Answer:
The answer is C.
Explanation:
Cu 2+ is reduced to Cu by gaining 2 electrons, so reduction occurs .
Proton cannot be gain or lose .
When ignited, solid ammonium dichromate decomposes in a fiery display. This is the reaction for a "volcano" demonstration. The decomposition produces nitrogen gas, water vapor, and chromium(III) oxide. The temperature is constant at 25°C.
Substance H0f (kJ/mol) S0 (kJ/mol . K)
a. Cr2O3 (g) -1140 0.0812
b. H2O (I) -242 0.1187
c. N2 (g) 0 0.1915
d. (NH4)2 Cr2O7 - 22.5 0.1137
Answer:
Your question is half unfinished, regarding the chromium III oxide the correct option that expresses the inorganic formula of said compound is "A"
Explanation:
In the reaction an initial salt reacts giving as product water vapor, nitrogen gas and an oxide that is chromium oxide.
Chromium oxide is an oxide that adopts the structure of corundum, compact hexagonal. It consists of an anion oxide matrix with 2/3 of the octahedral holes occupied by chromium. Like corundum, Cr2O3 is a tough, brittle material.
It is used as a pigment, green in color.
Which anion would bond with K+ in a 1:1 ratio to form a neutral ionic compound?
Select the correct answer below:
A. O2-
B. F-
C. N3-
D. S2-
The anion that would bond with K+ in a 1:1 ratio to form a neutral ionic compound is F- (fluoride ion)
Explanation:The anion that would bond with K+ in a 1:1 ratio to form a neutral ionic compound is B. F- (fluoride ion).
Potassium (K+) is a cation with a +1 charge, and to form a neutral ionic compound, it needs to bond with an anion with a -1 charge. The fluoride ion (F-) has a -1 charge, and therefore, it can form a 1:1 ratio with K+ to create a neutral compound.
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A 50.0 mL sample containing Cd 2 + and Mn 2 + was treated with 50.0 mL of 0.0400 M EDTA . Titration of the excess unreacted EDTA required 19.5 mL of 0.0270 M Ca 2 + . The Cd 2 + was displaced from EDTA by the addition of an excess of CN − . Titration of the newly freed EDTA required 17.1 mL of 0.0270 M Ca 2 + . What are the concentrations of Cd 2 + and Mn 2 + in the original solution?
Answer:
Check the explanation
Explanation:
VOLUME OF newly freed EDTA
=(VOLUME OF Ca2* STRENGTH OF Ca2)/ STRENGTH OF EDTA
=(24.9*0.0220)/ 0.0700
=7.825mL
STRENGTH OF Cd2
=( VOLUME OF newly freed EDTA * STRENGTH OF EDTA) / VOLUME OF SAMPLE
=(7.825*0.0700)/50
=0.0109 M
VOLUME OF excess unreacted EDTA
=(VOLUME OF Ca2* STRENGTH OF Ca2)/ STRENGTH OF EDTA
=(19.5*0.0270)/ 0.0400
=13.16mL
VOLUME OF EDTA REQUIRED FOR SAMPLE CONTAINING Cd2 AND Mn2= (17.1-13.16) mL
=3.94 mL
VOLUME OF EDTA REQUIRED FOR Mn2
= (VOLUME OF EDTA REQUIRED FOR SAMPLE CONTAINING Cd2 AND Mn2 - VOLUME OF newly freed EDTA )
=40.02-7.825 mL
=32.20 mL
STRENGTH OF Mn2
=( VOLUME OF EDTA REQUIRED FOR Mn2* STRENGTH OF EDTA) / VOLUME OF SAMPLE
=(32.20*0.0700)/50
=0.045 M
Draw the structure(s) of the product(s) of the Claisen condensation reaction between ethyl propanoate and ethyl formate. Draw only the condensation product, including the self-condensation product if applicable, do not draw the structure of the leaving group. Do not consider stereochemistry.
Answer:
See explanation below
Explanation:
A claisen reaction, is often used between two esters (One of them, usually having alpha hydrogens atoms) to form Beta Keto esters. Depending of the reagents, this reaction is taking place with base where one ester acts as nucleophile and the other as electrophile.
Now, we have here ethyl propanoate and ethyl formate. The ethyl propanoate has two alpha hydrogens, while ethyl formate do not have that. So, the base will react with the ethyl propanoate first, and then, it will react as nucleophile with the formate.
