Answer: The calculated molar enthalpy change of reaction for his second trial was -108 kJ.
Explanation:-
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]
Thus [tex]\text{no of moles}of HCl={1.0M}\times {0.05L}=0.05moles[/tex]
Thus [tex]\text{no of moles}of NaOH={1.0M}\times {0.05L}=0.05moles[/tex]
[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]
Given for second trial:
[tex]\text{no of moles}of HCl={1.0M}\times {0.1L}=0.1moles[/tex]
[tex]\text{no of moles}of NaOH={1.0M}\times {0.1L}=0.1moles[/tex]
0.05 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release heat = 54 kJ
0.1 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release heat =[tex]\frac{54}{0.05}\times 0.1=108kJ[/tex]
Thus calculated molar enthalpy change of reaction for his second trial was -108 kJ.
The electrolysis of molten AlCl 3 for 2.50 hr with an electrical current of 15.0 A produces ________ g of aluminum metal.
Describe the similarities between H3O and NH3. Compare/contrast their shapes and polarities within the context of your answer. These molecules are called isoelectronic. Why
Answer:
Explanation:
[tex]H_3O^+[/tex] also known as hydronium ion is formed as a result of the reaction between an hydrogen proton and a water molecules.
i.e [tex]\mathtt{H^+ + H_2O \to H_3O^+}[/tex]
(molecular geometry for the hydronium ion shows that the lewis structure of hydronium ion possess a three hydrogen ion bonded to a central atom known as oxygen. The oxygen possess a lone pair with a positive ion. So we have three hydrogen atoms and a lone pair attached to the oxygen. We can now say that there are four groups as the steric number in which one of them is a lone pair. This give rise to the trigonal pyramidal shape of the [tex]H_3O^+[/tex] (hydronium ion) with a bond angle of about 109,5°
Similarly, [tex]NH_3[/tex] on the other hand also known as ammonia has a shape that can be also determined by the Lewis structure.
IN ammonia, there are three hydrogen and a lone pairs of electron spreading out as far away from each other from the centre nitrogen. In essence, the valence shell electron pair around hydrogens tend to repel each other. Hence, giving it a trigonal pyramidal shape.
From above the similarities between H3O and NH3 is in their molecular geometry in which both H3O and NH3 have the same shape.
These molecules are called isoelectronic. Why?
Isoelectronic molecules are molecules having the same number of electrons and same electronic configuration structure. As a result H3O and NH3 possess the same number of electrons in the same orbitals and they also posses the same structure.
An atom of 120In has a mass of 119.907890 amu. Calculate the mass defect (deficit) in amu/atom. Use the masses: mass of 1H atom
Answer:
a
Explanation:
answer is a on edg
1.) A sample of neon gas at a pressure of 0.646 atm and a temperature of 242 °C, occupies a volume of 515 mL. If the gas is cooled at constant pressure until its volume is 407 mL, the temperature of the gas sample will be ________°C.
2.) A sample of argon gas at a pressure of 0.633 atm and a temperature of 261 °C, occupies a volume of 694 mL. If the gas is heated at constant pressure until its volume is 796 mL, the temperature of the gas sample will be___________°C.
3.) 0.962 mol sample of carbon dioxide gas at a temperature of 20.0 °C is found to occupy a volume of 21.5 liters. The pressure of this gas sample ismm ____________ Hg.
Answer:1 )T2=134°C 2) T2=339.48°C. 3)
P=817.59 mmHg.
Explanation:
1.Given ;
pressure, P1 of neon gas = 0.646 atm
temperature, T1 =242oC + 273=515oC
Volume, V1 =515ml
Volume V2= 407ml
temperature , T 2= ?
Solution;
And at constant pressure, the volume cools at V2=407 mL at T2=?
From ideal gas equation, PV=nRT
V/T=constant
therefore
V1/V2=T1/T2 = T2=(V2 xT1)/V1
T2=(407 mL x 515 K)/515 mL= 407K.
