass Red Dye 40 = 0.0696 mg
Explanation:
aas found by means of quantitative methods of analysis of dyes in sports drinks:
⇒ [Red Dze 40] = 5.842 E-7 mol/L
analysis performed at a wavelength of 504 nm; value that is below the allowable limits for these drinks.
∴ mass of Red Dye ingested:
⇒ mass Red Dye = (5.842 E-7 mol/L)(496.43 g/mol)(1000 mg/g)(0.240 L)
⇒ mass Red Dye = 0.0696 mg
The mass of Red Dye #40 consumed in a serving of Gatorade can be calculated using the formula m = mol solute/kg solvent, but we need the concentration of the dye in Gatorade (Molarity) to get the number of moles. Once we have that, we can calculate the mass using the molar mass provided. However, unfortunately, the question does not provide the concentration of Red Dye #40 in Gatorade, therefore, a precise answer can't be given.
Explanation:Unfortunately, it's not possible to answer this question without knowing the concentration of Red Dye #40 in Gatorade since that information is not provided. To calculate the mass of Red Dye #40 ingested per serving, you would need to use the following formula: m = mol solute/kg solvent. This equation means that the mass (m) in milligrams (mg) of Red Dye #40, equals the number of moles of Red Dye #40 per serving (mol solute), divided by the mass of the Gatorade serving (kg solvent).
To calculate the moles of Red Dye #40, you would need the concentration of it in Gatorade (Molarity, defined as moles per liter). Once you have the number of moles, you can directly calculate the mass of dye using the molar mass given. Thus, the mass (m) in grams (g), will be equals to the number of moles times the molar mass of Red Dye #40 (g/mol).
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A student researcher performed a chromatographic separation of caffeine and aspartame. The retention time for caffeine, t c , was found to be 216.7 s with a baseline peak width, w c , of 13.6 s. The retention time for aspartame, t a , was 267.1 s with a baseline peak width, w a , of 18.9 s. The retention time for the unretained solvent methanol was 45.2 s.
Calculate the average plate height, H , in micrometers for this separation, given that it was performed on a 24.1 cm long column.
H = μ m
Calculate the resolution, R , for this separation using the widths of the peaks.
R =
Calculate the resolution if the number of theoretical plates were to increase by a factor of 1.5 .
R 1.5 N =
To calculate the average plate height and resolution for a chromatographic separation, we need to find the number of theoretical plates and use specific formulas.
Explanation:To calculate the average plate height, H, we can use the formula: H = L/N, where L is the length of the column and N is the number of theoretical plates. Given that the column length is 24.1 cm, we first need to find the number of theoretical plates.
The number of theoretical plates can be calculated using the formula: N = (t_r / w)^2, where t_r is the retention time and w is the peak width at the base. For caffeine, N_c = (216.7 s / 13.6 s)^2 and for aspartame, N_a = (267.1 s / 18.9 s)^2.
After finding the values for N_c and N_a, we can calculate the average plate height using H = L / (N_c + N_a). Finally, to calculate the resolution, R, we can use the formula: R = 1.18 * (t_a - t_c) / (w_a + w_c).
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The answers are: 1). The average plate height (H) for the separation is 46.63 micrometers. 2). The resolution between caffeine and aspartame is 3.10. 3). The new resolution is 3.80.
1). To calculate the average plate height (H) in micrometers (µm) for this separation, we first determine the number of theoretical plates (N) using the formula:
[tex]\[ N = 16 \left( \frac{t_r}{w} \right)^2 \][/tex]For caffeine:
[tex]\[ N_c = 16 \left( \frac{216.7}{13.6} \right)^2 = 4010.66 \][/tex]For aspartame:
[tex]\[ N_a = 16 \left( \frac{267.1}{18.9} \right)^2 = 6323.53 \][/tex]The average number of theoretical plates [tex](N_{avg})[/tex] is:
[tex]\[ N_{avg} = \frac{4010.66 + 6323.53}{2} = 5167.1 \][/tex]The column length (L) is 24.1 cm (or 241,000 µm). Therefore, the plate height (H) is:
[tex]\[ H = \frac{L}{N_{avg}} = \frac{241,000 \, \mu m}{5167.1} = 46.63 \, \mu m \][/tex]2). Resolution (R):
The resolution (R) between caffeine and aspartame can be calculated using the formula:
[tex]\[ R = 2 \left( \frac{t_a - t_c}{w_a + w_c} \right) \][/tex]Substituting the given values:
[tex]\[ R = 2 \left( \frac{267.1 - 216.7}{18.9 + 13.6} \right) = 2 \left( \frac{50.4}{32.5} \right) = 3.10 \][/tex]3). Resolution with Increased Number of Plates:
If the number of theoretical plates increases by a factor of 1.5, the new average number of plates [tex](N_{new})[/tex] is:
[tex]\[ N_{new} = 5167.1 \times 1.5 = 7750.65 \][/tex]The new resolution [tex](R_{1.5N})[/tex] using the formula:
[tex]\[ R_{1.5N} = R \times \sqrt{1.5} = 3.10 \times 1.225 = 3.80 \][/tex]Complete Question: -
A student researcher performed a chromatographic separation of caffeine and aspartame. The retention time for caffeine, [tex]\( t_c \)[/tex], was found to be 216.7 s with a baseline peak width, [tex]\( w_c \)[/tex], of 13.6 s. The retention time for aspartame, [tex]\( t_a \)[/tex], was 267.1 s with a baseline peak width, [tex]\( w_a \)[/tex], of 18.9 s. The retention time for the unretained solvent methanol was 45.2 s.
1. Calculate the average plate height, H, in micrometers for this separation, given that it was performed on a 24.1 cm long column. [tex]\[ H = \, \mu m \][/tex]
2. Calculate the resolution, R, for this separation using the widths of the peaks. R = ?
3. Calculate the resolution if the number of theoretical plates were to increase by a factor of 1.5. [tex]\[ R_{1.5N} = \]?[/tex]
At elevated temperatures ammonium carbamate, NH2COONH4, is in equilibrium with NH3 and CO2 according to the equation:
NH2COONH4(s) --> 2 NH3(g) + CO2(g)
What is the equilibrium expression for this reaction?