Now, the base to be used in this case will have to be a relatively strong base such sodium ethanoate, because if we use a strong base such NaOH, this will cause the saponification of the ester and not the condensation. So, in order to promove the claisen condensation, we need to use a base not too strong.
In the picture attached you have the mechanism and structure of the final product.
Compared to the ideal angle, you would expect the actual angle between the bromine-oxygen bonds to be
The expected bond angle between bromine and oxygen, based on hybridization and molecular geometry, would likely be slightly less than the ideal tetrahedral angle of 109.5°, due to electron repulsion effects.
Explanation:The bond angle between bromine and oxygen can be predicted using the concepts of hybridization and molecular geometry. If bromine and oxygen were to form a molecule similar to water (H₂O), one might expect a tetrahedral geometry because the valence orbitals of oxygen consist of one 2s orbital and three 2p orbitals. These combine during hybridization to form four hybrid orbitals that point toward the corners of a tetrahedron. The ideal bond angle for a tetrahedron is 109.5°. However, due to the effects of lone pair repulsion, the actual bond angle would likely be less, similar to how the H-O-H bond angle in water is experimentally found to be 104.5°.
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The actual angle between the bromine-oxygen bonds in BrO3⁻ is expected to be less than 120° due to the presence of the lone pair on the central bromine atom.
The ideal bond angle for a sp³ hybridized central atom is 109.5°. However, the presence of lone pairs on the central atom can distort the bond angles. Lone pairs are more electron-rich than bonding pairs, so they repel the bonding pairs more strongly. This repulsion causes the bond angles to decrease.
In the case of BrO3⁻, the central bromine atom is sp³ hybridized and has one lone pair. The lone pair will repel the two bonding pairs between the bromine and oxygen atoms. This repulsion will cause the bond angles between the bromine and oxygen atoms to decrease from the ideal angle of 120°.
The actual bond angle between the bromine-oxygen bonds in BrO3⁻ has been measured to be 107.1°. This is slightly less than the ideal angle of 120°, as expected.
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The question probable may be:
Compared to the ideal angle 120 ,the central atom Br lone pair 1 How you would expect the actual angle between the bromine-oxygen bonds to be
Which compound contains a triple bond?
Answer: C2H2
Explanation: Because each of the lines represent one bond, and because there are three lines (bonds) between the carbons, it means that they are bonded by three bonds, also known as a triple bond.
The triple bond that we have is shown by image D.
What is a triple bond?Two atoms share three pairs of electrons in a triple bond, making a total of six electrons present in the bond. Multiple bonds are formed between the atoms as a result of the sharing of electrons, offering a high degree of stability.
Two pi (π) bonds and a sigma (σ) bond define a triple bond. Atomic orbitals overlap head-on to form the sigma bond, whereas side-to-side overlap produces the two pi bonds.
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Consider the reaction.
2 upper N upper o upper c l double-headed arrow 2 upper N upper O (g) plus upper C l subscript 2 (g).
At equilibrium, the concentrations are as follows.
[NOCl] = 1.4 ´ 10–2 M
[NO] = 1.2 ´ 10–3 M
[Cl2] = 2.2 ´ 10–3 M
What is the value of Keq for the reaction expressed in scientific notation?
Answer: The value of Keq for the reaction expressed in scientific notation is [tex]1.6\times 10^{-5}[/tex]
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
For the given chemical reaction:
[tex]2NOCl(g)\rightarrow 2NO(g)+Cl_2(g)[/tex]
The expression for [tex]K_{eq}[/tex] is written as:
[tex]K_{eq}=\frac{[NO]^2\times [Cl_2]}{[NOCl]^2}[/tex]
[tex]K_{eq}=\frac{(1.2\times 10^{-3})^2\times (2.2\times 10^{-3})}{(1.4\times 10^{-2})^2}[/tex]
[tex]K_{eq}=1.6\times 10^{-5}[/tex]
The value of Keq for the reaction expressed in scientific notation is [tex]1.6\times 10^{-5}[/tex]
Answer:
its A
Explanation:
In your lab you are studying the kinetics of the degradation of a pain killer in the human liver. You are monitoring the concentration of the pain killer over a period of time. The initial concentration of the pain killer in your experiment was 1.59 M. After 0.80881 hours the concentration was found to be 0.795 M. In another 0.80881 hours the concentration was found to be 0.3975 M (t=27.868 hours overall).