T2= 407K -273= 134°C. recall 0°C=273 K)
2..Given ;
pressure, P1 of neon gas = 0.633 atm
temperature, T1 =261oC + 273=534oC
Volume, V1 =694ml
Volume V2= 796ml
temperature , T 2= ?
Solution;
And at constant pressure, the volume expands at V2=796mL at T2=?
From ideal gas equation, PV=nRT
V/T=constant
therefore
V1/V2=T1/T2 = T2=(V2 xT1)/V1
T2=(796 mL x 534 K)/694mL= 612.48K.
T2= 612.48K -273= 339.48°C. recall 0°C=273 K
3
Given;
moles of CO2= n=0.962 mol,
temperature T=20°C=20+273 K =293 K,
volume V=21.5 L,
gas constant R at L·mmHg/mol·K= 62.3637 L mmHg mol^-1 K^-1
Using ideal gas equation PV=nRT
P=nRT/V
P=(0.962 mol)x(62.3637mmHg mol^-1 K^-1)x(293 K)/(21.5L)
P=817.59 mmHg.
The K sp for silver(I) phosphate is 1.8 × 10 –18. Determine the silver ion concentration in a saturated solution of silver(I) phosphate.
Answer:
[tex][Ag^+]=4.82x10^{-5}M[/tex]
Explanation:
Hello,
In this case, the dissociation reaction for silver phosphate is:
[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4^{3-}(aq)[/tex]
Therefore, the equilibrium expression is:
[tex]Ksp=[Ag^+]^3[PO_4^{3-}][/tex]
And in terms of the reaction extent [tex]x[/tex] is:
[tex]Ksp=1.8x10^{-18}=(3x)^3(x)[/tex]
Thus, [tex]x[/tex] turns out:
[tex]1.8x10^{-18}=27x^4\\\\x=\sqrt[4]{\frac{1.8x10^{-18}}{27} } \\\\x=1.61x10^{-5}M[/tex]
In such a way, the concentration of the silver ion is:
[tex][Ag^+]=3x=3*1.61x10^{-5}M=4.82x10^{-5}M[/tex]
Best regards.
Explain why only the lone pairs on the central atom are taken into consideration when predicting molecular shape
Answer:
Lone pairs cause more repulsion than bond pairs
Explanation:
A lone pair takes up more space around the central atom than bond pairs of electrons. This is because, a lone pair is attracted to only one nucleus while bond pairs are attracted to two nuclei.
Hence the repulsion between lone pairs is far greater than the repulsion between bond pairs or repulsion between a lone pair and a bond pair. The presence of a lone pair therefore distorts a molecule away from the ideal shape predicted on the basis of the valence shell electron pair repulsion theory.
Lone pairs are found to decrease the observed bond angles in a molecule.
what are the monomers of bakelite
Answer:
Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.
Answer: The monomers of bakelite are formaldehyde and phenol
Explanation:
In a reversible reaction, the endothermic reaction absorbs ____________ the exothermic reaction releases. A. less energy than B. None of these, endothermic reactions release energy C. the same amount of energy as D. more energy than
Answer: C. the same amount of energy as
Explanation:
A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back.
Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.
[tex]A+B\rightleftharpoons C+D[/tex]
Thus if forward reaction is exothermic i.e. the heat is released , the backward reaction will be endothermic i.e. the heat is absorbed and in same amount.
The amount of energy released will be equal and opposite in sign to the energy absorbed in that reaction.
Answer:
C.) the same amount of energy as
Explanation:
I got it correct on founders edtell
Determine whether the following statement about reaction rates is true or false. If the statement is false, select the reason why?
Increasing the temperature of a reaction system decreases the activation energy of the reaction.
Answer:
False
Explanation:
Reaction rates is a field under chemical kinetics that deals with the measure of speed of a chemical reaction. It is the change in the concentration of a reactant or product per unit time.
Activation Energy is a theory been put forward to explain why different chemical reactions proceed at different rates.
Activation Energy theory postulates that for a reactant to transform into a product , the colliding particles or molecules of the reactant must possess a certain amount of energy so as to overcome the reaction barrier.
