Answer:
C. Burning charcoal
Explanation:
Charcoal is a solid fuel. We normally expect a solid fuel to possess a very high activation energy. Activation energy refers to the energy barrier that must be crossed by reactants in order to be converted into products. A high activation energy implies that the reaction may not be spontaneous at ordinary temperatures and a large external energy must be supplied before the reaction proceeds.
Charcoal is not as spontaneously flammable as gasoline. This implies that a greater energy is required to initiate the combustion of charcoal compared to gasoline. This further buttresses the fact that the activation energy for the combustion of gasoline is less than that for the combustion of charcoal.
The reaction of metals with acid is quite spontaneous at ordinary temperatures for highly reactive metals hence it does not have a high activation energy. The melting of ice is not a chemical reaction hence we can't talk about its activation energy.
Draw the structure 2 butylbutane
Answer:
please look at the picture below.
Explanation:
A(n) _____ reaction occurs when an acid and a base are present in the same solution.
Answer:
The answer is Neutralization reaction
It occurs when an acid and a base are present in the same solution and react to form salt and water only
Hope this helps you
Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.74 oC compared to pure benzene. What is the molar mass of the unknown compound
Answer: The molar mass of the unknown compound is 200 g/mol
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=2.74^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for molecular compound)
[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (benzene)= 0.250 kg
Molar mass of solute = M g/mol
Mass of solute = 26.7 g
[tex]2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}[/tex]
[tex]M=200g/mol[/tex]
Thus the molar mass of the unknown compound is 200 g/mol
The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.
With the addition of the solute to the solution, there has been a depression in the freezing point. The depression in the freezing point can be expressed as:
Depression in freezing point = Van't Hoff factor × Freezing point constant × molality
The molality can be defined as the moles of solute per kg solvent
Molaity = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex]
The depression in freezing point can be given as:
Depression in freezing point = Van't Hoff factor × Freezing point constant × [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex] ......(i)
Given, the depression in freezing point = 2.74 [tex]\rm ^\circ C[/tex]
Van't Hoff factor = 1 (Molecular compound)
Freezing point constant (Kf) = 5.12 [tex]\rm ^\circ C[/tex]/m
Mass of solute = 26.7 g
Mass of solvent = 0.250 kg
Substituting the values in equation (i):
2.74 [tex]\rm ^\circ C[/tex] = 1 × 5.12
[tex]\rm \dfrac{2.74}{5.12}[/tex] = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{26.7}{0.250\;kg}[/tex]
0.535 = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;106.8[/tex]
Molecular mass of solute = [tex]\rm \dfrac{106.8}{0.535}[/tex] g/mol
Molecular mass of solute = 199.626 g/mol
The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.
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Based on the bond energies for the reaction below, what is the enthalpy of the reaction? H₂ (g) + N₂ (g) + 2 C (g) → 2 HCN (g)
Answer:
-1222 kj
Explanation:
You can calculate the bond dissociation energy for each species using the table. Subtract the energies of the bonds made from the energies of the bonds broken. Remember to use the coefficients from the balanced chemical reaction.
BDE = [(H-H) + (N≡ N)] - [2 * [(H-C) + (C≡N)]]
BDE = [(432 kJ) + (942 kJ] - [2 * [(411 kJ) + (887 kJ)]] = -1222 kJ
Based on the bond energies for the given reaction, the enthalpy of the reaction is:
-1222 kj
According to the given question, we are asked to calculate the enthalpy of the given reaction based on the bond energies given in H₂ (g) + N₂ (g) + 2 C (g) → 2 HCN (g).
As a result of this, we can see that bond dissociation energy for each of the species on the table need to be subtracted and then to make use of the coefficient of the balanced chemical reaction.
At the end, we would get 2*(411 kj) + (887 kj) which would give us
-1222 kj
Therefore, the correct answer is -1222 kj
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Draw a depiction of a gas sample, as described by kinetic molecular theory, containing equal molar amounts of helium, neon, and krypton. Use different color dots to represent each element. Give each atom a "tail" to represent its velocity relative to the others in the mixture.