K = 2[NH3] [CO2] / [NH2COONH4]
K = 2[NH3] [CO2]
K = [NH3]^2 [CO2]
K = [NH3]^2 [CO2] / [NH2COONH4]
Final answer:
The correct equilibrium expression for the reaction NH2COONH4(s) --> 2 NH3(g) + CO2(g) is Keq = [NH3]^2 [CO2], excluding the concentration of the solid ammonium carbamate. Option C
Explanation:
The correct equilibrium expression for the decomposition of ammonium carbamate into ammonia and carbon dioxide is based on the coefficients of the chemical reaction.
Since the ammonium carbamate is a solid, it does not appear in the equilibrium expression. Ammonia is produced in twice the amount of carbon dioxide, and the equilibrium constant expression reflects this stoichiometry.
The correct expression is:
Keq = [NH3]^2 [CO2]
This expression uses the concentrations of the gases because they are in the gaseous state and their amounts can change during the reaction. The solid ammonium carbamate does not appear in the expression, as its concentration is constant. Option C
Final answer:
The equilibrium expression for the decomposition of NH₄COONH₂(s) is K = [NH₃]^2[CO₂], representing the concentrations of gaseous products raised to the power of their stoichiometric coefficients. The correct answer is C) K = [NH₃]^2[CO₂]
Explanation:
The equilibrium expression for the decomposition of ammonium carbamate, NH₄COONH₂, into ammonia (NH₃) and carbon dioxide (CO₂) is based on the equilibrium constant (K).
In the balanced chemical equation NH₄COONH₂(s) → 2 NH₃(g) + CO₂(g), ammonium carbamate is a solid and thus its concentration does not appear in the expression because the concentration of solids and liquids are not included in K expressions.
We generate the equilibrium expression by raising the gaseous products' concentrations to the power of their stoichiometric coefficients in the balanced equation and then multiplying them together.
Therefore, the correct equilibrium expression is K = [NH₃]^2[CO₂] which includes the square of the ammonia concentration because two moles of NH₃ are produced per mole of NH₄COONH₂ that decomposes.
Balance the following redox equations by the half-reaction method:
(a) Mn^2 + H2O2 ---> MnO2 + H2O (in basic solution)
(b) Bi(OH)3 + SnO2^2- -----> SnO3^2- + Bi (in basic solution)
(c) Cr2O7^2- + C2O4^2-----> Cr^3+ + CO2 (in acidic solution)
(d) ClO^3- + Cl^− -----> Cl^2 + ClO^2 (in acidic solution)
(e) Mn^2 + BiO^3− ------> Bi^3 + MnO^4− (in acidic solution)
Answer:
a) Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O
b) 2Bi(OH)3 + 3SnO2^2- → 2Bi + 3SnO3^2- + 3H2O
c) Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2
d) 2ClO3- + 2Cl- + 4H+ → Cl2 + 2H2O +2ClO2
e) 5 BiO3^- + 14 H+ + 2Mn^2+ → 5Bi^3+ + 7H2O + 2MnO4^-
Explanation:
(a) Mn2 + H2O2 → MnO2 + H2O (in basic solution)
Step 1: The half reactions
Oxidation: Mn2+ + 4OH- → MnO2 + 2H2O + 2e-
Reduction: H2O2 + 2e- + 2H2O → 2H2O + 2OH-
Step 2: Sum of both half reactions
Mn2+ + 4OH- + H2O2 → MnO2 + 2H2O + 2OH-
Step 3: the netto reaction
Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O
(b) Bi(OH)3 + SnO2^2- → SnO3^2- + Bi (in basic solution)
Step 1: The half reactions
Reduction: Bi(OH)3 + 3e- → Bi
Oxidation : Sno2^2- → SnO3^2- +2e-
Step 2: Balance the half reactions
2* (Bi(OH)3 + 3e- → Bi + 3OH-)
3* (Sno2^2- +2OH- → SnO3^2- +2e- + H2O)
Step 3: The netto reaction
2Bi(OH)3 + 3SnO2^2- → 2Bi + 3SnO3^2- + 3H2O
(c) Cr2O7^2- + C2O4^2- → Cr^3+ + CO2 (in acidic solution)
Step 1: The half reactions
Reduction: Cr2O7^2- + 6e- → 2Cr+
Oxidation : C2O4^2- → 2CO2 + 2e-
Step 2: Balance the half reactions
Cr2O7^2- + 6e- +14H+ → 2Cr+ +7H2O
3*(C2O4^2- → 2CO2 + 2e-)
Step 3: The netto reaction
Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2
(d) ClO3^- + Cl^− → Cl^2 + ClO^2 (in acidic solution)
Step 1: The half reactions
Reduction: 2 ClO3^- + 10e- → Cl2
ClO3^- + e- → ClO2
2 Cl- + 2ClO3^- +8e- →2Cl2
Oxidation: 2Cl- → Cl2 + 2e-
Cl- → ClO2 + 5e-
Cl- +ClO3^- → 2ClO2 + 4e-
Step 2: Balance the reactions
2Cl- + 2ClO3^- + 8e- + 12H+ → 2Cl2 + 6H2O
2* (Cl- + ClO3^- + H2O → 2ClO2 + 4e- + 2 H+)
Step 3: The netto reaction
2ClO3^- + 2Cl- + 4H+ → Cl2 + 2H2O +2ClO2
(e) Mn^2 + BiO3^− → Bi^3 + MnO^4− (in acidic solution)
Step 1: The half reactions
Reduction: BiO3^- + 2e- → Bi^3+
Oxidation : Mn^2+ → MnO4^- +5e-
Step 2: Balanced the reactions
5* ( BiO3^- + 2e- + 6H+ → Bi^3+ + 3H2O)
2* ( Mn^2+ + 4H2O →MnO4^- + 5e- + 8H+)
Step 3: The netto reaction
5 BiO3^- + 14 H+ + 2Mn^2+ → 5Bi^3+ + 7H2O + 2MnO4^-
The given redox reactions are balanced through the half-reaction method, dividing each into oxidation and reduction steps. Hydrogen and oxygen balances are maintained through the addition of hydrogen ions (H^+) or hydroxide ions (OH^-) and water (H2O). Electrons are then added to balance charges.
