Answer: 72.4 kJ/mol
Explanation:
The balanced chemical reaction is,
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)[/tex] [tex]\Delta H=-2220.1kJ/mol[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})][/tex]
where,
n = number of moles
[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
[tex]-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})][/tex]
[tex]\Delta H_{C_3H_8}=72.4kJ/mol[/tex]
Therefore, the heat of formation of propane is 72.4 kJ/mol
The heat of formation of propane can be found using Hess's Law and the given enthalpy formations of CO2 and H2O, along with the enthalpy combustion of propane. Calculations reveal a heat of formation for propane of -103.85 kJ/mol.
Explanation:The heat of formation of propane can be calculated using Hess's Law, which states that the enthalpy change in a chemical reaction is independent of the pathway between the initial and final states. In this case, we know the enthalpy of combustion for propane (C3H8) is -2220.1kJ/mol, and that the standard enthalpy of formation (ΔHf°) for CO2 and H2O are -393.5kJ/mol and -241.8kJ/mol, respectively.
Therefore, by rearranging the equation ΔH(combustion) = Σ(ΔHf° products) - Σ(ΔHf° reactants), we can solve for the ΔHf° of propane.
Inserting the given values into the equation, -2220.1 kJ/mol = [ 3(-393.5 kJ/mol) + 4(-241.8 kJ/mol)] - [x + 5(0)], where x represents the ΔHf° of propane, and '5(0)' is the contribution from the oxygen, which is zero because the change in enthalpy formation for an element in its stable form is zero. Thus, solving for x, you get a heat of formation for propane of -103.85 kJ/mol.
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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.
A 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 MUse the partition function for a harmonic oscillator, q = 1 / (1- e^-hv/kBT) , to determine the ratio of populations in the lowest and first excited energy levels, P0/P1 , for the indicated systems at two temperatures, 50 K and 298 K :
(a) The 1700 cm−1 stretching vibrational mode for the carbonyl (C=O) group in the peptide bond of a polypeptide.
(b) the 150 cm−1 torsional vibrational mode for the C=C bond in trans-2-butadiene.
Answer
The answer and procedures of the exercise are attached in the following archives.
Explanation
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
why a solution is a home generous mixture?
The following reaction is exothermic.
2 S(s) + 3 O2(g) → 2 SO3(g)
What can we say about the spontaneity of this reaction?
(A) spontaneous at all temperatures
(B) spontaneous only at high temperatures
(C) spontaneous only at low temperatures
(D) non spontaneous at all temperatures
(E) more information is need to predict if the reaction is spontaneous
Answer : The correct option is, (C) spontaneous only at low temperatures.
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
As we know that:
[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
The given chemical reaction is:
[tex]2S(s)+3O_2(g)\rightarrow 2SO_3(g)[/tex]
As we are given that, the given reaction is exothermic that means the enthalpy change is negative.
In this reaction, the randomness of reactant molecules are more and as we move towards the formation of product the randomness become less that means the degree of disorderedness decreases. So, the entropy will also decreases that means the change in entropy is negative.
Now we have to determine the spontaneity of this reaction when ΔH is negative and ΔS is negative.
As, [tex]\Delta G=\Delta H-T\Delta S[/tex]
[tex]\Delta G=(-ve)-T(-ve)[/tex]
[tex]\Delta G=(+ve)[/tex] (at high temperature) (non-spontaneous)
[tex]\Delta G=(-ve)[/tex] (at low temperature) (spontaneous)
Thus, the reaction is spontaneous only at low temperatures.
The mass of a proton is 1.00728 amu andthat of a neutron is 1.00867 amu. What is the binding energy pernucleon (in J) of a Co nucleus? (The mass of a cobalt-60 nucleus is59.9338 amu.) a. 3.039× 10^-12 b. 2.487 × 10^-12 c. 7.009 × 10^-14 d. 1.368 × 10^-12 e. 9.432 × 10^-13
Answer:
The binding energy per nucleon = 1.368*10^-12 (option D)
Explanation:
Step 1: Data given
The mass of a proton is 1.00728 amu
The mass of a neutron is 1.00867 amu
The mass of a cobalt-60 nucleus is59.9338 amu
Step 2: Calculate binding energy
The mass defect = the difference between the mass of a nucleus and the total mass of its constituent particles.