If another experiment were set up where the initial concentration of the pain killer was 0.479 M, how long would it take for the pain killer concentration to reach 0.0161 M?
Answer:
Explanation:
From the data it is clear that after each .80881 hours , the concentration becomes half . so its decomposition appears to be first order .
k = 1 / t ln A₀ / A
= (1/.80881) ln 2
= .857
for the given case
A = .0161 , A₀ = .479
t = ?
t = 1/k ln( .479 / .0161)
= (1 / .857 )ln (.479 / .0161 )
= (1 / .857 ) x ln 29.75
= 3.959 hours
How much energy is released when 50.00 mol of glycerin melts? (Heat of fusion glycerin-91.7 KJ/mol)
Answer:
4585 KJ
Explanation:
Data obtained from the question. This includes following:
Number of mole (n) of glycerin = 50 moles
Heat of fusion glycerin (Hf) = 91.7 KJ/mol
Heat (Q) released =?
The heat released can be obtained as follow:
Q = n•Hf
Q = 50 x 91.7
Q = 4585 KJ
Therefore, 4585 KJ of heat is released.
Upon adding solid potassium hydroxide pellets to water the following reaction takes place: KOH(s) → KOH(aq) + 43 kJ/mol Answer the following questions regarding the addition of 14.0 g of KOH to water: Does the beaker get warmer or colder? Is the reaction endothermic or exothermic? What is the enthalpy change for the dissolution of the 14.0 grams of KOH?
Answer:
a) Warmer
b) Exothermic
c) -10.71 kJ
Explanation:
The reaction:
KOH(s) → KOH(aq) + 43 kJ/mol
It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.
Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.
The enthalpy change for the dissolution of 14 g of KOH is:
[tex] n = \frac{m}{M} [/tex]
Where:
m: is the mass of KOH = 14 g
M: is the molar mass = 56.1056 g/mol
[tex] n = \frac{m}{M} = \frac{14 g}{56.1056 g/mol} = 0.249 mol [/tex]
The enthalpy change is:
[tex] \Delta H = -43 \frac{kJ}{mol}*0.249 mol = -10.71 kJ [/tex]
The minus sign of 43 is because the reaction is exothermic.
I hope it helps you!
The rate constant of the elementary reaction C2H5CN(g) → CH2CHCN(g) + H2(g) is k = 7.21×10-3 s-1 at 634 °C, and the reaction has an activation energy of 247 kJ/mol. (a) Compute the rate constant of the reaction at a temperature of 741 °C. _________ s-1 (b) At a temperature of 634 °C, 96.1 s is required for half of the C2H5CN originally present to be consume. How long will it take to consume half of the reactant if an identical experiment is performed at 741 °C? (Enter numbers as numbers, no units. For example, 300 minutes would be 300. For letters, enter A, B, or C. Enter numbers in scientific notation using e# format. For example 1.43×10-4 would be 1.43e-4.)
Answer:
the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]
it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C
Explanation:
Given that :
[tex]k_1 = 7.21*10^{-3} s^{-1} \\ \\ E_a = 247 kJ/mol \ \ \ = 247*10^3 \ J/mol \\ \\ T_1 = 634 ^ {^0} C= (273 + 634) K = 907 \ K \\ \\ T_2 = 741^{^0 } C = (273+ 741) K = 1014 \ K \\ \\ R =8.314 \ \ J/mol/K[/tex]
a)
According to Arrhenius Equation ;
[tex]In\frac{k_2}{k_1} = -\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = 3.456412[/tex]
[tex]\frac{k_2}{7.21*10^{-3}} =e^{ 3.456412[/tex]
[tex]k_2 = 31.70302 * 7.21*10^{-3}[/tex]
[tex]k_2 = 0.22858 \ s^{-1}[/tex]
Therefore , the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]
b) Given that :
[tex]t_{(1/2)} = 96.1 \ s[/tex]
[tex]k_1 = \frac{0.693}{ t_{(1/2)}}[/tex]
[tex]k_1 = \frac{0.693}{ 96.1 \ s}[/tex]
[tex]k_1 = 7.211*10^{-3} \ s^{-1}[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = 3.456412[/tex]
[tex]\frac{k_2}{7.211*10^{-3}} =e^{ 3.456412[/tex]
[tex]k_2 =e^{ 3.456412 * 7.21*10^{-3}[/tex]
[tex]k_2 = 0.22862 \ s^{-1}[/tex]
[tex]t_{(1/2)_2}} = \frac{0.693}{k_2}[/tex]
[tex]t_{(1/2)_2}} = \frac{0.693}{0.22862}[/tex]
[tex]t_{(1/2)_2}} =3.0313 \ s[/tex]
it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C
When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of ammonium chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for ammonium chloride in . Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.
