The change in enthalpy, entropy, and free energy were calculated for the dehydrogenation reaction of ethylbenzene into styrene. The reaction was found to be endothermic and results in a decrease in overall disorder. Under the given conditions, the reaction will never be spontaneous.
Explanation:The processes involved in the production of styrene from ethylbenzene are fairly complex and require knowledge of thermodynamics. We'll begin with ΔH°rxn, which is found by subtracting the enthalpy (ΔH) of the reactants from that of the products: ΔH°rxn = [ΔH°f(styrene)] - [ΔH°f(ethylbenzene)] = 103.8 kJ/mol - (-12.5 kJ/mol) = 116.3 kJ/mol. This means the reaction is endothermic, as heat is absorbed.
The change in entropy ΔS°rxn, obtained likewise, is [S°(styrene) - S°(ethylbenzene)] = (238 J/mol·K - 255 J/mol·K) = -17 J/mol·K. This indicates a decrease in disorder in the system.
With these, we can calculate the change in free energy ΔG°rxn at a given temperature (T) using the equation ΔG°rxn = ΔH°rxn - TΔS°rxn. Substituting the known values at 298 K, ΔG°rxn = 116.3 kJ/mol - (298 K)(-17 J/mol·K) = 121.2 kJ/mol, indicating a non-spontaneous reaction.
For the reaction to be spontaneous, ΔG°rxn must be less than zero. Solving for T in the above equation with ΔG°rxn = 0, yields T = ΔH°rxn / ΔS°rxn = 116.3 kJ/mol / -17 J/mol·K ≈ -6840 K. This value is negative, implying the reaction is never spontaneous under the given conditions.
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At 40 C H2O2(aq) will decompose according to the following reaction:
2H2O2 (aq) -----> 2H2O (l) + O2 (g)
The following data were collected for the concentration of H2O2 at various times:
Time (s) H2O2 (mol/L)
1.000 2.16 x 10^4
0.500 4.32 x 10^4
(a) calculate the average rate of decomposition of H2O2 between0 and 2.16 x 104 s. Use this rate to calculate the averagerate of production of O2 (g) over the same time period.
(b) what are these rates for the time period 2.16 x 104 s to4.32 x 104 s ?
Answer:
a. 1,157x10⁻⁵M/s
b. 5,787x10⁻⁶M/s
Explanation:
For the reaction:
2H₂O₂(aq) → 2H₂O(l) + O₂(g).
a. The rate law of descomposition is:
[tex]rate=-\frac{1}{2} \frac{d[H_{2}O_{2}]}{dt}=\frac{d[O_{2}]}{dt}[/tex]
Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:
[tex]rate=-\frac{1}{2} \frac{0,500M}{-2,16x10^4s}[/tex]
[tex]rate=1,157x10^{-5}M/s[/tex]
As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s
b. Between 2,16x10⁴s and 4,32x10⁴s, rate law is:
[tex]rate=-\frac{1}{2} \frac{0,500M-0,250M}{2,16x10^4s-4,32x10^4s}[/tex]
[tex]rate=5,787x10^{-6}M/s[/tex]
The rates are 5,787x10⁻⁶M/s
I hope it helps!
The study of chemicals and bonds is called chemistry. There are two types of elements these rare metals and nonmetals.
Thus the rate is,
[tex]rate=1157\times10^{-5}[/tex]
What is rate law?The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters.
The balanced reaction is:-
[tex]2H_2O_2(aq)----->2H_2O(l)+o_2(g)[/tex]
The rate law of decomposition:-
[tex]rate=-\dfrac{1}{2}\dfrac{d[H_2O_2]}{dt}[/tex]
Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:
[tex]rate=-\dfrac{1}{2}\dfrac{0.5}{2.16\times10^{4}}[/tex]
[tex]rate=1157\times10^{-5}[/tex]
As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s
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Hydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH = -196 kJ Calculate the value of q (kJ) in this exothermic reaction when 2.50 g of hydrogen peroxide decomposes at constant pressure.
In this exothermic reaction, 2.50 g of hydrogen peroxide decomposes and generates -7.22 kJ of heat, as indicated by the negative sign.
Explanation:The question asks for the amount of heat (q) generated when 2.50 g of hydrogen peroxide decomposes at constant pressure to give water and oxygen. Given that the enthalpy change (ΔH) for the reaction 2H2O2(l) → 2H2O(l) + O2(g) is -196 kJ, it indicates that the decomposition of hydrogen peroxide is an exothermic reaction and heat is released in the process. The negative sign of ΔH confirms this.
Firstly, we will have to find out how many moles of H2O2 are there in 2.5 g. Using the molar mass of H2O2 (34 g/mol), we get about 0.074 moles of H2O2. In the balanced chemical equation, 2 moles of hydrogen peroxide generates -196 kJ of heat. Therefore, for 0.074 moles, we calculate it by (-196 kJ * 0.074)/2 = -7.22 kJ.
The <-strong>value of q in this exothermic reaction when 2.50 g of hydrogen peroxide decomposes at constant pressure is therefore -7.22 kJ. The negative sign indicates heat is being released, or in other words, the reaction is exothermic.
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The value of q, representing the thermal energy transferred in the decomposition of 2.50g of hydrogen peroxide at constant pressure, is approximately -7.16 kJ.
Explanation:The question asks for the value of q, which represents the thermal energy transferred during a reaction, in an exothermic reaction where 2.50 g of hydrogen peroxide decomposes into water and oxygen gas. In this given case, we know the enthalpy change (∆H) of the reaction is -196 kJ, indicating that this is an exothermic process where that much energy is being released to the surroundings.
We know from the balanced chemical equation 2H2O2(l) → 2H2O(l) + O2(g) that 2 moles of H2O2 decompose to release ∆H amount of energy. Let's find out the number of moles in 2.50 g of H2O2. The molar mass of H2O2 is approximately 34.01 g/mol. So, the number of moles = mass/molar mass = 2.50 g / 34.01 g/mol = 0.0735 moles.
Considering the stoichiometry of the reaction, ∆H of -196 kJ is associated with 2 moles of H2O2. Therefore, the energy associated with 0.0735 mol will be:
q ( energy ) = ∆H x (moles of H2O2 /2) = -196 kJ x (0.0735 mol / 2 mol) = -7.1625 kJ
Therefore, the quantity of thermal energy (q) released in this decomposition reaction is approximately -7.16 kJ.
Learn more about Thermal Energy in Chemical ReactionA sample of 0.495 grams of solid KHP is weighed into an Erlenmeyer flask. This sample is titrated with a sodium hydroxide solution, and 28.56 mL of NaOH are required to reach the endpoint. The sodium hydroxide solution is then used to titrate a sample of phosphoric acid of unknown concentration. It requires 29.88 mL of NaOH to react with 10.33 mL of H3PO4 solution. What is the concentration of the phosphoric acid?