Answer:
The lightest gas will have the greatest molecular speed as shown
Explanation:
According to the kinetic theory of gases, gas molecules are in constant motion and frequently collide with each other and the walls of the vessel. Gas molecules have negligible intermolecular forces of attraction between them hence they move at a very high speed.
However, the distribution of molecular velocities in a mixture of gases depends on the relative molecular masses of the gases in the mixture. Given helium, neon and krypton; the order of decreasing masses and increasing molecular velocities is ; krypton<neon<helium. Hence helium being the lightest gas will have the greatest molecular velocity.
The molecular velocities of each gas has been represented using the length of arrows with different colours in the image attached. The colour codes are;
Red- krypton
Blue- neon
Black-helium
It can be seen that helium has the highest molecular velocity followed by neon and lastly, krypton.
What is the main side reaction that competes with elimination when a primary alkyl halide is treated with alcoholic potassium hydroxide, and why does this reaction compete with elimination of a primary alkyl halide but not a tertiary alkyl halide
Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
A beaker has 0.2 M of Na2SO4. What will be the concentration of sodium and sulfate ions?
Answer:
Na+-0.0648M
SO4²-0.1352M
Explanation:
First find the R.A.M of the sodium sulphate
(23×2)+(32)+(16×4)=142
Find the portion of sodium ions=46/142×0.2
=0.0648M
Sulphate=
0.2-0.0648=0.1352M
Enter an equation for the formation of CaCO3(s) from its elements in their standard states. Enter any reference to carbon as C(s). Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
CaF2 + CO3- ----> CaCO3 + 2 F-
Explanation:
The chemical compounds found on the left side of the date are the reagents and those found on the right are the products, where calcium carbonate appears.
Calcium carbonate is a quaternary salt
For the reaction, 2SO2(g) + O2(g) <--> 2SO3(g), at 450.0 K the equilibrium constant, Kc, has a value of 4.62. A system was charged to give these initial concentrations, [SO3] = 0.254 M, [O2] = 0.00855 M, [SO2] = 0.500 M. In which direction will it go?
Answer:
To the left.
Explanation:
Step 1: Write the balanced reaction at equilibrium
2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)
Step 2: Calculate the reaction quotient (Qc)
Qc = [SO₃]² / [SO₂]² × [O₂]
Qc = 0.254² / 0.500² × 0.00855
Qc = 30.2
Step 3: Determine in which direction will proceed the system
Since Qc > Kc, the system will shift to the left to attain the equilibrium.
What is the pH of a solution prepared by dissolving 0.140 g of potassium hydroxide in sufficient pure water to prepare 250.0 ml of solution
Answer:
pH= 12
Explanation:
Potassium hydroxide (KOH) is a strong base, so it dissociates completely in water by giving OH⁻ anions as follows:
KOH⇒ K⁺ + OH⁻
Since dissociation is complete, it is assumed that the concentration of OH⁻ is equal to the initial concentration of KOH:
[OH⁻]= [KOH]
In order to find the initial concentration of KOH, we have to divide the mass (0.140 g) into the molecular weight of KOH (Mw):
Mw (KOH)= K + O + H = 39 g/mol + 16 g/mol + 1 g/mol = 56 g/mol
moles KOH: mass/Mw= 0.140 g/(56 g/mol) = 2.5 x 10⁻³ moles
The molality of the solution is the number of moles of KOH per liter of solution:
V= 250.0 ml x 1 L/1000 ml= 0.250 L
M = (2.5 x 10⁻³moles)/(0.250 L)= 0.01 M
Now, we calculate pOH:
pOH = -log [OH⁻]= - log [KOH]= -log (0.01) = 2
Finally, we calculate pH from pOH:
pH + pOH = 14
⇒pH = 14 - pOH= 14 -2 = 12
When a mixture of PH3, P2, and H2 comes to equilibrium at 873 K, PP2= 0.422 atm and PH2= 0.832 atm. What is PPH3? Express your answer with the appropriate units.