Answer:
The figures and parameters to answer this question are not clearly stated, so I will answer it in the closest best way I can.
Explanation:
Molality of alanine = mole / weight of solvent( kg)
= (170×1000)/(89×600)
= 3.18 molal
We know ∆ Tf = Kf × m
Kf = ∆ Tf / molality
= 7.9/3.18 = 2.48 °c.kg.mol-1
Now using the value of cryoscopic constant we calculate can't Hoff factor for Ammonium chloride.
i = ∆ Tf /( Kf × molality of NH4Cl)
= (24.7 × 53.5 × 600) /(2.48 × 170 × 1000)
= 1.8
Which process is a chemical change?
TIL
00
burning a match
boiling water
melting ice
O breaking glass
Answer:
Burning a match is a chemical change while the rest are physical changes.
Explanation:
Burning a match represents a chemical change, where a substance transforms into new substances with distinct chemical properties. In contrast, boiling water, melting ice, and breaking glass are examples of physical changes, altering the substance's form but not its identity.
Explanation:Among the given options, burning a match is a chemical change. A chemical change is a process that involves a substance changing into a new substance with different chemical properties. In this case, when a match is burned, it undergoes a chemical reaction leading to the formation of new substances such as heat, light, water vapor, and carbon dioxide, which were not there before the match was lit. On the other hand, boiling water, melting ice, and breaking glass are all examples of physical changes, which involve changes in the form of a substance but do not change the identity of the substance itself.
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Devin is watching two runners at the park. The runners are the same weight, and the speed that each of them is running is not changing.
Devin wants to know which runner has the most energy of motion. What would he need to find out about the runners in order to determine this?
Answer:
Reeeeeeeeeerereeeee
Reeeee reee reee ree ree re
Answer:
the speed of which they are running.
Explanation:
Write the balanced half‑reaction that occurs at the anode in a hydrogen‑oxygen fuel cell in which an acidic electrolyte is used. A hydrogen fuel cell. H 2 flows into the cell at the negative electrode. O 2 flows into the cell at the positive electrode. Electrons travel through an external circuit from the negative to the positive electrode. Between the two electrodes is an electrolyte solution. Arrows in the electrolyte point from the negative electrode to the positive electrode. anode half-reaction: Write the balanced half‑reaction that occurs at the cathode in a hydrogen‑oxygen fuel cell in which an acidic electrolyte is used. cathode half-reaction: Write the balanced overall cell reaction. overall cell reaction:
Answer:
Cathode: O2 + 4H+ +4e--------> 2H20
Anode: 2H2 -4e- ---------> 4H+
Overall: 2H2 + O2 → 2H2O
Explanation:
A hydrogen-oxygen fuel cell is an alternative cell to rechargeable cells and batteries. In this cell, hydrogen and oxygen is used to produce voltage and water is the only byproduct.
At the cathode (positive electrode):
O2 + 4H+ +4e--------> 2H2O
At the anode (negative electrode);
2H2 -4e- ---------> 4H+
Adding the two half reactions we have:
2H2 + O2 + 4H+ + 4e- -----------> 2H2O + 4H + 4e-
The overall reaction after cancelling out the like terms in the reaction is:
2H2 (g) + O2 (g) --------> 2H2O (l)
The hydrogen fuel cell is an electrochemical cell in which hydrogen is oxidized and oxygen is reduced.
A fuel cell is an electrochemical cell in which oxygen enters into the positive cathode and hydrogen flows into the negative cathode. Hydrogen is oxidized and oxygen is reduced during the electrochemical reaction.
At the negative electrode;
2H2(g) -----> 4H^+(aq)+ 4e
At the positive electrode;
O2(g) + 4H^+(aq) + 4e -----> 2H2O(l)
The overall reaction is;
2H2(g) + O2(g) -----> 4H^+(aq) + 2H2O(l)
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Which is an example of a beneficial mutation?