Answer:
The concentration of the H₃PO₄ solution is 0,245M
Explanation:
The first titration is:
KHP + NaOH → KP⁻ + Na⁺ + H₂O
0,495g of KHP are:
0,495g×[tex]\frac{1mol}{204,22g}[/tex]= 2,42x10⁻³ moles of KHP
As 1 mole of KHP reacts with 1 mole of NaOH, moles of NaOH are 2,42x10⁻³ moles.
As volume required was 28,56mL, the concentration of the NaOH solution is:
2,42x10⁻³ moles / 0,02856L = 0,0849M
The titration of the phosporic acid with NaOH occurs as follows:
H₃PO₄ + NaOH → H₂PO₄⁻ + Na⁺ + H₂O
If were required 29,88mL of NaOH, the moles of NaOH spent were:
0,0849M×0,02988L = 2,54x10⁻³ moles of NaOH that are the same than H₃PO₄ moles.
As the volume of the solution of H₃PO₄ was 10,33mL, the concentration of the H₃PO₄ solution is:
2,54x10⁻³ moles of H₃PO₄ / 0,01033L = 0,245M
I hope it helps!
Final answer:
The concentration of the phosphoric acid is calculated using the molarity of a standardized NaOH solution obtained from a titration with KHP. The determined NaOH molarity is then applied to find the moles of phosphoric acid reacted in its own titration, which leads to the final concentration of the acid.
Explanation:
To determine the concentration of the phosphoric acid (H3PO4), we first need to calculate the molarity of the sodium hydroxide (NaOH) solution using the titration with potassium hydrogen phthalate (KHP). Since 0.495 grams of KHP was titrated with 28.56 mL of NaOH, we can calculate the moles of KHP used:
Molar mass of KHP (KH2C8H4O4) ≈ 204.22 g/molMoles of KHP = 0.495 g / 204.22 g/mol = 0.00242 molThe reaction between KHP and NaOH is 1:1, so moles of NaOH = moles of KHP = 0.00242 molMolarity of NaOH = moles/volume (in liters) = 0.00242 mol / 0.02856 L = 0.0847 MNow, using the concentration of NaOH, we can find the concentration of phosphoric acid. Given that 29.88 mL of the NaOH solution is required to titrate the 10.33 mL of H3PO4, we have:
Moles of NaOH = 0.0847 M * 0.02988 L = 0.00253 molThe ratio of NaOH to H3PO4 in the balanced equation is 3:1, so moles of H3PO4 = 0.00253 mol / 3 = 0.000843 molConcentration of H3PO4 = moles/volume (in liters) = 0.000843 mol / 0.01033 L = 0.0816 MIf 55.0 mL of ethanol (density=0.789g/mL)) initially at 9.0 ∘C is mixed with 55.0 mL of water (density=1.0g/mL) initially at 28.6 ∘C in an insulated beaker, and assuming that no heat is lost, what is the final temperature of the mixture?
The final temperature of the mixture is approximately [tex]\(-57.97^\circ\text{C}\)[/tex].
To find the final temperature of the mixture, we can use the principle of conservation of energy, assuming no heat is lost or gained in the process.
The heat gained by the cold substance (water) will be equal to the heat lost by the hot substance (ethanol).
The heat gained or lost [tex](\(Q\))[/tex] can be calculated using the equation:
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\(m\)[/tex] is the mass of the substance,
- [tex]\(c\)[/tex] is the specific heat capacity of the substance,
- [tex]\(\Delta T\)[/tex] is the change in temperature.
The sum of the heats gained and lost is zero:
[tex]\[ Q_{\text{water}} + Q_{\text{ethanol}} = 0 \][/tex]
Since the final temperature is the same for both substances, we can write the equation:
[tex]\[ m_{\text{water}}c_{\text{water}}\Delta T_{\text{water}} + m_{\text{ethanol}}c_{\text{ethanol}}\Delta T_{\text{ethanol}} = 0 \][/tex]
Rearrange the equation to solve for the final temperature [tex](\(T_{\text{final}}\))[/tex]:
[tex]\[ \Delta T_{\text{water}} + \Delta T_{\text{ethanol}} = 0 \][/tex]
[tex]\[ T_{\text{final}} - T_{\text{initial, water}} + T_{\text{final}} - T_{\text{initial, ethanol}} = 0 \][/tex]
[tex]\[ 2T_{\text{final}} = T_{\text{initial, water}} + T_{\text{initial, ethanol}} \][/tex]
[tex]\[ T_{\text{final}} = \frac{T_{\text{initial, water}} + T_{\text{initial, ethanol}}}{2} \][/tex]
Now, substitute the given values:
[tex]\[ T_{\text{final}} = \frac{(55.0 \ \text{mL} \times 1.0 \ \text{g/mL} \times 4.18 \ \text{J/g}^\circ\text{C} \times (28.6^\circ\text{C} - T_{\text{final}})) + (55.0 \ \text{mL} \times 0.789 \ \text{g/mL} \times 2.44 \ \text{J/g}^\circ\text{C} \times (T_{\text{final}} - 9.0^\circ\text{C}))}{2} \][/tex]
[tex]\[ T_{\text{final}} = \frac{(55.0 \ \text{mL} \times 4.18 \ \text{J/g}^\circ\text{C} \times (28.6^\circ\text{C} - T_{\text{final}})) + (55.0 \ \text{mL} \times 0.789 \ \text{g/mL} \times 2.44 \ \text{J/g}^\circ\text{C} \times (T_{\text{final}} - 9.0^\circ\text{C}))}{2} \][/tex]
Combine like terms:
[tex]\[ T_{\text{final}} = \frac{(55.0 \times 4.18 \times 28.6) + (55.0 \times 0.789 \times 2.44 \times (T_{\text{final}} - 9.0))}{2} \][/tex]
[tex]\[ T_{\text{final}} = \frac{(6132.86 + 107.79 \times (T_{\text{final}} - 9.0))}{2} \][/tex]
[tex]\[ T_{\text{final}} = \frac{6132.86 + 107.79T_{\text{final}} - 971.1}{2} \][/tex]
[tex]\[ 2T_{\text{final}} = 7101.76 + 107.79T_{\text{final}} - 971.1 \][/tex]
Combine like terms:
[tex]\[ 2T_{\text{final}} - 107.79T_{\text{final}} = 7101.76 - 971.1 \][/tex]
[tex]\[ -105.79T_{\text{final}} = 6130.66 \][/tex]
[tex]\[ T_{\text{final}} \approx -57.97^\circ\text{C} \][/tex]
Therefore, the final temperature of the mixture is approximately [tex]\(-57.97^\circ\text{C}\)[/tex]. This negative value indicates that the final temperature is lower than the initial temperatures of water and ethanol, which is consistent with the fact that heat is transferred from the substances with higher initial temperatures to the one with a lower initial temperature.