Cobalt60 has 27 protons and 33 neutrons.
The mass of 27 protons = 27*1.00728 u = 27.19656 u
The mass of 33 neutrons = 33*1.00867 u = 33.28611 u
Total mass of protons + neutrons = 27.19656 u + 33.28611 u = 60.48267 u
Mass of a cobalt60 nucleus = 59.9338 amu
Mass defect = Δm = 0.54887 u
ΔE =c²*Δm
ΔE = (3.00 *10^8 m/s)² *(0.54887 amu))*(1.00 g/ 6.02 *10^23 amu)*(1kg/1000g)
Step 3: Calculate binding energy per nucleon
ΔE = 8.21 * 10^-11 J
8.21* 10^-11 J / 59.9338 = 1.368 *10^-12
The binding energy per nucleon = 1.368*10^-12 (option D)
To calculate the binding energy per nucleon of a Co nucleus, we calculate the mass defect, use Einstein's equation to calculate the binding energy, and then divide by the number of nucleons in the nucleus. The binding energy per nucleon can be converted to MeV using a conversion factor.
Explanation:To calculate the binding energy per nucleon (in J) of a Co nucleus, we first need to determine the total binding energy. The mass defect for a Co nucleus is the difference between the total mass of the nucleus and the sum of the masses of the individual protons and neutrons. The mass defect can be calculated by subtracting the actual mass of the Co nucleus (59.9338 amu) from the sum of the individual masses of protons (1.00728 amu) and neutrons (1.00867 amu). Once we have the mass defect, we can use Einstein's equation E=mc² to calculate the binding energy. Finally, to find the binding energy per nucleon, we divide the total binding energy by the number of nucleons in the nucleus.
Using the given values, we have:
Mass defect (Δm) = (59.9338 amu) - (27 * 1.00728 amu + 33 * 1.00867 amu)
Binding energy (E) = Δm * (c²) = Δm * (3 × 10^8 m/s)²
Binding energy per nucleon = E / number of nucleons in Co nucleus
Substituting the values and performing the calculations, we get the binding energy per nucleon in joules. We can then convert the binding energy per nucleon from joules to MeV by using the conversion factor, 1 MeV = 1.602 × 10^-13 J.
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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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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].
Learn more about bonds refer:
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.
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.
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 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.
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
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!
A 5.00 mL of a salt solution of unknown concentration was mixed with 35.0 mL of 0.523 M AgNO3. The mass of AgCl solid formed was 0.508 g. _ AgNO3(aq) + _ NaCl(aq) → _ AgCl(s) + _ NaNO3(aq) What is the chemical term for the silver chloride solid that is formed?
Answer:
The chemical term in the equation for the precipitate of AgCl(s) is n=3.54*10^-3
Explanation:
the quantity of AgCl(s) in moles is:
n = 0.508g / 143.32 g/mol = 3.54*10^-3 mol
to verify it the mass of AgNO3 involved in the reaction should be
n AgNO3 required = n = 3.54*10^-3 mol
the mass of n involved should be higher than n AgNO3
n existing = V*N = 0.523 mol/L * 35*10^-3 L = 18.305*10^-3 mol
The chemical term of the reaction
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Is when appears red brick in the solution indicating the formation of the precipitate
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
The thiocyanate polyatomic ion, SCN-, is commanly called a pseudohalogen because it acts very much like halide ions. For example, we know that pure halogens consists of diatiomic molecules, such as C12. Thiocanate ions form similar molecules in the following reaction2NaSCN +2H2SO4 + MnO2 -> (SCN)2 + 2H2O + MnSO4 + Na2SO4A) Write a conversion factor that could be used to convert between moles of NaSCN and moles of (SCN)2B) How many moles of (SCN)2 form when 0.05 moles of NaSCN react completely?C) What is the maximum number of moles of (SCN)2 that could form in the combination of 4 moles of NaSCN and 3 moles of MnSO4?D) Write a convertion factor that could be used to convert between moles of sulfuric acid, H2SO4 and moles manganese (II) sulfate, MnSO4.E) What is the minimum number of moles of H2SO4 that must react to form 1.7752 moles of manganese (II) sulfate?