Answer:
b.one that results in lighter flower petal colors without changing the plant’s ability to reproduce
Explanation:
on edge
Beneficial mutations improve the survival and reproductive chances of an organism. Examples include mosquitos' resistance to insecticides, the color change in peppered moths during the Industrial Revolution, and the Antennapedia mutation in Drosophila.
Explanation:A beneficial mutation can be understood as the type of mutation that improves an organism's chances of survival and reproduction. For instance, consider the case of mosquitoes. The mutation that has provided them with resistance to some insecticides is an example of a beneficial mutation. This resistance has allowed them to survive even when faced with these chemicals, thus enhancing their chances of propagation.
Similarly, during the Industrial Revolution, peppered moths underwent a beneficial mutation where their coloration turned from light to dark. As industrial pollution darkened the environment, the dark coloration assisted in camouflage, aiding their survival.
The mutant allele can also interfere with the normal gene's function or its distribution in the body, which can result in beneficial mutations. For example, the Antennapedia mutation in Drosophila that allowed for the development of legs in place of antennae. However, whether a mutation is beneficial or not primarily depends on its effects on the organism's ability to mature sexually and reproduce successfully.
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substances change temperature at different rates because of their
density
specific heat
melting point
boiling point
Answer:
ok
Explanation:
I have the same question
The rust that appears on steel surfaces is iron(III) oxide. If the rust found spread over the surfaces of a steel bicycle frame contains a total of 9.62×1022 oxygen atoms, how many grams of rust are present on the bicycle frame?
Answer:
The grams of rust present in the bicycle frame is [tex]x = 8.50 g[/tex]
Explanation:
From the question we are told that
The number of oxygen atom contained is [tex]n = 9.62 *10^{22} \ atoms[/tex]
The molar mass of the compound is [tex]M_{Fe_3O_3} = 159.69 g[/tex]
At standard temperature and pressure the number of oxygen atom in one mole of iron(III) oxide is mathematically evaluated as
[tex]N_o = 3 * Ne[/tex]
Where Ne is the avogadro's constant with a value [tex]N_e = 6.023 *10^{23} \ atoms[/tex]
So
[tex]N_o= 3 * 6.023*10^{23} \ atoms[/tex]
[tex]N_o= 1.8069*10^{24} \ atoms[/tex]
So
[tex]1.8069*10^{24} \ atoms[/tex] is contained in [tex]M_{Fe_3O_3} = 159.69 g[/tex]
[tex]9.62 *10^{22} \ atoms[/tex] is contained in x
So
[tex]x = \frac{159.69 * 9.62 *10^{22}}{1.8069 *10^{24}}[/tex]
[tex]x = 8.50 g[/tex]
Final answer:
To calculate the grams of rust on the bicycle frame, we need to use stoichiometry. By converting the number of oxygen atoms to moles and then to grams using the molar mass of iron(III) oxide, we find that there are approximately 5.37x10^22 grams of rust on the bicycle frame.
Explanation:
To determine the grams of rust present on the bicycle frame, we need to calculate the molar mass of iron(III) oxide and then use stoichiometry to convert the number of oxygen atoms to grams of rust. The molar mass of Fe2O3 is 159.69 g/mol. Given that 1 mole of Fe2O3 contains 3 moles of oxygen atoms, we can calculate the moles of Fe2O3 using the given number of oxygen atoms and then convert it to grams using the molar mass.
9.62x10^22 oxygen atoms x (1 mol Fe2O3 / 3 mol O) x (159.69 g Fe2O3 / 1 mol Fe2O3) = 5.37x10^22 g Fe2O3
Therefore, there are approximately 5.37x10^22 grams of rust present on the bicycle frame.
Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is liberated as indicated. Calculate the heats of combustion of these alcohols in kJ/mol. (a) methanol (CH3OH), −22.6 kJ kJ/mol (b) ethanol (C2H5OH), −29.7 kJ Enter your answer in scientific notation. × 10 kJ/mol (c) n−propanol (C3H7OH), −33.4 kJ Enter your answer in scientific notation. × 10 kJ/mol
The heats of combustion for methanol, ethanol, and n-propanol, when expressed in kJ/mol, are -0.710 × 10³ kJ/mol, -0.644 × 10³ kJ/mol, and -0.56 × 10³ kJ/mol, respectively. This shows that the amount of heat liberated decreases as the molecular weight of the alcohol increases.