Which of the following statements is true regarding sodium and chlorine?A.) Sodium has greater electronegativity and a larger first ionization energyB.) Sodium has a larger first ionization energy and a larger atomic radiusC.) Chlorine has a larger atomic radius and greater electronegativityD.) Chlorine has greater electronegativity and a larger first ionization energy
Answer:
D.) Chlorine has greater electronegativity and a larger first ionization energy
Explanation:
first we have to consider where sodium and chlorine are in the atomic table. Both are in the third row and both are nearly opposite ends of third row. sodium have high tendency to lose an electron while chlorine have higher tendency to gain an electron. so because of this fact chlorine is more electronegative than sodium.
As we move left to right in the periodic table, there is increase in the number of protons in atom and therefore there is increase in attraction between the nucleus and electrons and shrinking down the atom which is general trend when we move from left to right in the periodic table. so chlorine will be smaller than sodium because of the shrinking of the atom and smaller atoms have higher ionization energy because their electrons are close to the nucleus. They require more energy to remove. so this will make chlorine have both a smaller atomic radius and larger ionization energy. so D is the right answer.
Final answer:
The correct statement regarding sodium and chlorine is that chlorine has a larger atomic radius and greater electronegativity.
Explanation:
The correct statement regarding sodium and chlorine is Option C: Chlorine has a larger atomic radius and greater electronegativity.
Chlorine, with 17 electrons, has a strong attraction for electrons and tends to gain an electron to form a chloride ion (Cl-). Sodium, with 11 electrons, has a lower ionization potential and tends to lose an electron to form a sodium ion (Na+). Therefore, chlorine has a greater electronegativity and a larger atomic radius compared to sodium.
Additionally, when sodium and chlorine react, sodium donates an electron to chlorine, forming an ionic bond and creating the compound sodium chloride (NaCl).
Under which of the following sets of conditions will a sample of a real gas deviate most from ideal behavior? a. strong interparticle attractions, low temperature, high pressure b. strong interparticle attractions, high temperature, high pressure c. weak interparticle attractions, low temperature, high pressure d. weak interparticle attractions, high temperature, high pressure e. strong interparticle attractions, low temperature, low pressure
Answer:
a. strong interparticle attractions, low temperature, high pressure.
Explanation:
A gas behaves as an ideal gas when it fulfills the following conditions:
The interparticle attractions are neglectable.The volume of the gas is neglectable in comparison with the volume of the container.The gas deviates from the ideal behavior:
When there are strong interparticle attractions.At low temperatures and high pressures, the particles are closer to each other and interparticle attractions are stronger.A real gas deviates most from ideal behavior under the conditions of strong interparticle attractions, low temperature, and high pressure.
Explanation:A real gas deviates most from ideal behavior under the following set of conditions: strong interparticle attractions, low temperature, high pressure.
When a real gas has strong interparticle attractions, the gas particles are more likely to stick together and deviate from ideal behavior. Low temperature also contributes to the deviation as it slows down the gas particles and makes the attractive forces between them more prominent. High pressure further increases the deviation due to the decrease in empty space between the particles.
In summary, a real gas deviates most from ideal behavior when it has strong interparticle attractions, low temperature, and high pressure.
A major component of gasoline is octane . When octane is burned in air, it chemically reacts with oxygen gas to produce carbon dioxide and water . What mass of oxygen gas is consumed by the reaction of of octane? Be sure your answer has the correct number of significant digits.
Answer:
mass of O₂ = 800 g
Explanation:
Formula of Octane = C₈H₁₈
Formula of Oxygen = O₂
Balanced Chemical Reaction:
Octane react with oxygen and produce water and carbon dioxide.
2C₈H₁₈ + 25O₂ ---------> 16CO₂+ 18H₂O
Moles calculation:
From the number of moles of the reactant used in a chemical reaction specific number of product produced. we can find the amount of any reactant and product from the mole ratio of a chemical reaction.
Mole formula =
no. of moles = mass in grams / molecular mass ............... (1)
Given Data:
Molecular Weight of O₂ = (16 x2 ) = 32 g/mol
Mass of O₂ = To be find
Calculations:
2C₈H₁₈ + 25O₂ ---------> 16CO₂ + 18H₂O
2mole 25 mole 16 mole 18 mole
From the above balanced chemical equation it is know that 2 mole of octane consume 25 mole of oxygen.
so we have to calculate the mass of oxygen that is consumed
by using mole formula (1) we can fine the mass of oxygen
we know
Molecular Weight of O₂ = 32 g/mol
number of moles of Oxygen molecule = 25 mol
putting the value in the below formula
no. of moles of O₂ = mass of O₂ / molecular mass of O₂
25 mole = mass of O₂ / 32 g/mol ....... (2)
By rearragming the equation (2)
mass of O₂ = 25 mole x 32 g/mol
mass of O₂ = 25 mole x 32 g/mol
mass of O₂ = 800 g
So in the octan reaction with oxygen 800g of oxygen will use.
Determine the volume fraction of fibers required to obtain a longitudinal tensile strength of 812 MPa in a unidirectional carbon fiber reinforced composite. You are given that the tensile strength of the carbon fibers are 1240 MPa and that the stress in the matrix at fiber failure will be 300 MPa. Answer Format X.XX Unit: Unitless (fraction, example: 50 vol% fiber would be entered as 0.50)
Answer:
V= 0.544
Explanation:
Given that
σ=812 MPa
σ₁=1240 MPa
σ₂=300 MPa
Lets take volume fraction V
The strength of composite fiber given as
σ = σ₁ V + σ₂(1-V)
By putting the values
812 = 1240 V + 300 (1-V)
812 = 1240 V +300 - 300 V
812 - 300 = 1240 V- 300 V
512 = 940 V
V= 0.544
The percentage volume = 54.4 %
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 S2O3^2-. The end point was reached following the addition of 13.02 mL of 0.03247 M Na2S2O3. Calculate the weight percent of Ce^4+ in the sample?
Answer:
The weight percent in the sample is 17,16%
Explanation:
The dissolution of the Ce(IV) salt provides free Ce⁴⁺ that reacts, thus:
2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻
I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻
The moles in the end point of S₂O₃⁻ are:
0,01302L×0,03247M Na₂S₂O₃ = 4,228x10⁻⁴ moles of S₂O₃²⁻.
As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:
4,228x10⁻⁴ moles of S₂O₃⁻×[tex]\frac{1molI_{3}^-}{2molS_{2}O_{3}^{2-}}[/tex] = 2,114x10⁻⁴ moles of I₃⁻
As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:
2,114x10⁻⁴ moles of I₃⁻× [tex]\frac{2molCe^{4+}}{1molI_{3}^-}[/tex] = 4,228x10⁻⁴ moles of Ce(IV).