An important factor which may influence the attainment of activation energy by colliding particles of reactants is the temperature at which the reaction is carried out. The higher the temperature, the greater is the fraction of the reactant particles which possess the activation energy and thus the faster the reaction becomes. SO , in essence increasing the temperature of a reaction system do not decreases the activation energy of the reaction but rather also increases the activation energy of the reaction.
A runner can cover 2.0 miles in 31 minutes, how long would it take for this runner to cover 6.0 Km. Hint (1 mile= 1.609 Km)
The answer to this question is approximately equal to 57.8
Pentanone was treated with excess sodium cyanide in HCl (aq) followed by hydrogen gas has over Pd. This produced:________
A. 2-amino-1-hexanol
B. 1-amino-2-methylpentan-2-ol
C. 1-cyano-1-pentanol
D. 2-aminomethylpentan-1-ol
Answer:
B. 1-amino-2-methylpentan-2-ol
Explanation:
In this case, the first step, we have the attack of the nucleophile cyanide ([tex] CN^-[/tex] produced by sodium cyanide to the carbon on the carbonyl group (C=O) producing a negative charge in the oxygen.
Then HCl protonates the molecule to produce a cyanohydrin. This cyanohydrin can be reduced by the action of hydrogen gas ([tex]H_2[/tex]) in the presence of a catalyst ([tex]Pd[/tex]), producing an amino group. With this in mind, the final molecule is: 1-amino-2-methylpentan-2-ol.
See figure 1 to further explanations
I hope it helps!
2NO + 2H2 ⟶N2 + 2H2O What would the rate law be if the mechanism for this reaction were: 2NO + H2 ⟶N2 + H2O2 (slow) H2O2 + H2 ⟶2H2O (fast)
Answer:
rate = [NO]²[H₂]
Explanation:
2NO + H2 ⟶N2 + H2O2 (slow)
H2O2 + H2 ⟶2H2O (fast)
From the question, we are given two equations.
In chemical kinetics; that is the study of rate reactions and changes in concentration. The rate law is obtained from the slowest reaction.
This means that our focus would be on the slow reaction. Generally the rate law is obtained from the concentrations of reactants in a reaction.
This means our rate law is;
rate = [NO]²[H₂]
Two elements represents by the letter Q and R atomic number 9 and 12 respectively. Write the electronic configuration of R
Answer:
Atomic no = 12 = Mg
Explanation:
It is given that,
The atomic number of two elements that are represented by letter Q and R are 9 and 12.
We need to write the electronic configuration of R. Atomic number shows the number of protons in atom.
For R, atomic number = 12
Its electronic configuration is : 2,8,2
It has two valance electrons in its outermost shell. The element is Magnesium (Mg).
How has the work of chemists affected the environment over the years?
Answer:
Chemistry is one of the causes for global warming, and in some cases it can even cause certain illnesses.
Answer:
Chemists have both hurt the environment and helped the environment by their actions.
Explanation:
<3
The surface temperature on Venus may approach 753 K. What is this temperature in degrees Celsius?
Answer:
461.85 degrees Celsius
Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i? Al2(SO4)3, i = 4 NH4NO3, i = 2 Mg(NO3)2, i = 3 Na2SO4, i = 3 Sucrose, i = 1
Answer:
- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).
Explanation:
Hello,
In this case, since the van't Hoff factor is related with the species that result from the ionization of a chemical compound, we can see that that
- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).
- Ammonium nitrate NH4NO3 dissociates in one ammonium ions and one nitrate ion, therefore, van't Hoff factor is 2 (correct).
- Sodium sulfate Na2SO4 dissociates in two sodium ions and one sulfate, therefore, van't Hoff factor is 3 (correct).
- Sucrose is not ionized, therefore, van't Hoff factor is 1 (correct).
Best regards.