Answer:
0.0243 atm
Explanation:
1. Chemical equation
2PH₃ ⇌ P₂ + 3H₂
E/atm: x 0.417 0.826
2. Kₚ expression
[tex]K_{\text{p}} = \dfrac{p_{\text{P}_{2}}p_{\text{H}_{2}}^{3}}{p_{\text{PH}_{3}}^{2}} = 398[/tex]
Step 3. Calculate the equilibrium concentrations
[tex]\begin{array}{rcl}\\\dfrac{p_{\text{P}_{2}}p_{\text{H}_{2}}^{3}}{p_{\text{PH}_{3}}^{2}}&=&398\\\\\dfrac{0.417 \times 0.826^{3}}{p_{\text{PH}_{3}}^{2}}&=&398\\\\0.2350&=&398 \times{p_{\text{PH}_{3}}^{2}}\\p_{\text{PH}_{3}}^{2}&=& \dfrac{0.2350}{398}\\\\&=&5.905 \times 10^{-4}\\p_{\text{PH}_{3}}&=& \textbf{0.0243 atm}\end{array}\\\text{The equilibrium pressure of PH$_{3}$ is $\large \boxed{\textbf{0.0243} \textbf{ atm}}$}[/tex]
Check:
[tex]\begin{array}{rcl}\dfrac{0.417 \times 0.826^{3}}{0.0243^{2}}&=&398\\\\398 & = & 398\\\end{array}[/tex]
It checks.
Do you think you could go a week without causing any chemical reactions?
yes yes yes yes
yes
yes
yes
yes
yes
Which of the following are meso compounds? A) trans-1,4-dimethylcyclohexane B) cis-1,3-dimethylcyclohexane C) trans-1,3-dimethylcyclohexane D) cis-1,4-dimethylcyclohexane E) trans-1,2-dimethylcyclohexane
Answer:
See explanation
Explanation:
For this question, we have to remember the definition of a meso-compound. In a meso-compound, we will have chiral carbons but we don't optical activity. This is due to the symmetry, if we have symmetry in a substance with chiral carbons the optical activity is nullified. So, if we want to find the meso-compounds we have to find symmetry planes in the molecule.
A symmetry plane is an imaginary cut that can divide the molecule in two equal parts. We have to draw the molecule first (see figure 1) and then we can try to find the symmetry planes.
With this in mind, the only compounds with symmetry planes are:
b) cis-1,3-dimethylcyclohexane
d) cis-1,4-dimethylcyclohexane
See figure 2 to more explanations
I hope it helps!
What happened to the limewater in
the experiment? What does this
prove?
Explanation:
In the reaction above, we clearly see that a chemical reaction took place but what kind of reaction you might ask?
Looks like it is Thermal decomposition of copper carbonate because decomposition reaction takes place due to the added heat. Decomposition in simple terms in chemical breakdown and here this result is obtained by adding heat.
1) What happened to the lime water?
Although not pictured above, but my assumption is that lime water turned milky or turbid because when CO2 comes in presence of limewater they react to form a percipitate of Calcium carbonate which is the milky color that you get.
2) What does this prove?
- My understanding would be that it proves that CO2 was formed, and that most metal carbonates undergo thermal decomposition into metal oxide and carbon dioxide, and also that a reaction took place since new products were made.
Which substance is a base? HCOOH RbOH H2CO3 NaNO3
Answer:
RbOH
Explanation:
For this question, we have to remember what is the definition of a base. A base is a compound that has the ability to produce hydroxyl ions [tex]OH^-[/tex], so:
[tex]AOH~->~A^+~+~OH^-[/tex]
With this in mind we can write the reaction for each substance:
[tex]HCOOH~->~HCOO^-~+~H^+[/tex]
[tex]RbOH~->~Rb^+~+~OH^-[/tex]
[tex]H_2CO_3~->~CO_3^-^2~+~2H^+[/tex]
[tex]NaNO_3~->~Na^+~+~NO_3^-[/tex]
The only compound that fits with the definition is [tex]RbOH[/tex], so this is our base.