These moles are:
4,228x10⁻⁴ moles of Ce(IV)×[tex]\frac{140,116g}{1mol}[/tex] = 0,05924 g of Ce(IV)
As was taken an aliquot of 25,00mL from the solution of 250,0mL:
0,05924 g of Ce(IV)×[tex]\frac{250,0mL}{25,00mL}[/tex] =0,5924g of Ce(IV) in the sample
As the sample has 3,452g, the weight percent is:
0,5924g of Ce(IV) / 3,452g × 100 = 17,16 wt%
I hope it helps!
Suppose a 250.0 mL flask is filled with 1.3mol of I2 and 1.0mol of HI. The following reaction becomes possible:
H2 (g) +I2 (g)=2HI (g)
The equilibrium constant K for this reaction is 0.983 at the temperature of the flask.
Calculate the equilibrium molarity of HI . Round your answer to one decimal place.
Answer:
The molarity of HI at the equilibrium is 2.8M
Explanation:
Step 1: Data given
Volume of the flask = 250.0 mL = 0.250L
Number of moles I2 = 1.3 mol
Number of moles HI = 1.0 mol
Kc = 0.983
Step 2: The balanced equation
H2(g) +I2(g) ⇆ 2HI(g)
For 1 mole I2 consumed, we need 1 mole H2 to produce 2 moles HI
Step 3: Calculate initial concentrations
Initial concentration I2 = 1.3mol / 0.25L
Initial concentration I2 = 5.2 M
Initial concentration HI = 1.0 mol / 0.25L
Initial concentration HI = 4.0 M
Step 4: Calculate concentrations at equilibrium
The concentration at equilibrium is:
[I2] = (5.2+x)M
[HI] = (4.0 - x)M
[H2] = xM
Kc = [HI]²/[H2][I2]
0.983 = (4-x)²/ (x*(5.2+x))
0.983 = (4-x)²/ (5.2x +x²)
5.1116x + 0.983 x² = 16 -8x +x²
-0.017x² +13.1116x -16 = 0
x = 1.222 = [H2]
[HI] = 4.0 - 1.222 = 2.778M ≈ 2.8 M
[I2] = 5.2 + 1.222 = 6.422 M ≈ 6.4 M
To control we can calculate:
[2.778]² / [1.222][6.422] = 0.983 = Kc
The molarity of HI at the equilibrium is 2.8M
What is the concentration in molarity of a solution made using 50.0 grams of C6H12O6 in 300.0 mL of water?
Please help ASAP!!! :(
Answer: concentration in Molarity it's actually Molarity
Molarity=concentration/molar mass
Concentration=50g/0.3dm³=166.67g/dm³
Molar mass=180g/mol
Molarity=0.926mol/dm³
Answer:
Answer is =0. 925 mol L−1
Explanation:
C₆H₁₂O₆ is formula of Glucose
For Molarity, we must know the following things
• the number of moles of solute present in solution
• the total volume of the solution
we know the mass of one mole of Glucose = 180.156 g/mol
Number of moles = Given Mass of substance / Mass of one mole
No of moles = 50g / 180.156 g/mol
= 0.277 moles
Now we know that molarity is expressed per liter of solution. Since you dissolve 0.277 moles of potassium chloride in 300. mL of solution, you can say that 1.0 L will contain
For 300 ml of solution, no of moles are = 0.277 moles
For 1 ml of solution, no of moles are = 0.277/300 moles
For 1(1000) ml of solution, no of moles are= 0. 277/300 x 1000
= 0.925 moles/ L
Answer is =0. 925 mol L−1
Which statement is FALSE? a. ΔHrxn is the heat of reaction. b. ΔErxn is a measure of heat. c. An exothermic reaction gives heat off heat to the surroundings. d. Endothermic has a positive ΔH. e. Enthalpy is the sum of a system's internal energy and the product of pressure and volume.
Answer:
b. ΔE rxn is a measure of heat
Explanation:
a. ΔHrxn is the heat of reaction. TRUE. ΔHrxn or change in enthalpy of reaction is per definition the change in heat that is involved in a chemical reaction.
b. ΔErxn is a measure of heat. FALSE. Is the change in internal energy of a reaction
c. An exothermic reaction gives heat off heat to the surroundings. TRUE. An exothermic reaction is a chemical reaction that releases heat.
d. Endothermic has a positive ΔH. TRUE. When a process is exothermic ΔH<0 and when the process is endothermic ΔH>0
e. Enthalpy is the sum of a system's internal energy and the product of pressure and volume. TRUE. Under constant pressure and volume the formula is ΔH = ΔE + PV
I hope it helps!
The statement that is FALSE is b. ΔErxn is a measure of heat.
Explanation:The statement that is FALSE is b. ΔErxn is a measure of heat.
ΔErxn is the change in internal energy of a system, not a direct measure of heat. It is related to heat through the equation ΔErxn = q + w, where ΔErxn is the change in internal energy, q is the heat transferred, and w is the work done on or by the system. Enthalpy, on the other hand, is a measure of heat at constant pressure, and is given by ΔHrxn = q (heat of reaction).
Therefore, option b is the false statement.
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A chemist prepares a solution of silver(I) nitrate (AgNO3) by measuring out 269. micromol of silver(I) nitrate into a 300. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mmol/L of the chemist's silver(I) nitrate solution. Be sure your answer has the correct number of significant digits.
Answer:
The concentration of the chemist's silver(I) nitrate solution is 0.897 mmol/L
Explanation:
Step 1: Data given
Number of moles AgNO3 = 269 micromol = 269 * 10^-6 mol
Volume of AgNO3 = 300 mL = 0.3 L
Molar mass of AgNO3 = 169.87 g/mol
Step 2: Calculate molarity of AgNO3
Molarity = number of moles per volume (in Liters)
Molarity AgNO3 = 269 *10^-6 mol / 0.3 L
Molarity AgNO3 = 8.97 * 10^-4 M
8.97 *10^-4 mol/L = 897 micromol/L = 0.897 mmol/L
The concentration of the chemist's silver(I) nitrate solution is 0.897 mmol/L
Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C. Zn(s) + 2H+ (aq) → Zn2+(aq) + H2(g) Assume the volume of reactants is negligible compared to that of the vapor produced.
Answer:
19,26 kJ
Explanation:
The work done when a gas expand with a constant atmospheric pressure is:
W = PΔV
Where P is pressure and ΔV is the change in volume of gas.
Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:
500,0g Zn(s)×[tex]\frac{1molZn}{65,38g}[/tex]×[tex]\frac{1molH_{2}(g)}{1molZn}[/tex] = 7,648 moles of H₂
At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:
V = nRT/P
V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm
V = 190,1L
That means that ΔV is:
190,1L - 0L = 190,1L
And the work done is:
W = 1atm×190,1L = 190,1atmL.
In joules:
190,1 atmL×[tex]\frac{101,325}{1atmL}[/tex] = 19,26 kJ
I hope it helps!
The amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C is approximately 19601.44 joules.
The amount of work done against an atmospheric pressure when a gas is produced can be calculated using the ideal gas law, which is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to determine the number of moles of hydrogen gas (H2) produced from the dissolution of 500.0 g of zinc. The balanced chemical equation is:
[tex]\[ \text{Zn}(s) + 2\text{H}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{H}_2(g) \][/tex]
From the stoichiometry of the reaction, 1 mole of zinc produces 1 mole of hydrogen gas. The molar mass of zinc is approximately 65.38 g/mol. Therefore, the number of moles of zinc (n_Zn) is:
[tex]\[ n_{\text{Zn}} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}} = \frac{500.0 \text{ g}}{65.38 \text{ g/mol}} \approx 7.647 \text{ mol} \][/tex]
Since the stoichiometry is 1:1, the number of moles of hydrogen gas produced (n_H2) is also approximately 7.647 moles.
Next, we convert the temperature from degrees Celsius to Kelvin:
[tex]\[ T = 30.0^\circ\text{C} + 273.15 = 303.15 \text{ K} \][/tex]
Now, we can use the ideal gas law to find the volume of hydrogen gas produced at 1.00 atm of pressure:
[tex]\[ PV = nRT \][/tex]
[tex]\[ V = \frac{nRT}{P} \][/tex]
Plugging in the values:
[tex]\[ V = \frac{(7.647 \text{ mol})(0.0821 \text{ L·atm/mol·K})(303.15 \text{ K})}{1.00 \text{ atm}} \][/tex]
[tex]\[ V \approx 193.57 \text{ L} \][/tex]
Finally, the work done (W) against the atmospheric pressure to produce this volume of hydrogen gas is given by:
[tex]\[ W = P \cdot V \][/tex]
[tex]\[ W = 1.00 \text{ atm} \cdot 193.57 \text{ L} \][/tex]
[tex]\[ W \approx 193.57 \text{ L·atm} \][/tex]
To express this work in joules, we use the conversion factor[tex]1 L*atm[/tex] = [tex]101.325 J:[/tex]
[tex]\[ W \approx 193.57 \text{ L·atm} \times \frac{101.325 \text{ J}}{1 \text{ L·atm}} \][/tex]
[tex]\[ W \approx 19601.44 \text{ J} \][/tex]
Therefore, the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C is approximately 19601.44 joules.
The answer is: [tex]19601.44 \text{ J}.[/tex]
Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing copper and silver. Cu ( s ) ∣ ∣ Cu 2 + ( aq ) ∥ ∥ Ag + ( aq ) ∣ ∣ Ag ( s ) anode: cathode: net cell reaction:
Answer: The half reactions and net cell reaction of the cell is written below.
Explanation:
The given cell is:
[tex]Cu(s)/Cu^{2+}(aq.)||Ag^{+}(aq.)/Ag(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
Half reactions for the given cell follows:
Oxidation half reaction (anode): [tex]Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-[/tex]
Reduction half reaction (cathode): [tex]Ag^{+}(aq.)+e^-\rightarrow Ag(s)[/tex] ( × 2)
Net cell reaction: [tex]Cu(s)+2Ag^+(aq.)\rightarrow Cu^{2+}(aq.)+2Ag(s)[/tex]
Hence, the half reactions and net cell reaction of the cell is written above.
The half-reactions at each electrode and the net cell reaction for the electrochemical cell containing copper and silver are provided.
Explanation:The half-reactions at each electrode in this electrochemical cell are:
Anode: Cu (s) → Cu2+ (aq) + 2e−
Cathode: 2Ag+ (aq) + 2e− → Ag (s)
The net cell reaction for this electrochemical cell is:
Cu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s)
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A mixture of nitrogen and hydrogen gases, at a total pressure of 663 mm Hg, contains 3.46 grams of nitrogen and 0.156 grams of hydrogen. What is the partial pressure of each gas in the mixture? PN2 = mm Hg PH2 = mm Hg
Answer: The partial pressure of nitrogen gas is 405.76 mmHg and that of hydrogen gas is 257.24 mmHg
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For nitrogen gas:Given mass of nitrogen gas = 3.46 g
Molar mass of nitrogen gas = 28 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of nitrogen gas}=\frac{3.46g}{28g/mol}=0.123mol[/tex]
For hydrogen gas:Given mass of hydrogen gas = 0.156 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of hydrogen gas}=\frac{0.156g}{2g/mol}=0.078mol[/tex]
Mole fraction of a gas is calculated by using the formula:
[tex]\chi_{A}=\frac{n_{A}}{n_{A}+n_{B}}[/tex] ......(1)
For nitrogen gas:Putting values in equation 1, we get:
[tex]\chi_{\text{nitrogen gas}}=\frac{0.123}{0.123+0.078}=0.612[/tex]
For hydrogen gas:Putting values in equation 1, we get:
[tex]\chi_{\text{hydrogen gas}}=\frac{0.078}{0.123+0.078}=0.388[/tex]
The partial pressure of a gas is given by Raoult's law, which is:
[tex]p_A=p_T\times \chi_A[/tex] ......(2)
where,
[tex]p_A[/tex] = partial pressure of substance A
[tex]p_T[/tex] = total pressure = 663 mmHg
[tex]\chi_A[/tex] = mole fraction of substance A
For nitrogen gas:[tex]p_{\text{Nitrogen gas}}=663mmHg\times 0.612\\\\p_{\text{Nitrogen gas}}=405.76mmHg[/tex]
For hydrogen gas:[tex]p_{\text{Hydrogen gas}}=663mmHg\times 0.388\\\\p_{\text{Hydrogen gas}}=257.24mmHg[/tex]
Hence, the partial pressure of nitrogen gas is 405.76 mmHg and that of hydrogen gas is 257.24 mmHg
Which of the following describes a system that CANNOT be thermodynamically favored?