Explanation:Balance of Redox ReactionsTo balance redox equations, we'll take each reaction and divide it into half-reactions, one for oxidation and one for reduction, before balancing these separately.
(a) Mn^2+ + H2O2 -> MnO2 + H2O
Oxidation: 2Mn^2+ -> MnO2 + 4e-Reduction: H2O2 + 2e- -> 2H2O(b) Bi(OH)3 + SnO2^2- -> SnO3^2- + Bi
Oxidation: 2Bi(OH)3 -> 2Bi + 6OH- + 6e-Reduction: SnO2^2- + 2e- -> SnO3^2-(c) Cr2O7^2- + C2O4^2- -> Cr^3+ + CO2
Oxidation: C2O4^2- -> 2CO2 + 2e-Reduction: Cr2O7^2- + 6e- + 14H+ -> 2Cr^3+ + 7H2O(d) ClO^3- + Cl^− -> Cl^2 + ClO^2
Oxidation: 2Cl^− -> Cl2 + 2e-Reduction: ClO3^- + 2e- -> ClO2 + O2(e) Mn^2+ + BiO3^− -> Bi^3+ + MnO4^−
Oxidation: Mn^2+ -> MnO4^- + 8H^+ + 5e-Reduction: BiO3^- + 6H^+ + 3e- -> Bi^3+ + 3H2OLearn more about Redox Reactions here:https://brainly.com/question/38346693
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A 24.201 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 72.053 g of water. A 13.774 g aliquot of this solution is then titrated with 0.1045 M HCl. It required 28.02 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.
Answer:
1.44%
Explanation:
Initially, there is 24.201g of the waste. Let this waste contain 'x' mols of ammonia([tex]NH_{3}[/tex]). Now, this waste is dissolved in 72.053g of water, thus making the weight of the solution to 24.201+72.053 = 96.254g. Now, from this, 13.774g is taken. Note that, when the aqueous waste is dissolved in water, the ammonia particles are uniformly distributed, i.e, the 'x' mols of ammonia is uniformly present in the 96.254g of the solution. Hence, when we take 13.774g of the solution, only a fraction of the ammonia particles is taken. This fraction is equal to [tex]\frac{13.774}{96.254}[/tex] of x, which is equal to 0.1431 times 'x'.
For the HCl solution, 28.02mL of 0.1045M solution contains 28.02x0.1045 = 2.9281mmols of HCl is present in it.
The basic titration reaction that occurs is : [tex]HCl + NH_{3}[/tex]→[tex]NH_{4}Cl[/tex] , i.e, one mol of ammonia requires one mol of HCl for neutralization. Therefore, for the above solution of HCl containing 2.9281mmols, same amount of ammonia, i.e, 2.9281mmols is required for complete neutralization.
Therefore, 0.1431x = 2.9281 mmols ⇒ x = 20.4619 mmols.
The molecular weight of ammonia is 17g/mol. Therefore, the weight of 20.4619mmols of ammonia has a weight(w) = 20.4619 x [tex]10^{-3}[/tex] x 17 = 0.3478 g.
Therefore the weight of ammonia in the initial aqueous waste is 0.3478 g. The total weight of the waste is 24.201g, hence, the percent weight of ammonia is given by [tex]\frac{0.3478}{24.201}[/tex]×100 = 1.44%
Final answer:
To calculate the weight percent of ammonia in the waste sample, the moles of HCl used in titration are converted to mass of NH3. This value is then adjusted for dilution and related to the original mass of the waste to find the weight percent of NH3, which is calculated to be 1.442%.
Explanation:
To calculate the weight percent of ammonia (NH3) in the aqueous waste sample from the fertilizer manufacturer, we follow a series of steps involving dilution and titration. The aliquot of the solution (13.774 g) titrated with 0.1045 M HCl required 28.02 mL to reach the endpoint, which indicates the amount of ammonia that was present in the aliquot.
We first need to calculate the number of moles of HCl that reacted with NH3 using the molarity of HCl and the volume used in the titration:
Moles of HCl = Molarity of HCl imes Volume of HCl in litersMoles of HCl = 0.1045 mol/L imes 0.02802 L = 0.00293 molSince each mole of NH3 reacts with one mole of HCl, the moles of NH3 in the aliquot will also be 0.00293 mol. To find the mass of NH3, we multiply the moles by the molar mass of NH3 (17.031 g/mol).
Mass of NH3 = moles of NH3 imes molar mass of NH3Mass of NH3 = 0.00293 mol imes 17.031 g/mol = 0.04988 gTo calculate the weight percent of NH3 in the original waste sample, we need to adjust the mass of NH3 found for the dilution and relate that to the original mass of the waste sample:
Total mass of the solution = mass of waste sample + mass of water addedTotal mass of the solution = 24.201 g + 72.053 g = 96.254 gWeight percent NH3 = (Mass of NH3 in original sample / mass of waste sample) imes 100%Since only a portion of the diluted sample was titrated, we calculate the proportion of the total solution that was titrated.Proportion of solution titrated = Mass of aliquot / Total mass of the solutionProportion of solution titrated = 13.774 g / 96.254 gMass of NH3 in original sample = Mass of NH3 found from titration / Proportion of solution titratedMass of NH3 in original sample = 0.04988 g / (13.774 g / 96.254 g)Mass of NH3 in original sample = 0.3489 gWeight percent NH3 = (0.3489 g / 24.201 g) imes 100% = 1.442%Therefore, the weight percent of ammonia in the initial aqueous waste sample is 1.442%.
An open flask sitting in a lab refrigerator looks empty, but it is actually filled with a mixture of gases called air. If the flask volume is 3.50 L, and the air is at standard temperature and pressure, how many gaseous molecules does the flask contain?
Answer:
9.39 × 10²² molecules
Explanation:
We can find the moles of gases (n) using the ideal gas equation.
P . V = n . R . T
where,
P is the pressure (standard pressure = 1 atm)
V is the volume
R is the ideal gas constant
T is the absolute temperature (standard temperature = 273.15 K)
[tex]n=\frac{P.V}{R.T} =\frac{1atm.3.50L}{(0.08206atm.L/mol.K).273.15K} =0.156mol[/tex]
There are 6.02 × 10²³ molecules in 1 mol (Avogadro's number). Then,
[tex]0.156mol.\frac{6.02\times10^{23}molecules}{mol} =9.39\times10^{22}molecules[/tex]
Final answer:
The number of gaseous molecules in the flask can be determined using Avogadro's number and the ideal gas law. For a 3.50 L flask at standard temperature and pressure, the number of molecules is approximately 2.40 x 10^24.