A student ran the following reaction in the laboratory at 242 K: 2NOBr(g) 2NO(g) Br2(g) When she introduced 0.143 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of NOBr(g) to be 0.108 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc
Answer:
1.84 × 10⁻³
Explanation:
Step 1: Write the balanced equation
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
Step 2: Calculate the initial concentration of NOBr
0.143 moles of NOBr(g) are introduced into a 1.00 liter container. The molarity is:
M = 0.143 mol / 1.00 L = 0.143 M
Step 3: Make an ICE chart
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
I 0.143 0 0
C -2x +2x +x
E 0.143-2x 2x x
Step 4: Find the value of x
The equilibrium concentration of NOBr(g) was 0.108 M. Then,
0.143-2x = 0.108
x = 0.0175
Step 5: Calculate the concentrations at equilibrium
[NOBr] = 0.108 M
[NO] = 2x = 0.0350 M
[Br₂] = x = 0.0175 M
Step 6: Calculate the equilibrium constant (Kc)
Kc = [0.0350]² × [0.0175] / [0.108]²
Kc = 1.84 × 10⁻³
Calculate the energy required to heat of 1.50 kg silver from -7.8 C to 15.0 C . Assume the specific heat capacity of silver under these conditions is .0235 J*g^-1*K^-1 . Be sure your answer has the correct number of significant digits.
Answer:
804 J
Explanation:
Step 1: Given data
Mass of silver (m): 1.50 kgInitial temperature: -7.8 °CFinal temperature: 15.0 °CSpecific heat capacity of silver (c): 0.0235J·g⁻¹K⁻¹Step 2: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.0235J·g⁻¹K⁻¹ × (1.50 × 10³g) × [15.0°C-(-7.8°C)]
Q = 804 J
When balancing redox reactions under basic conditions in aqueous solution, the first step is to:________.
a. balance oxygen
b. balance hydrogen
c. balance the reaction as though under acidic conditions
d. none of the above
Answer:
When balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.
Explanation:
Oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.
The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.
The oxidation and reduction half-reactions, in a basic medium, adjust the oxygens and hydrogens as follows:
In the member of the half-reaction that presents excess oxygen, you add as many water molecules as there are too many oxygen. Then, in the opposite member, the necessary hydroxyl ions are added to fully adjust the half-reaction. Normally, twice as many hydroxyl ions, OH-, are required as water molecules have previously been added.
In short, you first adjust the oxygens with OH-, then you adjust the H with H₂O, and finally you adjust the charge with e-
So, when balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.
Answer:
c. balance the reaction as though under acidic conditions
Explanation:
When balancing redox reactions under basic conditions, a good technique is to first balance the reaction as though under acidic conditions. We then adjust the result to reflect the basic conditions.
Calculate the [H+] and pH of a 0.0010 M acetic acid solution. The Ka of acetic acid is 1.76×10−5. Use the method of successive approximations in your calculations.
Answer:
[tex][H^+]=0.000123M[/tex]
[tex]pH=3.91[/tex]
Explanation:
Hello,
In this case, dissociation reaction for acetic acid is:
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
For which the equilibrium expression is:
[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
Which in terms of the reaction extent [tex]x[/tex] could be written as:
[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]
Thus, solving by using a solver or quadratic equation we obtain:
[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]
And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:
[tex][H^+]=0.000123M[/tex]
Now, the pH is computed as follows:
[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]
Best regards.
What volume (in mL) needs to be added to 69.6 mL of 0.0887 M MgF2 solution to make a 0.0224 M MgF2 solution
Answer:
The correct answer is 206 ml.
Explanation:
Based on the given information, the molarity or M₁ of MgF₂ solution is 0.0887 M, the molarity or M₂ of the final solution given is 0.0224 M. The initial volume of V₁ of the solution is 69.6 ml, for finding the final volume of V₂ of the solution, the formula to be used is,
M₁V₁ = M₂V₂
Now putting the values in the formula we get,
0.0887 × 69.6 = 0.0224 M × V₂
V₂ = 0.0887 × 69.6 / 0.0224
V₂ = 275.6 ml
Therefore, the volume in ml added to the initial volume of 69.6 ml to make the molarity of the solution 0.0224 will be,
= 275.6 ml - 69.6 ml = 206 ml
Enter the balanced chemical equation for the reaction of each of the following carboxylic acids with KOH.Part Aacetic acidExpress your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).Part B2-methylbutanoic acid (CH3CH2CH(CH3)COOH)Express your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).Part C4-chlorobenzoic acid (ClC6H4COOH)Express your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).