Answer:
A) [tex]\frac{1mol(SCN)_{2}}{2molNaSCN}[/tex]
B) 0.025 mol (SCN)₂
C) 2 mol (SCN)₂
D) [tex]\frac{1molMnSO_{4}}{2molH_{2}SO_{4}}[/tex]
E) 3.5504 mol H₂SO₄
Explanation:
2NaSCN +2H₂SO₄ + MnO₂ → (SCN)₂ + 2H₂O + MnSO₄ + Na₂SO₄
A) A conversion factor could be [tex]\frac{1mol(SCN)_{2}}{2molNaSCN}[/tex] , as it has the units that we want to convert to in the numerator, and the units that we want to convert from in the denominator.
B) 0.05 mol NaSCN * [tex]\frac{1mol(SCN)_{2}}{2molNaSCN}[/tex] = 0.025 mol (SCN)₂
C) With 4 moles of NaSCN and 3 moles of MnSO₄, the reactant is NaSCN so we use that value to calculate the moles of product formed:
4 mol NaSCN * [tex]\frac{1mol(SCN)_{2}}{2molNaSCN}[/tex] = 2 mol (SCN)₂
D) [tex]\frac{1molMnSO_{4}}{2molH_{2}SO_{4}}[/tex]
E) 1.7752 mol MnSO₄ * [tex]\frac{2molH_{2}SO_{4}}{1molMnSO_{4}}[/tex] = 3.5504 mol H₂SO₄
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
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.
If 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).
An experimental spacecraft consumes a special fuel at a rate of 353 L/min . The density of the fuel is 0.700 g/mL and the standard enthalpy of combustion of the fuel is − 57.9 kJ/g . Calculate the maximum power (in units of kilowatts) that can be produced by this spacecraft. 1 kW = 1 kJ/s
Explanation:
The given data is as follows.
Space craft fuel rate = 353 L/min
As 1 liter equals 1000 ml and 1 min equals 60 seconds.
So, [tex]353 \times \frac{1000 ml}{60 sec}[/tex]
= 5883.33 ml/sec
It is also given that density of the fuel is 0.7 g/ml and standard enthalpy of combustion of fuel is -57.9 kJ/g.
Fuel rate per second is 5883.33 ml.
[tex]5883.33 ml \times 0.7 g/ml[/tex]
= 4118.33 g
Hence, calculate the maximum power as follows.
Power = Fuel consumption rate × (-enthalpy of combustion)
= 4118.33 g/s \times 57.9 kJ/g
= 238451.36 kJ/s
or, = 238451.36 kW
Thus, we can conclude that maximum power produced by given spacecraft is 238451.36 kW.
A 6.22-kg piece of copper metal is heated from 20.5 °C to 324.3 °C. The specific heat of Cu is 0.385 Jg-1°C-1.a.Calculate the heat absorbed (in kJ) by the metal.b.How close is this heat capacity to the expected heat capacity at the classical limit?
Answer:
a) 727.5 kJ
Explanation:
Step 1: Data given
Mass of the piece of copper = 6.22 kg
Initial temperature of the copper = 20.5 °C
Final temperature of the copper = 324.3 °C
Specific heat of copper = 0.385 J/g°C
Step 2:
Q = m*c*ΔT
⇒ with Q = heat transfer (in J)
⇒ with m = the mass of the object (in grams) = 6220 grams
⇒ with c = the specific heat capacity = 0.385 J/g°C
⇒ with ΔT = T2 -T1 = 324.3 - 20.5 = 303.8
Q = 6220 grams * 0.385 J/g°C * 303.8 °C
Q = 727509.9 J = 727.5 kJ
b) This heat capacity is the heat capacity given for a copper at a temperature of 25°C
The heat absorbed by the copper metal is 727.5 kJ.
Explanation:To calculate the heat absorbed by the copper metal, we can use the formula Q = mcΔT, where Q is the heat absorbed, m is the mass of the copper, c is the specific heat of copper, and ΔT is the change in temperature. Plugging in the given values, we get Q = 6.22 kg * 0.385 J/g°C * (324.3°C - 20.5°C).