Explanation:
The number of gaseous molecules in the flask can be determined using Avogadro's number and the ideal gas law. Avogadro's number (6.022 x 10^23) represents the number of molecules in one mole of any substance. The ideal gas law equation (PV = nRT) relates the pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T) of a gas.
Since the air in the flask is at standard temperature and pressure, we can assume the temperature is 273 K and the pressure is 1 atm. Plugging in these values, along with the flask volume, into the ideal gas law equation allows us to calculate the number of moles of air in the flask. The moles can then be converted to the number of molecules by multiplying by Avogadro's number.
For a flask volume of 3.50 L, the number of gaseous molecules is approximately 2.40 x 10^24.
Sulfur hexafluoride, a dense gas, is held in two separate containers in a storage room at an atmospheric pressure of 755 mmHg and 20.3 °C. The volume of container 1 is 2.09 L, and it contains 7.61 mol of the gas. The volume of container 2 is 4.46 L. Determine the moles of F atoms in container 2 and the density of the gas at the conditions in the room
Explanation:
As it is given that both the given containers are at same temperature and pressure, therefore they have the same density.
So, mass of [tex]SF_{6}[/tex] in container- 1 is as follows.
5.35 mol x molar mass of [tex]SF_{6}[/tex]
= 7.61 mol x 146.06 g/mol
= 1111.52 g
Therefore, density of [tex]SF_{6}[/tex] will be calculated as follows.
Density = [tex]\frac{mass}{volume}[/tex]
density = [tex]\frac{1111.52 g}{2.09 L \times 1000 ml/L}[/tex]
= 0.532 g/mL
Now, mass of [tex]SF_{6}[/tex] in container- 2 is calculated as follows.
4.46 L x 1000 mL/L x 0.532 g/mL
= 2372.72 g
Hence, calculate the moles of moles [tex]SF_{6}[/tex] present in container 2 as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{2372.72 g}{146.06 g/mol}[/tex]
= 16.24 mol
Since, 1 mol [tex]SF_{6}[/tex] contains 6 moles F atoms .
So, 16.24 mol [tex]SF_{6}[/tex] contains following number of atoms.
= [tex]16.24 mol \times 6[/tex]
= 97.46 mol
Thus, we can conclude that moles of F atoms in container 2 are 97.46 mol.
A sample of 0.300 mg pure chromium was added to excess hydrochloric acid to form a 10.0 mL aqueous solution of a chromium (III) salt, which has a violet hue. Exactly 1.00 mL of the resulting solution was analyzed using a spectrophotometer in a 1.00-cm cell at 575 nm, and the percent transmittance for the solution was 62.5%. What is the extinction coefficient?
extinction coefficient (ε) = 347 L·mol⁻¹·cm⁻¹
Explanation:
The chemical reaction between chromium (Cr) and hydrochloric acid (HCl):
2 Cr + 6 HCl → 2 CrCl₃ + 3 H₂
number of moles = mass / molar weight
number of moles of Cr = 0.3 × 10⁻³ (g) / 52 (g/mole)
number of moles of Cr = 5.77 × 10⁻⁶ moles
From the chemical reaction we see that 2 moles of Cr will produce 2 moles of CrCl₃ so 5.77 × 10⁻⁶ moles of Cr will produce 5.77 × 10⁻⁶ moles of CrCl₃.
molar concentration = number of moles / volume (L)
molar concentration of CrCl₃ = 5.77 × 10⁻⁶ / 10 × 10⁻³
molar concentration of CrCl₃ = 5.77 × 10⁻⁴ moles / L
Now we need to transform percent transmittance (%T) in absorbance (A) using the following formula:
A = 2 - log (%T)
A = 2 - log (62.5)
A = 2 - 1.8
A = 0.2
We know that absorbance (A) is defined in respect with extinction coefficient (ε), cell length (l) and concentration (c):
A = εlc
ε = A / lc
ε = 0.2 / (1 × 5.77 × 10⁻⁴)
ε = 0.0347 × 10⁴
ε = 347 L·mol⁻¹·cm⁻¹
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The extinction coefficient for a chromium (III) salt solution can be determined using spectrophotometry, if the absorbance, path length, and concentration are known. The formula A = εcl, integrates these variables, but without the concentration of the chromium (III) in the analysed solution, it is impossible to accurately calculate the extinction coefficient.
Explanation:The question pertains to the measurement of the extinction coefficient for a chromium (III) salt solution using spectrophotometry. The percent transmittance provided allows us to calculate the absorption of light by the solution, which is related to the extinction coefficient. Absorbance can be calculated using the formula A = -log10(T), where T represents transmittance.
The formula for the extinction coefficient, ε, is formulated as A = εcl and includes absorbance (A), path length (l), and concentration (c) variables. However, we don't have the concentration of chromium (III) in the 1.00 mL of the solution we're examining. Therefore, we can't provide an accurate answer to this question without that piece of information.
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Calculate the number of Carbon atoms in 1.50g of C6H12O6
Answer:
3.01 ·10↑22
Explanation:
First you want to convert the grams of Glucose to moles of Glucose.
[tex]\frac{1.5 grams of glucose}{180.15588grams/molesglucose} =.008326 Moles of Glucose[/tex]
Next find the formula units of glucose.
.008326Moles of Glucose · 6.022 · 10↑23Forumula Units*Moles↑-1 =
5.01 ·10↑21 Formula Units of Glucose
Now multiply the formula units of glucose by the amount of each element in the molecule.