I hope it helps!
RbOH or rubidium hydroxide is a base.
• RbOH or rubidium hydroxide is the inorganic compound.
• It comprises hydroxide anions and an equal number of rubidium cations.
• Rubidium hydroxide, like other strong bases is highly corrosive.
• The formation of rubidium hydroxide takes place when the metal rubidium reacts with water.
• The bases refers to the substance, which gets dissociate in an aqueous solution to produce OH- or hydroxide ions.
• Rubidium hydroxide is a base, when it is dissolved in water it give rise to an alkali, when reacts with acid it generates a rubidium salt.
Thus, RbOH is a base.
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Write the condensed electron configurations for the Ca atom. Express your answer in condensed form as a series of orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2.
Answer:
[Ar] 4s²
Explanation:
Ca is the symbol for Calcium. It is the 20th element and it has 20 electrons.
The full electronic configuration for calcium is given as;
1s²2s²2p⁶3s²3p⁶4s²
The condensed electronic configuration is given as;
[Ar] 4s²
The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 63 s for the gas to effuse, whereas nitrogen gas (N2) required 48 s. The molar mass of the unknown gas is-
Answer:
8.13 g/mol.
Explanation:
The following formula gives us the relationship between the effusion rates of two gases and their molar masses:
[tex]\sqrt{\frac{MM_{x} }{MM_{y} } } = \frac{rate_{y} }{rate_{x} }[/tex]
where x and y are respective sample gases and MM and rate are molar mass and rate of effusion respectively.
⇒[tex]\sqrt{\frac{14}{y} } = \frac{63}{48}[/tex]
[tex]\frac{14}{y} = 1.3125^{2}[/tex]
y= 14 / [tex]1.3125^{2}[/tex] = 8.13 g/mol.
2.
Name the following compounds:
a. Rb20
Answer:
Rubidium oxide
Explanation:
Attractive forces between molecules in a solid are ______ than bonds between atoms in a molecule.
Answer:
Stronger
Explanation:
Attractive forces between molecules in a solid are weaker than bonds between atoms in a molecule.
What do you mean by attractive forces in solid?Solid is a state of matter in which the force of attraction between the particles is very high and the space between the particles is negligible. Solids are hard and have fixed shape, size, and volume .
Due to high stiffness and toughness with less intermolecular force in solid the attractive force in solids are very high.
Hence, attractive forces between molecules in a solid are weaker than bonds between atoms in a molecule.
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What is the standard cell potential for the spontaneous voltaic cell formed from the given half-reactions
Answer:
because it is
differentiate between satured and unsatured fats
Answer:
...
Explanation:
in saturated fats there is no double bond between the acids and are tightly packed and unsaturated fats arent tight and loosely packed/put together
saturated- solid at room temperature
unsaturated= liquid at room temperature
two types of unsaturated fats, Polyunsaturated fats and Monounsaturated fats
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The Ka of hydrazoic acid = 1.9 x 10-5.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
[tex]HN_3 + OH- ---> N^-_{3} + H_2O[/tex]
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of [tex]HN_3[/tex] = [tex]\dfrac{0.001755}{0.0383}[/tex] = 0.0458 M
Concentration of [tex]N^{-}_3[/tex] = [tex]\dfrac{ 0.001995 }{0.0383}[/tex] = 0.0521 M
GIven that :
Ka = [tex]1.9 x 10^{-5}[/tex]
Thus; it's pKa = 4.72
[tex]pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})[/tex]
[tex]pH =4.72 + log(1.1376)[/tex]
[tex]pH =4.72 + 0.05598[/tex]
[tex]pH =4.77598[/tex]
pH ≅ 4.80
The pH of the solution 0.150 M hydrazoic acid after 13.3 mL of NaOH base is added is 4.80.