Simplifying the equation, we get Q = 6.22 kg * 0.385 J/g°C * 303.8°C. Converting grams to kilograms, and simplifying further, we get Q = 727.5 kJ.
The expected heat capacity at the classical limit is not provided in the information given, so we cannot determine how close the calculated heat capacity is to the expected value.
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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
Use the Gizmo to answer the following questions.
A. How many electrons are in a neutral atom of lithium? ___________
B. How many neutrons are in an atom of Mg-25? ___________
C. What is the mass number of an atom with 5 protons and 7 neutrons? ________
D. An ion is a charged atom. How many electrons are in O2- ? ___________
E. How many electrons are in Mg2+?
Answer:
a. 3 electrons
b. 13 neutrons
c. 12 mass number
d. 10 electrons
e. 10 electrons
Explanation:
A. For an element , the atomic number of a given element is equal to the number of protons , and
For a neutral atom , the number of electrons equals the number of protons ,
Hence ,
The atomic number of Li = 3 ,
Hence ,
Number of electron for Li = 3 .
B.
Number of neutrons = mass number - atomic number ,
For Mg-25
For Magnesium , atomic number = 12 ,
Number of neutrons = 25 - 12 = 13 neutrons
C. Mass number of an element is equal to the sum of number of protons and number of neutrons ,
Hence , mass number = 5 + 7 = 12 .
D. For a negatively charged ion , add the negative charge value with the number of electrons of the neutral atom ,
For Oxygen , atomic mass = 8 ,
hence , number of electrons = 8
For O²⁻ = 8 + 2 = 10 electrons
E. For a positively charged ion , subtract the positive charge value with the number of electrons of the neutral atom ,
For Magnesium , atomic number = 12 ,
hence , number of electrons = 12 ,
for , Mg²⁺ , number of electrons = 12 - 2 = 10 electrons ,
A neutral atom of Lithium has 3 electrons. Mg-25 has 13 neutrons. An atom with 5 protons and 7 neutrons has a mass number of 12. O2- has 10 electrons and Mg2+ has 10 electrons.
Explanation:A. A neutral atom of lithium contains 3 electrons, equal to its atomic number. B. An atom of Mg-25 has 12 protons (since magnesium's atomic number is 12), and as Mg-25 indicates a mass number of 25, subtracting the atomic number from the mass number gives us 13 neutrons. C. The mass number of an atom is the total number of protons and neutrons in its nucleus. So for an atom with 5 protons and 7 neutrons, the mass number is 12. D. O2- is an ion of oxygen that has gained 2 electrons. Oxygen has 8 electrons in its neutral state so O2- has 10 electrons. E. Mg2+ means that magnesium has lost 2 electrons, so Mg2+ has 10 electrons (since in its neutral state it has 12).
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During an exothermic chemical reaction,
A. a system becomes warmer, and the chemical substances undergo a decrease in potential energy.
B. a system becomes warmer, and the chemical substances undergo an increase in potential energy.
C. a system becomes cooler, and the chemical substances undergo a decrease in potential energy.
D. a system becomes cooler, and the chemical substances undergo an increase in potential energy.
E. a system becomes warmer, and additional heat is gained from the surroundings.
An exothermic chemical reaction occurs when a system becomes warmer and the potential energy of the chemical substances increases.
What is Exothermic chemical reaction?An exothermic reaction is one in which energy is released in the form of light or heat. Thus, in an exothermic reaction, energy is transferred into the surroundings rather than taken from the surroundings as in an endothermic reaction. The change in enthalpy ( ΔH ) in an exothermic reaction is negative. Exothermic reactions characterise all combustion reactions. A substance burns as it combines with oxygen during a combustion reaction. When substances burn, they typically emit energy in the form of heat and light. Wood combustion is an exothermic reaction that releases a lot of energy in the form of heat and light.The net amount of energy required to start an exothermic reaction is less than the net amount of energy released by the reaction. The net amount of heat energy that flows through a calorimeter, a device used to measure the heat released by a chemical reaction, is equal to the negative of the total energy change of the system. The absolute total of energy in a given chemical system is extremely difficult to measure or even calculate. As a result, the energy change (or enthalpy change, denoted by ΔH) is used instead. The following equation describes the relationship between the value of ΔH and the reaction's bond energies.ΔH = (energy used in the bond formation that yields products) – (energy released when the reactant bonds are broken)To learn more about exothermic chemical reaction refer :
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Final answer:
The correct answer is A, wherein during an exothermic reaction, the system warms up due to a decrease in potential energy of the substances involved.