So for Carbon:
6carbon · 5.01 · 10↑21 = 3.01 · 10↑22
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Answer:
Number of C-atoms = 3.01 *10^22
Explanation:
Step 1: Data given
Mass of glucose = 1.50 grams
Molar mass of Glucose = 180.156 g/mol
Molar mass of carbon = 12.01 g/mol
Step 2: Calculate number of moles glucose
Number of moles glucose = mass glucose / molar mass glucose
Moles glucose = 1.50 grams / 180.156 g/mol
Moles glucose = 0.00833 moles
Step 3: Calculate formula units of glucose
0.00833 moles glucose * 6.02 *10^23 = 5.01 *10^21
Step 4: Calculate number of carbon atoms
Glucose has 6 C'atoms in it's molecular formule
Number of C-atoms = 5.01 *10^21 * 6 = 3.01 *10^22
At 700 K, Kp for the following equilibrium is (5.6 x 10-3) 2HgO(s)--> 2Hg(l) + O2(g) Suppose 51.2 g of mercury(II) oxide is placed in a sealed 3.00-L vessel at 700 K. What is the partial
pressure of oxygen gas at equilibrium? (R = 0.0821 Lxatm/(Kxmol))
A) 0.075 atm
B) 0.0056 atm
C) 4.5 atm
D) 19 atm
E) 2.3 atm
Answer: Option (B) is the correct answer.
Explanation:
According to the given reaction equation, formula to calculate [tex]\Delta n[/tex] is as follows.
[tex]\Delta n[/tex] = coefficients of gaseous products - gaseous reactants
= 1 - 0
= 1
Also we know that,
[tex]K_{p} = K_{c} \times (RT)^{\Delta n}[/tex]
[tex]5.6 \times 10^{-3} = K_{c} \times (0.0821 \times 700)^{1}[/tex]
[tex]K_{c} = 0.097 \times 10^{-3}[/tex]
For the equation, [tex]2HgO(s) \rightarrow 2Hg(l) + O_{2}(g)[/tex]
Activity of solid and liquid = 1
As, [tex]K_{p} = \frac{P^{2}_{Hg} \times P_{O_{2}}}{P^{2}_{HgO}}[/tex]
[tex]5.6 \times 10^{-3} = P_{O_{2}}[/tex]
Hence, [tex]P_{O_{2}}[/tex] = 0.0056 atm
Thus, we can conclude that partial pressure of oxygen gas at equilibrium is 0.0056 atm.
Final answer:
To calculate the partial pressures of the three gases at equilibrium, we can use the equilibrium constant equation. Plugging in the given initial partial pressures into the equation gives a Kp value of 0.0135.
Explanation:
To solve this problem, we can use the equation for calculating the equilibrium constant (Kp):
Kp = ([NO]²)/([N₂O][O₂])
We are given the initial partial pressures of each gas, so we can plug these values into the equation. The initial partial pressures are as follows: [NO] = 0.08 atm, [N₂O] = 0.62 atm, and [O₂] = 0.24 atm. Plugging these values into the equation:
Kp = (0.08²)/(0.62·0.24) = 0.0135
Therefore, the equilibrium constant (Kp) for this reaction at the given temperatures and pressures is 0.0135.
Lets say the reaction H2SO4 + 2KOH -> K2SO4 + 2H2O, is known to proceed by a 20.85% yield via a particular procedure.
How many grams of KOH must be reacted with an excess of H2SO4, in order to collect 67.5g of H2O?
Answer:
There should react 1008.3 grams KOH
Explanation:
Step 1: Data given
yield = 20.85 %
actual yield of H2O = 67.5 grams
Molar mass of H2SO4 = 98.08 g/mol
Molar mass of H2O = 18.02 g/mol
Molar mass of KOH = 56.11 g/mol
Step 2: The balanced equation
H2SO4 + 2KOH ⇒ K2SO4 + 2H2O
Step 3: Calculate moles of H2O
Moles H2O = Mass H2O / Molar mass H2O
Moles H2O = 67.5 grams / 18.02 g/mol
Moles H2O = 3.746 moles
Step 4: Calculate theoretical yield
% yield= (actual yield/theoretical yield)*100%
Theoretical yield= (Actual yield/Percent yield) * 100%
Theoretical yield = (3.746 moles /20.85) *100%
Theoretical yield = 17.97 moles of H2O
Step 5: Calculate moles of KOH
For 2 moles of H2O produced, we need 2 moles of KOH
For 17.97 moles of H2O produced, we need 17.97 moles of KOH
Step 6: Calculate mass of KOH
Mass KOH =moles KOH * Molar mass KOH
Mass KOH = 17.97 moles * 56.11 g/mol
Mass KOH = 1008.3 grams
There should react 1008.3 grams KOH
Ferris & Mona used the ORP sensor to titrate a ferrous ammonium sulfate solution, (NH4)2Fe(SO4)2 with KMnO4 titrant.
They titrated a 15.00 mL aliquot of the Fe+2 solution with 0.0250 M MnO4- solution and determined that the equivalence point was at 20.2 mL.
What is the molarity of the Fe+2 solution?5 Fe+2(aq) + MnO4-(aq) + 8 H+(aq) â 5 Fe+3(aq) + Mn+2(aq) + 4 H2OSelect one:a. 0.168 Mb. 0.0928 Mc. 0.0337 Md. 0.673 M
Answer:
The correct option is: A. 0.168 M
Explanation:
Chemical reaction involved:
5 Fe²⁺ (aq) + MnO₄⁻ (aq) + 8 H⁺ (aq) → 5 Fe³⁺ (aq) + Mn²⁺ (aq) + 4 H₂O
Given: For MnO₄⁻ solution-
Number of moles: n₁ = 1, Volume: V₁ = 20.2 mL, Concentration: M₁ = 0.0250 M;
For Fe²⁺ solution:
Number of moles: n₂ = 5, Volume: V₂ = 15 mL, Concentration: M₂ = ?M
To find out the concentration of Fe²⁺ solution (M₂), we use the equation:
[tex]\frac{M_{1}\times V_{1}}{n_{1}} = \frac{M_{2}\times V_{2}}{n_{2}}[/tex]
[tex]\frac{0.0250 M\times 20.2 mL}{1} = \frac{M_{2}\times 15 mL}{5}[/tex]
[tex]0.505 = M_{2}\times 3[/tex]
[tex]M_{2} = \frac{0.505}{3} = 0.168M[/tex]
Therefore, the concentration or molarity of Fe²⁺ solution: M₂ = 0.168 M
The molarity of the Fe⁺₂ solution is 0.168 M, calculated by taking the stoichiometric ratio into account and using the volume of MnO₄⁻ solution used at the equivalence point.