How we calculate the pH?pH of the given solution will be used by using the following equation:
pH = pKa + log[conjugate base] / [weak acid]
Given chemical reaction will be represented as:
HN₃ + OH⁻ → N₃⁻ + H₂O
Moles will be calculated as:
n = M×V, where
M = molarity
V = volume
Moles of 0.150 M hydrazoic acid = (0.150M)(0.025L) = 0.00375 mol
Moles of 0.150 M NaOH = (0.0133)(0.150) = 0.001995 mol
From the above calculation it is clear that moles of hydrazoic acid is present in excess and it will be:
0.00375 - 0.001995 = 0.001755 mol
And 0.001995 mol of N₃⁻ is preduced by the reaction.
Total volume of the solution = 0.025 + 0.0133 = 0.0383 L
To calculate the pH after titration, first we have to calculate the concentration in terms of molarity of N₃⁻ and HN₃ as:
[N₃⁻] = 0.001995 mol / 0.0383 L = 0.0521 M
[HN₃] = 0.001755 mol / 0.0383 L = 0.0458 M
Ka for HN₃ = 1.9 × 10⁻⁵
pKa = -log( 1.9 × 10⁻⁵ ) = 4.72
On putting all these values on the above equation, we get
pH = 4.72 + log (0.0521) / (0.0458)
pH = 4.80
Hence, pH of the solution is 4.80.
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Think about what you know about science today. How do you think scientific knowledge will be different in 100 years?
Answer:
I think we will know a lot more about the universe and the things around us. We may also know a lot more about other planets, such as Mars and Saturn, and we might also know a lot more about other stars in the universe.
Explanation:
An aqueous solution of glucose (C6H12O6), called D5W, is used for intravenous injection. D5W contains 54.30 g of glucose per liter of solution. What is the molar concentration of glucose in D5W
Answer:
The correct answer is 0.30 M
Explanation:
The molar concentration or molarity of a solution is defined as moles of solute per liter of solution. We found the moles of solute (glucose) by dividing the mass (54.30 g) into the molecular weight (MW) of glucose (C₆H₁₂O₆):
MW(C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol
Moles of glucose= mass/MW= 54.30 g/(180 g/mol)= 0.30 mol
There is 0.30 mol of solute per liter of solution, thus the molarity is:
M= moles solute/L solution= 0.30 mol/1 L = 0.30 M
2) Which type movement do pivot joints allow?
Determine the the mass of one molecule of hydrogen sulfide gas.
Answer:
the molecular mass of hydrogen sulphide, which contains two atoms of hydrogen and one atom of sulphur is = 2 — 1 + 1 — 32 = 34 a.m.u.
Based on the properties of the compounds in the interactive, predict whether the given compounds behave as electrolytes or as nonelectrolytes.
1. LioH
2. C4H2O4
3. LiBr
4. HNo3
Explanation:
Before proceeding we have to understand what electrolytes and non electrolytes are;
An electrolyte is a substance that produces an electrically conducting solution when dissolved. An electrolyte is a compound that can dissociate into ions.
Non electrolytes: A substance whose molecules in solution do not dissociate to ions and thus do not conduct an electric current
Going through the options;
1. LiOH
This is a compound of hat would dissociate into Li+ and OH-. This is an electrolyte.
2. C4H2O4
This is an organic compound. Gnerally organic acids are non electrolytes, with the exception og the acids. This is a nonelectrolyte.
3. LiBr
This is an electrolyte because it would dissociate into Li+ and Br- ions.
4. HNO3
HNO3 is a strong acid. Because it is a strong acid it will dissociate completely into its ions (H+ and NO3-). Therefore we consider HNO3 to be a strong electrolyte.
Select the oxidation reduction reactions??
Answer:
Explanation:
1 ) Cl₂ + ZnBr₂ = ZnCl₂ + Br₂
In this reaction , oxidation number of Cl decreases from 0 to -1 so it is reduced and oxidation number of Br increases from -1 to 0 so it is oxidised . Hence this reaction is oxidation - reduction reaction .