A.) ΔS is negative and ΔH is positive
B.) ΔS is positive and ΔH is positive
C.) ΔS is negative and ΔH is negative
D.) ΔS is positive and ΔH is negative
Answer: d) ΔS is positive and ΔH is negative
Explanation:
According to Gibb's equation:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
[tex]\Delta G[/tex] = Gibbs free energy
[tex]\Delta H[/tex] = enthalpy change
[tex]\Delta S[/tex] = entropy change
T = temperature in Kelvin
[tex]\Delta G[/tex]= +ve, reaction is non spontaneous
[tex]\Delta G[/tex]= -ve, reaction is spontaneous
[tex]\Delta G[/tex]= 0, reaction is in equilibrium
a) ΔS is negative and ΔH is positive
[tex]\Delta G=(-ve)-T(+ve)[/tex]
[tex]\Delta G=(-ve)(-ve)=-ve[/tex]
Reaction is spontaneous at all temperatures.
b) ΔS is positive and ΔH is positive
[tex]\Delta G=(+ve)-T(+ve)[/tex]
[tex]\Delta G=(+ve)(-ve)=-ve[/tex]
Reaction is spontaneous at high temperatures.
c) ΔS is negative and ΔH is negative
[tex]\Delta G=(-ve)-T(-ve)[/tex]
[tex]\Delta G=(-ve)(+ve)=-ve[/tex]
Reaction is spontaneous at low temperatures.
d) ΔS is positive and ΔH is negative
[tex]\Delta G=(+ve)-T(-ve)[/tex]
[tex]\Delta G=(+ve)(+ve)=+ve[/tex]
Reaction is non spontaneous or thermodynamically unfavored at all temperatures.
Consider a galvanic cell based on the reaction Al^3+_(aq) + Mg_(s) rightarrow Al_(s) + Mg^2+ _(aq) The half-reactions are Al^3+ + 3 e^- rightarrow Al E degree = - 1.66 V Mg^2+ + 2 e^- rightarrow Mg E degree = - 2.37 V Give the balanced cell reaction and calculate E degree for the cell.
Answer: The standard cell potential of the cell is -0.71 V
Explanation:
The half reactions follows:
Oxidation half reaction: [tex]Mg\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V[/tex] ( × 3)
Reduction half reaction: [tex]Al^{3+}(aq.)+3e^-\rightarrow Al(s);E^o_{Al^{3+}/Al}=-1.66V[/tex] ( × 2)
The balanced cell reaction follows:
[tex]2Al^{3+}(aq.)+3Mg(s)\rightarrow 2Al(s)+3Mg^{2+}(aq.)[/tex]
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation, we get:
[tex]E^o_{cell}=-2.37-(-1.66)=-0.71V[/tex]
Hence, the standard cell potential of the cell is -0.71 V
A chemist prepares repares a solution of magnesium chloride (MgCl2) by measuring out 48. mg of MgCl, into a 300 ml. volumetric flask and filling to the mark with distilled water.
Calculate the molarity of Cl^- anions in the chemist's solution.
Be sure your answer is rounded to 2 significant digits.
Answer:
The molarity of the Cl- anions is 0.0033 M
Explanation:
Step 1: Data given
Mass of MgCl2 = 48 mg = 48 *10^-3 grams
volume = 300 mL = 0.3L
Molar mas of MgCl2 = 95.21 g/mol
Step 2: The balanced equation
MgCl2 → Mg2+ + 2Cl-
This means for 1 mol MgCl2, we'll have 2 moles of Cl-
Step 3: Calculate moles MgCl2
Number of moles of MgCl2 = mass of MgCl2 / Molar mass of MgCl2
Moles MgCl2 = 48*10^-3 grams / 95.21 g/mol
Moles MgCl2 = 5.04 *10-4 moles
Step 4: Calculate moles of Cl-
This means for 1 mol MgCl2, we'll have 2 moles of Cl-
For 5.04 *10-4 moles MgCl2, we will have 2* 5.04 *10-4 moles = 0.001 moles Cl-
Step 5: Calculate molarity of the Cl- anion
Molarity = Moles / volume
Molarity Cl- = 0.001 moles / 0.3 L
Molarity Cl- = 0.0033 M
The molarity of the Cl- anions is 0.0033 M
A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO3−(aq). What was the concentration of Pb(ClO3)2 in the original undiluted solution? 3.60 × 10−2 M 7.19 × 10−2 M 0.238 M 0.156 M 0.477 M
Answer:
0.238 M
Explanation:
A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.
C₁.V₁ = C₂.V₂
C₁ × 24.00 mL = 0.220 M × 52.00 mL
C₁ = 0.477 M
The concentration of Pb(ClO₃)₂ is:
[tex]\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M[/tex]
The freezing point of a 1.00 molal aqueous solution of calcium chloride (CaCl2) is lower than the freezing point of a 1.00 molal aqueous solution of ethylene glycol (C2H6O2). Why are the freezing points of the two solutions different?
Answer:
The calcium chloride is an electrolyte salt, so the frezzing point of solution must be higher than ethylene glycol.
Explanation:
This is the colligative property for this question: Frezzing point depression
ΔTf = Kf · molal · i
ΔTf = T° fussion solvent pure - T° fussion solution
As both solutions are the same in molality and the solvent is water, the formula stands the same but the Calcium Chloride is a salt which is dissociated in water like this:
CaCl₂ → Ca²⁺ + 2Cl⁻
We have 3 moles of ions, so this value modiffy the formula with the Van't Hoff Factor (number of ion particles per individual molecule of solute).
Ethyleneglycol is a non-electrolytic organic compound (It is often used 1 as the i).
Let's see the formula in both:
0° - T° fussion solution = Kf · molal · 3 → CaCl2
0° - T° fussion solution = Kf · molal · 1 → C2H6O2
Identify which sets of quantum numbers are valid for an electron. Each set is ordered (n,l,ml,ms).a. 3,1,1,-1/2b. 4,3,1,-1/2c. 2,0,0,-1/2d. 1,3.0,1/2e. 3,2,-1,-1/2f. 3,3,-1,1/2g. 3,2,1,-1h. 1,-1,-1,-1/2i. 3,3,1,1/2j. 3,0,0,1/2k. 4,3,4,-1/2l. 0,2,1,1/2
Answer:
The valid set of quantum numbers are:
a. 3,1,1,-1/2
b. 4,3,1,-1/2
c. 2,0,0,-1/2
e. 3,2,-1,-1/2
j. 3,0,0,1/2
Explanation:
Quantum numbers (n, l, mℓ, ms) are the set of numbers that describe the state of an electron in an atom.