Explanation:To find the molarity of the Fe⁺₂ solution, we first need to realize that the reaction between Fe⁺₂ and MnO₄⁻ is a 5:1 stoichiometric ratio. This means for every 1 mole of MnO₄⁻, 5 moles of Fe⁺₂ react. Given that the MnO₄⁻ solution is 0.0250 M and it took 20.2 mL to reach the equivalence point, we can convert that to liters (0.0202 L) and use the molarity (mol/L) to find moles. This results in 0.000505 moles of MnO₄⁻. Now, remembering our 5:1 ratio, there would be 0.002525 moles of Fe⁺₂ in our 15.00 mL or 0.01500 L of solution. Therefore, the molarity (M) of our Fe⁺₂ solution is moles (mol) / liters (L) = 0.002525 mol / 0.01500 L = 0.168 M.
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A cylinder and piston assembly (defined as the system) is warmed by an external flame. The contents of the cylinder expand, doing work on the surroundings by pushing the piston outward against the external pressure.
If the system absorbs 563 J of heat and does 498 J of work during the expansion, what is the value of ΔE?
Explanation:
According to the first law of thermodynamics, energy can neither be created nor it can be destroyed. It can only be converted from one form to another.
Formula given by first law of thermodynamics is as follows.
U = q + w
The given values are as follows.
q = +563 J
w = -498 J (as work done by the system is negative)
Hence, putting the given values into the above formula and calculate the internal energy as follows.
U = q + w
= 563 J + (-498 J)
= 65 J
As internal energy is represented by [tex]\Delta U[/tex] or [tex]\Delta E[/tex]. Therefore, the value of [tex]\Delta E[/tex] for the given system is 65 J.
Consider a closed flask containing a liquid and its vapor. Which statement is incorrect?
A.) The vapor exerts a pressure called the vapor pressure
B.) Increasing the temperature of the liquid would lead to a great vapor pressure
C.) Evaporation and condensation will eventually cease after a constant pressure has been attained
D.) Increasing the volume of the container at a constant temperature would cause increased condensation until the pressure of the vapor was once eagain the same as it had been
Answer:
C). Evaporation and condensation will eventually cease after a constant pressure has been attained
Explanation:
A) The vapor exerts a pressure called the vapor pressure is correct statement B) On increasing the temperature more vapor will be formed as thus leading to a higher vapor pressure.
C) This is an incorrect statement, Evaporation and condensation never stops what ever may be the pressure. Although they may achieve an equilibrium that is the rate of evaporation may be equal to the rate of condensation.
d) This is again a correct statement as vapor pressure is an extrinsic property that it depends upon the volume or mass of liquid. Greater the volume greater will be the vapor pressure.
Final answer:
Increasing the temperature of the liquid would lead to a greater vapor pressure. Evaporation and condensation continue until a dynamic equilibrium is reached. Increasing the volume of the container causes increased condensation until the pressure of the vapor returns to its original value.
Explanation:
When the liquid in a closed container is heated, more molecules escape the liquid phase and evaporate. This increase in the number of vapor molecules leads to an increase in pressure, known as the vapor pressure (A). Increasing the temperature of the liquid would actually lead to a greater vapor pressure (B), as more molecules would have enough energy to escape from the liquid phase. Evaporation and condensation processes continue in a closed flask until a constant pressure has been attained, referred to as a dynamic equilibrium (C). Increasing the volume of the container at a constant temperature would cause increased condensation until the pressure of the vapor was once again the same as it had been (D). Therefore, the incorrect statement is option B.
Complete the description of the pentose phosphate pathway by moving the correct term to each blank.
The pentose phosphate pathway is a two‑stage pathway that generates__________ which is a reductant in many biosynthetic reactions and takes part in detoxifying reactive oxygen species, and__________ which is a nucleotide (DNA and RNA) precursor. The substrate for the pentose phosphate pathway is__________.
The__________ phase of the pentose phosphate pathway interconverts phosphorylated monosaccharides, some of which can feed into the __________pathway.
Answer:
Detailed in explanation.
Explanation:
The pentose phosphate pathway is a two‑stage pathway that generates NADPH which is a reductant in many biosynthetic reactions and takes part in detoxifying reactive oxygen species, and ribose 5-phosphate which is a nucleotide (DNA and RNA) precursor. The substrate for the pentose phosphate pathway is glucose-6-phosphate.
The isomerization phase of the pentose phosphate pathway interconverts phosphorylated monosaccharides, some of which can feed into the glycolysis pathway.
Answer:
1) NADPH .
2) Ribose 5-phosphate .
3) G6P .
4) Nonoxidative .
5) glycolytic.
Explanation:
Hello,
In this case, each statement turn out into:
-The pentose phosphate pathway is a two‑stage pathway that generates NADPH (it is a donating electrons cofactor and a hydrogens provider to reactions catalyzed by enzymes) which is a reductant in many biosynthetic reactions and takes part in detoxifying reactive oxygen species, and ribose 5-phosphate (it is a fundamental constituent of the pyridine-based nucleotides such as nicotinamide adenine dinucleotide phosphate and nicotinamide adenine dinucleotide phosphate and the purine nucleotides adenosine diphosphate and ATP) which is a nucleotide (DNA and RNA) precursor.
-The substrate for the pentose phosphate pathway is G6P (glucose 6-phosphate, sometimes called the Robison ester, is a glucose based sugar phosphorylated at the hydroxyl group on carbon 6) .
-The nonoxidative (not carrying out oxidation) phase of the pentose phosphate pathway interconverts phosphorylated monosaccharides, some of which can feed into the glycolytic (referred to glycolysis as the metabolic pathway that converts glucose into pyruvate and a hydrogen ion) pathway.
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Interstellar space has an average temperature of about 10 K and an average density of hydrogen atoms of about one hydrogen atom per cubic meter. Calculate the mean free path of hydrogen atoms in interstellar space. Take d = 100 pm for a hydrogen atom.