Answer:
Explanation:
Answer in attached file .
The tosylate of (2R,3S)-3-phenylbutan-2-ol undergoes an E2 elimination on treatment with sodium ethoxide. Draw the structure of the alkene that is produced.
Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.
The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.
See figure 1
I hope it helps!
What's the mass in grams of 0.442 moles of calcium bromide, CaBr2? The atomic
weight of Ca is 40.1 and the atomic weight of Br is 79.9.
A) 452.3 g
B) 53.04 g
C) 44.2 g
D) 88.4 g
Answer:
Below
Explanation:
Let n be the quantity of matter in the Calcium Bromide
● n = m/ M
M is the atomic weight and m is the mass
M of CaBr2 is the sum of the atomic wieght of its components (2 Bromes atoms and 1 calcium atom)
M = 40.1 + 2×79.9
● 0.422 = m/ (40.1+2×79.9)
●0.422 = m/ 199.9
● m = 0.422 × 199.9
● m = 84.35 g wich is 88.4 g approximatively
88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 ,therefore option (d) is correct.
What do you mean by mass ?Mass is the amount of matter that a body possesses. Mass is usually measured in grams (g) or kilograms (kg) .
To calculate mass in grams of 0.442 moles of calcium bromide, CaBr2,
Let n be the quantity of matter in the Calcium Bromide
M is the atomic weight and m is the mass
n = m/ MM of CaBr2 is the sum of the atomic weight of its components
Mass of Ca = 40.1 , Mass of Br = 79.9
M = 40.1 + 2×79.9
0.422 = m/ (40.1+2×79.9)
0.422 = m/ 199.9
m = 0.422 × 199.9
m = 84.35 g which is 88.4 g approximatively .
Thus ,88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 , hence option (d) is correct .
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A 50.0 L cylinder of oxygen gas is stored at 150. atm. What volume would the oxygen gas occupy if the cylinder were opened into a hot air balloon (completely deflated) until the final pressure is 735 torr
Answer:
THE VOLUME OF THE OXYGEN GAS AFTER DEFLATION TILL A PRESSURE OF 735 TORR IS ATTAINED IS 7836.99 L
Explanation:
Using Boyle's law,
P1V1 = P2V2
P1 = 150 atm
V1 = 50 L
P2 = 735 Torr
V2 = unknown
We must first convert the pressures into the same SI unit for easy calculation
1torr = 1/760 atm
So converting 735 torr to atm; we have:
1 torr = 1/ 760 atm
735 torr = 735 * 1 / 760 atm
= 0.967 atm
In other words, P2 = 0.957 atm
So rearranging the formula by making V2 the subject of the equation, we have:
V2 = P1 V1 / P2
V2 = 150 * 50 / 0.957
V2 = 7836.99 L
The volume of the oxygen cylinder after deflation to a final pressure of 735 torr or 0.967 atm pressure is 7836.99 L.
What is the energy of a photon of electromagnetic radiation with a wavelength of 963.5 nm? (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J · s
Answer:
[tex]E=2.06\times 10^{-19}\ J[/tex]
Explanation:
Given that,
The wavelength of electromagnetic radiation is 963.5 nm.
We need to find the energy of a photon with this wavelength.
The formula used to find the energy of a photon is given by :
[tex]E=\dfrac{hc}{\lambda}\\\\E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{963.5\times 10^{-9}}\\\\E=2.06\times 10^{-19}\ J[/tex]
So, the energy of a photon is [tex]2.06\times 10^{-19}\ J[/tex].
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many
The given question is incomplete.
The complete question is:
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction?