The four quantum numbers and their rules are:
Principal: n ≥ 1 Azimuthal: ℓ ≤ (n-1)Magnetic: mℓ = (- ℓ) to (+ ℓ)Spin: ms = (- s), (+ s)Therefore,
a. 3,1,1,-1/2: Valid
b. 4,3,1,-1/2: Valid
c. 2,0,0,-1/2: Valid
d. 1,3,0,1/2: NOT Valid
Reason: ℓ ≤ (n-1). Therefore, ℓ can not be greater than n, .
e. 3,2,-1,-1/2: Valid
f. 3,3,-1,1/2: NOT Valid
Reason: ℓ ≤ (n-1). Therefore, ℓ can not be equal to n. (ℓ ≠n)
g. 3,2,1,-1: NOT Valid
Reason: the only allowed values of ms = (- s), (+ s) = (- 1/2), (+ 1/2)
h. 1,-1,-1,-1/2: NOT Valid
Reason: ℓ ≤ (n-1) and n ≥ 1. Therefore, ℓ can not be equal to n and also ℓ can't be negative.
i. 3,3,1,1/2: NOT Valid
Reason: ℓ ≤ (n-1). Therefore, ℓ can not be equal to n. (ℓ ≠n)
j. 3,0,0,1/2: Valid
k. 4,3,4,-1/2: NOT Valid
Reason: mℓ = (- ℓ) to (+ ℓ). Therefore, mℓ can't be greater than ℓ.
l. 0,2,1,1/2: NOT Valid
Reason: n ≥ 1. Therefore, value of n can't be 0 (n≠0)
The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?
A.) 2.0x10^-3
B.) 1.25x10-3
C.) 5.0x10^-4
D.) 8.9x10^-5
Answer:
The molar solubility of distilled water is [tex]$1.25 \times 10-3$[/tex] (Option B)
Explanation:
Given:
[tex]Pbl_2 -7.9 \times 10^-^9.[/tex]
Expression for solubility constant is,
[tex]$K_{s p}=\left[P b^{2+}\right]\left[I^{-}\right]^{2}$[/tex]
The given equation is,
[tex]$K_{s p}=7.9 \times 10^{-9}$[/tex]
Equation is to calculated as,
[tex]$7.9 \times 10^{-9}=(s) \times(2 s)^{2}$[/tex]
[tex]$7.9 \times 10^{-9}=4 s^{3}$[/tex]
[tex]$s=1.25 \times 10^{-3} \mathrm{~mol} / L$[/tex]
The solubility product in distilled water is [tex]$s=1.25 \times 10^{-3} \mathrm{~mol} / L$[/tex].
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https://brainly.com/question/819068The correct answer is C.)[tex]5.0x10^-^4.[/tex]
To determine the molar solubility of PbI2 in distilled water, we need to use the solubility product constant (Ksp) for PbI2, which is given as 7.9x10^-9.
The chemical equation for the dissociation of PbI2 in water is:
[tex]\[ PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq) \][/tex]
The solubility product expression for this equilibrium is:
[tex]\[ Ksp = [Pb^{2+}][I^-]^2 \][/tex]
Let's denote the molar solubility of PbI2 as s. This means that the concentration of Pb^2+ ions will be s, and the concentration of I^- ions will be 2s (since 2 moles of I^- are produced for every mole of PbI2 that dissolves). We can now express the Ksp equation in terms of s:
[tex]\[ Ksp = s \cdot (2s)^2 \][/tex]
[tex]\[ Ksp = s \cdot 4s^2 \][/tex]
[tex]\[ Ksp = 4s^3 \][/tex]
Now we can solve for s using the given Ksp value:
[tex]\[ 7.9 \times 10^{-9} = 4s^3 \][/tex]
[tex]\[ s^3 = \frac{7.9 \times 10^{-9}}{4} \][/tex]
[tex]\[ s^3 = 1.975 \times 10^{-9} \][/tex]
[tex]\[ s = \sqrt[3]{1.975 \times 10^{-9}} \][/tex]
[tex]\[ s \approx 5.82 \times 10^{-4} \][/tex]
Since we are looking for the molar solubility, we can round this value to the appropriate number of significant figures, which is two significant figures based on the Ksp value given [tex](7.9x10^-9[/tex] has two significant figures). Therefore, the molar solubility of PbI2 in distilled water is approximately:
[tex]\[ s \approx 5.8 \times 10^{-4} \][/tex]
The closest answer choice to this calculated value is C.) 5.0x10^-4.
A buret is filled with 0.1517 M naoh A 25.0 mL portion of an unknown acid and two drops of indicator are added to an Erlenmeyer flask, and the titration experiment is carried out. If the initial buret reading was 0.55 mL, and the buret reading at the end point was 22.50 mL, what is the molarity of the unknown acid?
Answer:
molarity of acid =0.0132 M
Explanation:
We are considering that the unknown acid is monoprotic. Let the acid is HA.
The reaction between NaOH and acid will be:
[tex]NaOH+HA--->NaA+H_{2}O[/tex]
Thus one mole of acid will react with one mole of base.
The moles of base reacted = molarity of NaOH X volume of NaOH
The volume of NaOH used = Final burette reading - Initial reading
Volume of NaOH used = 22.50-0.55= 21.95 mL
Moles of NaOH = 0.1517X21.95=3.33 mmole
The moles of acid reacted = 3.33 mmole
The molarity of acid will be = [tex]\frac{mmole}{volumne(mL)}=\frac{0.33}{25}=0.0132M[/tex]
The problem is about acid-base titration used in Chemistry. A 1:1 reaction between NaOH and the unknown acid is assumed. The molarity of the unknown acid is calculated as 0.1332 M.
Explanation:This problem pertains to acid-base titration, a common method in Chemistry for determining the concentration of an unknown acid or base. The initial and final buret readings indicate the volume of NaOH used in the reaction: 22.50 mL - 0.55 mL = 21.95 mL, which is 0.02195 L. Using the molarity of NaOH (0.1517 M), we can find the moles of NaOH used: 0.1517 M * 0.02195 L = 0.00333 mol. Since the reaction between NaOH and the unknown acid is assumed to be 1:1, the moles of the unknown acid should also be 0.00333. Therefore, to obtain the molarity of the unknown acid, divide the moles of the acid by the volume of the acid in liters (25.0 mL = 0.025 L): 0.00333 mol / 0.025 L = 0.1332 M. Therefore, the molarity of the unknown acid is 0.1332 M.
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What is the concentration of ammonia in a solution if 23.4 mL of a 0.117 M solution of HCl are needed to titrate a 100.0 mL sample of the solution?
Answer: 0.0274 M
Explanation:-
The balanced chemical solution is:
[tex]NH_4OH(aq)+HCl(aq)\rightarrow NH_4Cl(aq)+H_2O(l)[/tex]
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of [tex]HCl[/tex] solution = 0.117 M
[tex]V_1[/tex] = volume of [tex]HCl[/tex] solution = 23.4 ml
[tex]M_2[/tex] = molarity of [tex]NH_4OH[/tex] solution = ?
[tex]V_2[/tex] = volume of [tex]NH_4OH[/tex] solution = 100.0 ml
[tex]n_1[/tex] = valency of [tex]HCl[/tex] = 1
[tex]n_2[/tex] = valency of [tex]NH_4OH[/tex] = 1
[tex]1\times 0.117M\times 23.4=1\times M_2\times 100.0[/tex]
[tex]M_2=0.0274[/tex]
Therefore, the concentration of ammonia in a solution will be 0.0274 M
For which of these processes is the value of ΔH expected to be negative?