Answer:
The mean free path is [tex]0.0000373631 m[/tex]
Explanation:
The formula for mean free path is :
λ = [tex]\frac{V}{\sqrt{2}\pi d^{2}N }[/tex]
where,
λ - is the mean free path distance
V - volume of the gas
d - the diameter of the molecule
N - number of molecules.
now ,
density [tex]D[/tex] = [tex]\frac{mass}{volume}[/tex] = [tex]\frac{M}{V}[/tex] ;
mass of the gas = (number of molecules)[tex]*[/tex](mass of one molecule) ;
as it's atomic hydrogen
[tex]M = N*m \\m=1.66*10^{-24}\\M=N*1.66*10^{-24}[/tex]
∴
[tex]D[/tex] = [tex]\frac{N*1.66*10^{-24}}{V}[/tex]
∴
[tex]\frac{V}{N*1.66*10^{-24}} = \frac{1}{ D}[/tex]
⇒ λ = [tex]\frac{1}{\sqrt{2}\pi d^{2}D }[/tex]
⇒ λ = [tex]\frac{1.66*10^{-24}}{\sqrt{2}\pi (100*10^{-12})^{2}*1 }[/tex]
⇒ λ = [tex]0.0000373631 m[/tex]
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is analyzed by adding KI and titrating the I3– that forms with S2O32–. The end point was reached following the addition of 13.02 mL of 0.03428 M Na2S2O3. Calculate the weight percent of Ce4 in the sample.When should the indicator be added to this titration? (at the beginning or just before the end point?)
Answer:
1,812 wt%
Explanation:
The reactions for this titration are:
2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻
I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻
The moles in the end point of S₂O₃⁻ are:
0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:
4,463x10⁻⁴ moles of S₂O₃⁻×[tex]\frac{1molI_{3}^-}{2molS_{2}O_{3}^-}[/tex] = 2,2315x10⁻⁴ moles of I₃⁻
As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:
2,2315x10⁻⁴ moles of I₃⁻×[tex]\frac{2molCe^{4+}}{1molI_{3}^-}[/tex] = 4,463x10⁻⁴ moles of Ce(IV). These moles are:
4,463x10⁻⁴ moles of Ce(IV)×[tex]\frac{140,116g}{1mol}[/tex] = 0,0625 g of Ce(IV)
As the sample has a 3,452g, the weight percent is:
0,0625g of Ce(IV) / 3,452g × 100 = 1,812 wt%
I hope it helps!
Here is the information: A 1.2516 gram sample of a mixture of CaCO3 and Na2SO4 was analyzed by dissolving the sample and adding C2O42- to completely precipitate the Ca2+ as CaC2O4. The CaC2O4 was dissolved in sulfuric acid and the resulting H2C204 was titrated with a standard KMnO4 solution.
This is the balanced equation: 3H2C2O4 + 2H+ + MnO4- ---> Mn2+ + 4H20 + 6CO2
Another part: The titration of the H2C2H4 obtained required 35.62 milliliters of .1092 molar MnO4- solution. Calculate the number of moles of H2C2O4 that reacted with the MnO4-. i think the correct answer is .01167 mol H2C2O4. This is not my question.
Next Part: Calculate the number of moles of CaCO3 in the original sample. Based on the answer to the part above, the answer is .003883 mol CaCO3 (i think this is correct, however this is not my question)
Calculate the percentage by weight of CaCO3 in the original sample.
FInd this based off of the .01167 mol H2C2O4 and the .003883 mol CaCO3. Basically just use the answers above to solve for this.
Answer:
93,32 % (w/w)
Explanation:
The Ca²⁺ of CaCO₃ was completely precipitate to CaC₂O₄ that results in H₂C₂O₄. The titration of this one is:
3H₂C₂O₄ + 2H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O + 6CO₂
As you required 35,62mL of 0,1092M MnO₄⁻, the moles of H₂C₂O₄ are:
0,03562L×[tex]\frac{0,1092M}{L}[/tex] =
3,890x10⁻³mol of MnO₄⁻×[tex]\frac{3molH_{2}C_{2}O_{4}}{1mol MnO_{4}^-} [/tex] = 0,01167 mol of H₂C₂O₄
The number of moles of CaCO₃ are the same number of moles of H₂C₂O₄ because every Ca²⁺ was converted in CaC₂O₄ that was converted in H₂C₂O₄. That means: 0,01167 mol of CaCO₃
0,01167 mol of CaCO₃ are:
0,01167 mol of CaCO₃×[tex]\frac{100,0869g}{1mol}[/tex] = 1,168 g of CaCO₃
As the mass of the initial mixture is 1,2516 g, the percentage by weight of CaCO₃ is:
[tex]\frac{1,168g}{1,2516g}[/tex]×100 = 93,32 % (w/w)
I hope it helps!
Consider an electrochemical cell based on the reaction: 2H+ (aq) + Sn (s) → Sn2+ (aq) + H2 (g) Which of the following actions would not change the measured cell potential? Consider an electrochemical cell based on the reaction: 2H+ (aq) + Sn (s) Sn2+ (aq) + H2 (g) Which of the following actions would not change the measured cell potential? addition of more tin metal to the anode compartment increasing the pressure of hydrogen gas in the cathode compartment lowering the pH in the cathode compartment increasing the tin (II) ion concentration in the anode compartment Any of the above will change the measured cell potential. Request Answer
Answer:
The only thing that will not affect the potential is the adition of solid Sn.
Explanation:
The potencial of a cell is linked to the concentration of the substances involved in the reactions by the equation of Nernst. So a change of one of them would change the cell potential.
[tex]E=E^{\circ}-\frac{R*T}{n*F}*ln(Keq)[/tex]
The Keq for this reaction is:
[tex]K_{eq}=\frac{[Sn^{2+}]*[H_2]}{[H^+]}[/tex]
Sn is not included because it's in solid state.
As can be seen, changing the concentrations of H2 (increasing the pressure), H+ (lowering the pH) or Sn2+ will affect the potential.
The only thing that will not affect it is the adition of solid Sn.
Based on the equilibrium constant equation, addition of Sn will not affect cell potential.
What is cell potential?Cell potential refers to the potential difference that exists between two points in an electrochemical cell.
The Nerst equation shows the relationship between the cell potential and the concentration of the substances involved in the reactions.