Explanation:
The correct answer to this question is option A: a system becomes warmer, and the chemical substances undergo a decrease in potential energy. During an exothermic chemical reaction, energy is released into the surroundings in the form of heat, causing the temperature of the surrounding system to increase. This energy release is due to the reactants possessing higher potential energy than the products; as the reaction occurs, the potential energy stored within the chemical bonds of the reactants is converted into kinetic energy (heat), significantly reducing the potential energy within the resulting products. This principle is foundational in understanding energy changes during chemical reactions, which is crucial in fields ranging from chemical engineering to environmental science.
When 0.103 g of Zn (s) is combined with enough HCl to make 50.0 mL of solution in a coffee cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.5 degree C to 23.7 degree C. Find Delta H_rxn per mole of Zn. (Use 1.02 glmL for the density of the solution and 4.18 J/g degree C for the specific heat capacity.)
Zn (s) + 2 HCl (aq) rightarrow ZnCl_2 (aq) + H_2 (g)
Delta H_rxn/mol Zn = ____kJ/mol.
First, calculate the total heat absorbed by the solution, then divide this value by the number of moles of Zn reacted. The resulting ΔH_rxn is -162.839 kJ/mol.
Explanation:The question asks us to calculate the heat of reaction, or ΔH_rxn, per mole of Zinc (Zn) in the reaction.
First, we calculate the total heat absorbed by the solution using the formula q=mcΔT, where q is the heat absorbed, m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature.
The mass of the solution can be calculated by multiplying the volume of solution by its density, hence m = V . d = 50.0 mL x 1.02 g/mL = 51.0 g. The change in temperature is 23.7°C - 22.5°C = 1.2°C. Therefore, the heat absorbed by the solution, q, can be calculated as q = (51.0 g) * (4.18 J/g°C) * (1.2°C) = 256.536 J.
Then, to find the heat per mole, ΔH_rxn, we divide this value by the number of moles of Zn reacted. The molar mass of Zn is 65.38 g/mol, so 0.103 g is 0.103 g / 65.38 g/mol = 0.001575 mol of Zn. Therefore, ΔH_rxn = (256.536 J) / (0.001575 mol) = -162,839 J/mol = -162.839 kJ/mol. We get a negative value because the reaction is exothermic.
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ΔHrxn per mole of Zn is approximately -162.27 kJ/mol.
To find the enthalpy change (ΔHrxn) per mole of Zn, we need to follow several steps:
Calculate the mass of the solution: The density of the solution is given as 1.02 g/mL for 50.0 mL, so the mass is 1.02 g/mL × 50.0 mL = 51.0 g.Calculate the heat absorbed or released (q) using the formula: q = mass × specific heat capacity × change in temperature.The temperature change is from 22.5 °C to 23.7 °C, so ΔT = 23.7 °C - 22.5 °C = 1.2 °C.Using the specific heat capacity of 4.18 J/g°C, the heat released is: q = 51.0 g × 4.18 J/g°C × 1.2 °C = 255.70 J.Convert the heat absorbed to kJ: 255.70 J × (1 kJ / 1000 J) = 0.2557 kJ.Calculate the moles of Zn: The molar mass of zinc (Zn) = 65.38 g/mol, so moles of Zn = 0.103 g / 65.38 g/mol = 0.001576 moles.Determine ΔHrxn per mole of Zn: ΔHrxn per mole = 0.2557 kJ / 0.001576 moles = 162.27 kJ/mol.Therefore, ΔHrxn per mole of Zn is approximately -162.27 kJ/mol.
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
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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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 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.
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