2 )
Pb( ClO₄)₂ + 2KI = PbI₂ + 2KClO₄
In this reaction oxidation number of none is changing so it is not an oxidation - reduction reaction.
3 )
CaCO₃ = CaO + CO₂
In this reaction also oxidation number of none is changing so it is not an oxidation - reduction reaction.
So only first reaction is oxidation - reduction reaction.
2nd option is correct.
Use the following reactions and given Δh values to find standard enthalpies of reactions (in kilojoules) given below.C(s)+O2(g)→CO2(g) ΔH= -393.5 kJ 2CO(g)+O2(g)→2CO2(g) ΔH= -566.0 kJ2H2(g)+O2(g)→2H2O(g) ΔH= -483.6 kJ
Answer:
The heat of the reaction or standard enthalpy of the reaction CO(g) + H₂O(g) → CO₂(g) + H₂(g) is ΔH(rxn) = -41.2 kJ
Explanation:
The reaction whose standard enthalpy is required, as obtained from the internet is
CO(g) + H₂O(g) → CO₂(g) + H₂(g)
The formation reaction for some of the reactants and products are given as
C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ
2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ
2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -483.6 kJ
To find the standard enthalpies of the given reaction, we need the heat of formation of each of the species involved in the reaction
ΔH (CO₂(g)) = -393.5 kJ
ΔH (H₂O(g)) = -483.6 kJ ÷ 2 = -241.8 kJ (Because 2 moles of H₂O(g) are formed in the given formation reaction)
ΔH (O₂(g)) = ΔH (H₂(g)) = 0 kJ (No heat of formation for elements)
ΔH (CO(g)) = ? (This isn't given)
But it can be calculated from the second given reaction
2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ
Heat of reaction = ΔH(products) - ΔH(reactants)
Heat of reaction = -566.0 kJ
ΔH (products) = 2 × ΔH (CO₂(g)) = 2 × -393.5 = -787 kJ
ΔH (reactants) = [2 × ΔH (CO(g))] + [1 × ΔH (O₂(g))] = 2 × ΔH (CO₂(g))
Hence, we have
-566 = -787 - [2 × ΔH (CO₂(g))]
2 × ΔH (CO₂(g)) = -787 + 566 = -221 kJ
ΔH (CO₂(g)) = -221 ÷ 2 = -110.5 kJ
CO(g) + H₂O(g) → CO₂(g) + H₂(g)
Heat of reaction = ΔH(products) - ΔH(reactants)
ΔH (products) = [ΔH (CO₂(g))] + [ΔH (H₂(g))]
= -393.5 + 0 = -393.5 kJ
ΔH (reactants) = [1 × ΔH (CO(g))] + [1 × ΔH (H₂O(g))] = -110.5 - 241.8 = -352.3 kJ
Heat of reaction = -393.5 - (-352.3) = -41.2 kJ
Hope this Helps!!!
The constant pressure molar heat capacity of argon, C_{p,m}C
p,m
, is
20.79\text{ J K}^{-1}\text{ mol}^{-1}20.79 J K
−1
mol
−1
at 298\text{ K}298 K. What
will be the value of the constant volume molar heat capacity of argon,
C_{V,m}C
V,m
, at this temperature?
Answer:
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹
Explanation:
Argon is a monoatomic gas that behaves as an ideal gas at 298K.
Using the first law of thermodinamics you can obtain:
Work, Q, for constant pressure molar heat capacity,CP:
CP = (5/2)R
For constant-volume molar heat capacity,CV:
CV = (3/2)R
That means:
2CP/5 = 2CV/3
3/5 = CV / CP
As CP of Argon is 20.79 J K ⁻¹mol⁻¹, CV will be:
3/5 = CV / CP
3/5 = CV / 20.79 J K ⁻¹mol⁻¹
12.47 J K ⁻¹mol⁻¹ = CV
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