Answer: 4 grams of methane were needed for the reaction
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
{tex]CH_4+2O_2\rightarrow CO_2+H_2O[/tex]
Given: mass of oxygen = 16 g
Mass of carbon dioxide = 11 g
Mass of water = 9 g
Mass of products = Mass of carbon dioxide + mass of water = 11 g +9 g = 20 g
Mass or reactant = mass of methane + mass of oxygen = mass of methane + 16 g
As mass of reactants = mass of products
mass of methane + 16 g= 20 g
mass of methane = 4 g
Thus 4 grams of methane were needed for the reaction
g Use the References to access important values if needed for this question. A researcher took 2.592 g of a certain compound containing only carbon and hydrogen and burned it completely in pure oxygen. All the carbon was changed to 7.851 g of CO2, and all the hydrogen was changed to 4.018 g of H2O . What is the empirical formula of the original compound
Answer:
Empirical formula is: C₂H₅
Explanation:
The chemical equation of burning of a compound that conatins only Carbon and Hydrogen is:
CₓHₙ + O₂ → XCO₂ + n/2H₂O
That means the moles of CO₂ produced are the moles of Carbon in the compound and moles of hydrogen are twice moles of water. Empirical formula is the simplest ratio between moles of each element in the compound. Thus, finding molse of C and moles of H we can find empirical formula:
Moles C and H:
Moles C = Moles CO₂:
7.851g CO₂ ₓ (1mol / 44g) = 0.1784 moles CO₂ = Moles C
Moles H = 2 Moles H₂O
4.018g H₂O ₓ (1mol / 18.01g) = 0.2231 * 2 = 0.4417 moles H
Ratio C:H
The ratio between moles of hydrogen and moles of Carbon are:
0.4417 moles H / 0.1784 moles C = 2.5
That means there are 2.5 moles of H per mole of Carbon. As empirical formula must be given only in whole numbers,
Empirical formula is: C₂H₅A sample of ice absorbs 15.6kJ of heat as it undergoes a reversible phase transition to form liquid water at 0∘C. What is the entropy change for this process in units of JK? Report your answer to three significant figures. Use −273.15∘C for absolute zero.
Answer:
Entropy change of ice changing to water at 0°C is equal to 57.1 J/K
Explanation:
When a substance undergoes a phase change, it occurs at constant temperature.
The entropy change Δs, is given by the formula below;
Δs = q/T
where q is the quantity of heat absorbed or evolved in Joules and T is temperature in Kelvin at which the phase change occur
From the given data, T = 0°C = 273.15 K, q = 15.6 KJ = 15600 J
Δs = 15600 J / 273.15 K
Δs = 57.111 J/K
Therefore, entropy change of ice changing to water at 0°C is equal to 57.1 J/K
The entropy change of ice changing to water will be "57.1 J/K".
Entropy changeThe shift in what seems like a thermodynamic system's condition of confusion is caused by the transformation of heat as well as enthalpy towards activity. Entropy seems to be greater mostly in a network with a high quantity or measure of chaos.
According to the question,
Temperature, T = 0°C or,
= 273.15 K
Heat, q = 15.6 KJ or,
= 15600 J
We know the formula,
Entropy change, Δs = [tex]\frac{q}{T}[/tex]
By substituting the values, we get
= [tex]\frac{15600}{273.15}[/tex]
= 57.11 J/K
Thus the above answer is correct.
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Which of the following processes have a ΔS < 0? Which of the following processes have a ΔS < 0? carbon dioxide(g) → carbon dioxide(s) water freezes propanol (g, at 555 K) → propanol (g, at 400 K) methyl alcohol condenses All of the above processes have a ΔS < 0.
Answer:
All of the above processes have a ΔS < 0.
Explanation:
ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.
The question requests us to identify the process that has a negative change of entropy.
carbon dioxide(g) → carbon dioxide(s)
There is a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.
water freezes
There is a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.
propanol (g, at 555 K) → propanol (g, at 400 K)
Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.
This reaction highlights a drop in temperature which means a negative change in entropy.
methyl alcohol condenses
Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.