I. The temperature (of water) increases when calcium chloride dissolves in water
II. Steam condenses to liquid water
III. Water freezes
IV. Dry ice sublimes
A.) IV only
B.) I, II, and III
C.) I only
D.) II and III only
The value of ΔH is expected to be negative in processes where heat is released, which includes the condensation of steam to liquid water (II) and the freezing of water (III), both exothermic processes. Therefore, the correct answer is option D.) II and III only.
Explanation:The question asks for which processes the value of ΔH (enthalpy change) is expected to be negative, indicating an exothermic reaction where heat is released. We can analyze each process one by one:
I. When calcium chloride dissolves in water and the temperature increases, this process is exothermic. However, it indicates that heat is being released due to the dissolution, not the enthalpy change of water itself.II. When steam condenses to liquid water, it is an exothermic process as it releases heat.III. When water freezes, heat is released during the phase transition from liquid to solid, thus it is also an exothermic process.IV. Sublimation of dry ice is an endothermic process because it requires heat to transition from solid to gas.Therefore, the processes where ΔH is expected to be negative (exothermic) are options II and III, where steam condenses and water freezes.
In addition to running aspartic acid, phenylalanine, aspartame and hydrolyzed aspartame, let's assume that you also ran leucine on your TLC plate in this lab. Would you expect leucine to travel farther than or less than aspartic acid, and why?
A. Leucine is nonpolar, thus it should travel farther than aspartc acid on my TLC plate
B. Leucine is polar, thus i should travel less than aspartic acid on my TLC plate
C. Leucine is polar, thus it should travel farther than aspartic acid on my TLC plate
D. Leucine is nonpolar, thus should travel less than aspartic acid on my TLC plate.
Final answer:
Leucine, being nonpolar, should travel farther than the polar amino acid aspartic acid on a TLC plate because nonpolar substances have a higher affinity for the stationary phase and a lower affinity for the polar mobile phase.
Explanation:
On a thin-layer chromatography (TLC) plate, the distance that a compound travels is closely related to its polarity. In this lab experiment, leucine is a nonpolar amino acid, while aspartic acid is polar due to its acidic side chain. Therefore, leucine should travel farther on the TLC plate than aspartic acid because the nonpolar amino acids have a higher affinity for the nonpolar stationary phase and a lower affinity for the polar mobile phase, so they're carried less distance by the solvent front.
The correct answer to this question is A. Leucine is nonpolar, thus it should travel farther than aspartic acid on my TLC plate.
Atoms of Mg combine with atoms of F to form a compound. Atoms of which of the following elements combine with atoms of F in the same ratio?
Al
Li
Cl
Ba
Answer:
Ba
Explanation:
Magnesium is the element of second group and third period. The electronic configuration of magnesium is - 2, 8, 2 or [tex]1s^22s^22p^63s^2[/tex]
There are 2 valence electrons of magnesium.
Fluorine is the element of seventeenth group and second period. The electronic configuration of fluorine is - 2, 7 or [tex]1s^22s^22p^5[/tex]
There is 1 valence electron of fluorine.
They will combine and form [tex]MgF_2[/tex]
The same type of compound is formed by the other members of the group 2 since they will have 2 valence electron.
Group 2 includes:- beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra).
Hence, correct options is:- Ba
The atom that will combine with F in the same ratio as it combines with Mg is Ba.
We must recall that F combines with atoms of elements group 2 to form ionic compounds of the sort MX2 where M is the metal.
Mg and Ba all belong to group 2 hence they will form compounds with F in the ratio of 1:2
Therefore, the atom that will combine with F in the same ratio as it combines with Mg is Ba.
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24.711 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 79.741 g of water. A 11.169 g aliquot of this solution is then titrated with 0.1045 M HCl . It required 28.42 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.
Final answer:
The weight percent of NH3 in the aqueous waste was calculated from the titration data to be 0.2047%. The moles of NH3 were found to be 0.0029711 mol, corresponding to 0.050585g NH3 in the original waste sample.
Explanation:
The weight percent of NH3 in the aqueous waste is calculated using titration analysis. First, determine the moles of HCl that reacted:
# mol HCl = 0.02842 L × 0.1045 M = 0.0029711 mol HCl.
The reaction between HCl and NH3 is:
NH3(aq) + HCl(aq) → NH4Cl(aq).
Hence, 1 mole of HCl reacts with 1 mole of NH3. Therefore, the moles of NH3 in the aliquot are also 0.0029711 mol. To find the mass of NH3:
Mass of NH3 = moles × molar mass
= 0.0029711 mol × 17.031 g/mol
= 0.050585 g NH3.
Since the aliquot is part of the diluted waste, and assuming the ammonia is evenly distributed, we calculate the weight percent in the original waste sample:
Weight percent = (mass of NH3 in aliquot / mass of waste sample) × 100
= (0.050585 g / 24.711 g) × 100
= 0.2047%.
When 0.485 g of compound X is burned completely in a bomb calorimeter containing 3000 g of water, a temperature rise of 0.285°C is observed. What is ΔU of the reaction for the combustion of compound X? The hardware component of the calorimeter has a heat capacity of 3.81 kJ/°C. The specific heat of water is 4.184 J/g·°C, and the MW of X is 56.0 g/mol.
-538 kJ/mol
4660 kJ/mol
-4660 kJ/mol
538 kJ/mol
Answer:
ΔU = -538kJ/mol
Explanation:
Step 1: Data given
Mass of compound X = 0.485 grams
Mass of water = 3000 grams
Temperature rise = 0.285 °C
Heat capacity of the calorimeter = 3.81 kJ/°C
Specific heat of water = 4.184 J/g°C
MW of X = 56.0 g/mol
Step 2: Calculate q
∆U = ΔH - PΔV
Since the bomb calorimeter has a constant volume ΔV = 0
⇒ ΔU = ΔH
qrxn = - (q(water) + q(bomb))
q(bomb) = 3810 J/°C * 0.285 = 1085.85 J
q(water) = 3000g * 4.184 J/g°C * 0.285°C = 3577.32 J
qrxn = q(water) + q(bomb)
qrxn = 4663.17 J = 4.66 kJ (Since this is an exothermic, the heat is released. (q is positive, ΔH is negative).
Step 3: Calculate moles of compound
Moles = mass / molar mass
Moles = 0.485 grams / 56.0 g/mol
Moles = 0.00866 moles
Step 4: Calculate ΔU
= 4663.17 J /0.00866 moles = 538472 j/mol = 538.5 kJ/mol
Since the reaction is exothermic, ΔU is negative
ΔU = -538kJ/mol