[tex]E=E^{\circ}-\frac{R*T}{n*F}*ln(Keq)E=E
∘
−
n∗F
R∗T
∗ln(Keq)[/tex]
Any change of one values results in a change in the cell potential.
The equilibrium constant for this reaction is given as follows:
[tex]K_{eq}=\frac{[Sn^{2+}]*[H_2]}{[H^+]}[/tex]
Based on equilibrium constant equation, changing the amount of Sn will not affect cell potential since it is not included in the equilibrium constant equation.
Therefore, addition of Sn will not affect cell potential.
Learn more about cell potential at: https://brainly.com/question/19036092
Given that the antacid milk of magnesia contains 400. mg of Mg(OH)2 per teaspoon, calculate the number of milliliters of stomach acid that are neutralized by 1 tablespoon of milk of magnesia. (1 tablespoon 3 teaspoons.) Express the volume in millilite rs to two significant figures.
Answer:
1.5 ml
Explanation:
Assuming that the stomach acid is HCl then:
Mg(OH)₂ + 2HCl → MgCl₂ + H₂O
since
number of moles of Mg(OH)₂ = mass / molecular weight of Mg(OH)₂ = 3*400 mg / 58.3 gr/mol = 20.583 m mol
thus
number of moles of HCl required = number of moles of Mg(OH)₂*2 = 41.166 m mol = 41.166 m moles
knowing that
density = mass / volume = (molecular weight* moles) / volume
volume =(molecular weight* moles)/ density
thus for HCl
volume = (36.46 gr/mol * 41.166*10^-3 moles)/( 1 gr/cm³)= 1.5 cm³= 1.5 ml
One tablespoon of milk of magnesia can neutralize approximately 210 mL of stomach acid, which is calculated from the neutralization reaction between Mg(OH)2 and stomach acid (HCl), taking into account molar ratios and concentrations.
Explanation:To calculate the volume of stomach acid neutralized by 1 tablespoon of milk of magnesia, we need to determine the amount of Mg(OH)2 provided by the antacid and how it reacts with hydrochloric acid (HCl), which is the main component of stomach acid. According to the information provided, milk of magnesia contains 400 mg of Mg(OH)2 per teaspoon. Since one tablespoon is equal to three teaspoons, one tablespoon of milk of magnesia contains 1200 mg (or 1.2 g) of Mg(OH)2.
The neutralization reaction between Mg(OH)2 and HCl is given by:
Mg(OH)2 + 2HCl → MgCl2 + 2H2O.
This shows that one mole of Mg(OH)2 reacts with 2 moles of HCl. The molar mass of Mg(OH)2 is approximately 58.3 g/mol, so 1.2 g corresponds to 1.2 g / 58.3 g/mol = 0.0206 moles of Mg(OH)2.
Since one mole of Mg(OH)2 neutralizes two moles of HCl, 0.0206 moles of Mg(OH)2 will neutralize 0.0206 moles × 2 = 0.0412 moles of HCl. If we assume that the stomach acid has the same concentration as mentioned in the provided information, 0.20 M, then we can find the volume of stomach acid neutralized by dividing the number of moles of HCl by the concentration. Hence, the volume of stomach acid that can be neutralized is 0.0412 moles / 0.20 mol/L = 0.206 L or 206 mL. We should state the volume in milliliters to two significant figures: 210 mL.
The vapor pressure of chloroform is 173.11 mm Hg at 25°C. How many grams of aspirin, C9H8O4, a nonvolatile, nonelectrolyte (MW = 180.1 g/mol), must be added to 244.0 grams of chloroform to reduce the vapor pressure to 171.05 mm Hg ? chloroform = CHCl3 = 119.40 g/mol. The boiling point of water, H2O, is 100.000 °C at 1 atmosphere. Kb(water) = 0.512 °C/m In a laboratory experiment, students synthesized a new compound and found that when 11.55grams of the compound were dissolved in 243.4 grams of water, the solution began to boil at100.071 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound?
a) There must be added 4.465 grams of aspirin
b) The unknown has a molecular weight of 342.2 g/mol
Step 1: Data given
The vapor pressure of chloroform is 173.11 mm Hg at 25°C
a nonvolatile, nonelectrolyte (MW = 180.1 g/mol), must be added
Mass of chloroform = 244.0 grams
The vapor pressure is reduced to 171.05 mm Hg
chloroform = CHCl3 = 119.40 g/mol.
The boiling point of water, H2O, is 100.000 °C
Step 2: Calculate moles chloroform
244.0g CHCl3/ 119.40g/ mol = 2.0436 mol CHCl3
step 3: Calculate moles aspirin
Moles aspirin = x grams / 180.1 g/mol
(p0 - p)/p0
⇒ p0 is the initial pressure = 173.11 mm Hg
⇒ p = the reduced pressure = 171.05 mm Hg
(p0 - p)/p0 = (173.11-171.05)/173.11 = (xg/180.1 g/mol) / (2.0436 mol + (xg/180.1g/mol))
0.0119 mol = (xg/180.1 g/mol) / (2.0436 mol + (xg/180.1g/mol))
0.02432 + (0.0119x/180.1) = (x/180.1)
0.02432 =(( x - 0.0119x)/180.1)
0.9981x = 4.38
x = 4.465 grams
There must be added 4.465 grams of aspirin
b) What is the molecular weight they determined for this compound?
Step 1: Define molar mass of the compound:
Molar mass = mass / moles
Molar mass compound = 11.55 grams / moles
Step 2: Calculate ΔT
ΔT = Tsoln - Tpure
ΔT = 100.071 - 100.00 = 0.071 °C
Step 3: Calculate m
ΔT = m*Kb
0.071°C = m*0.512 °C/m
m = 0.071 °C / 0.512 °C/m
m = 0.13867 m = 0.13867 mol/kg
Step 4: Calculate number of moles of the compound
Moles = mass of water * m
Moles = 0.2434 kg * 0.13867 mol/kg
Moles = 0.03375 moles
Step 5: Calculate molar mass
Molar mass = mass / moles
Molar mass compound = 11.55 grams / 0.03375 moles
Molar mass compound = 342.2 g/mol
The unknown has a molecular weight of 342.2 g/mol (and can be sucrose. It has a molar mas of 342.2 g/mol, is nonvolatile and non-electrolyte.).