Explanation:The heat of combustion of a substance is defined as the amount of heat energy released when 1 mole of a substance is completely burned in oxygen. Here, the heats of combustion for methanol (CH3OH), ethanol (C2H5OH), and n−propanol (C3H7OH) are given in kJ/g, and we need to convert them into kJ/mol.
To do this conversion, you first need to find the molar mass of each alcohol. You can do this by adding up the atomic masses of each element in the molecule.
Using the atomic masses in g/mol from the periodic table, we have: Methanol (CH3OH): 12.01 + 3(1.01) + 16 + 1.01 = 32.05 g/mol; Ethanol (C2H5OH): 2(12.01) + 5(1.01) +16 + 1.01=46.07 g/mol; n-Propanol (C3H7OH): 3(12.01) + 7(1.01) + 16 + 1.01 = 60.1 g/mol.
Next, divide the heat of combustion given in kJ/g by the molar mass calculated in g/mol to obtain the heat of combustion in kJ/mol:
Methanol: -22.6 kJ/g ÷ 32.05 g/mol = -0.710 × 10³ kJ/mol (scientific notation)
Ethanol: -29.7 kJ/g ÷ 46.07 g/mol = -0.644 × 10³ kJ/mol (scientific notation)
n-Propanol: -33.4 kJ/g ÷ 60.1 g/mol = -0.56 × 10³ kJ/mol (scientific notation)
So, the heats of combustion for the alcohols are: Methanol: -0.710 × 10³ kJ/mol, Ethanol: -0.644 × 10³ kJ/mol, and n-Propanol: -0.56 × 10³ kJ/mol. As you can see, the amount of heat liberated per mole decreases as the molecular mass of the alcohol increases.
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Which option is a solution that is 20% (v/v) methanol in water?
Answer:
Explanation:
1.The solution contains 20 mL of methanol and 100 mL of water.
2.The solution contains 20 mL of methanol and 80 mL of water.
3.The solution contains 20 g of methanol and 100 mL of water.
4.The solution contains 20 g of methanol and 80 mL of water.
Answer:
20 % volume/volume methanol in water contains 20 mL of methanol and 80 mL of water.
Explanation:
Total volume of the solution= 20 + 80 = 100 mL
How much 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl?
Answer:
54 L
Explanation:
Given data
Initial concentration (C₁): 9.0 MInitial volume (V₁): 450 mLFinal concentration (C₂): 0.075 MFinal volume (V₂): to be calculatedWe have a concentrated NaCl solution and we will add water to it to obtain a diluted NaCl solution. We can find the volume of the final solution using the dilution rule.
[tex]C_1 \times V_1 = C_2 \times V_2\\V_2 = \frac{C_1 \times V_1}{C_2} = \frac{9.0M \times 450mL}{0.075M} = 5.4 \times 10^{4} mL = 54 L[/tex]
The volume of the final solution should be 54L.
Calculation of the volume:Since
Initial concentration (C₁): 9.0 M
Initial volume (V₁): 450 mL
Final concentration (C₂): 0.075 M
We know that
C1 * V1 = C2 * V2
V2 = C1 * V1/ C2
= 9.0M * 450 mL/ 0.075 M
= 54 L
hence, The volume of the final solution should be 54L.
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The energy of any one-electron species in its nth state (n = principal quantum number) is given by E = –BZ2 /n2 where Z is the charge on the nucleus and B is 18 2.180 10 J. a) Find the ionization energy of the Be3+ ion in its first excited state in kilojoules per mole. b) Find the wavelength of light given off from the Be3+ ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels.
Explanation:
(a) The given data is as follows.
B = [tex]2.180 \times 10^{-18} J[/tex]
Z = 4 for Be
Now, for the first excited state [tex]n_{f}[/tex] = 2; and [tex]n_{i} = \infinity[/tex] if it is ionized.
Therefore, ionization energy will be calculated as follows.
I.E = [tex]\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})[/tex]
= [tex]8.72 \times 10^{-18} J[/tex]
Converting this energy into kJ/mol as follows.
[tex]8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol[/tex]
= 5249 kJ/mol
Therefore, the ionization energy of the [tex]Be^{3+}[/tex] ion in its first excited state in kilojoules per mole is 5249 kJ/mol.
(b) Change in ionization energy is as follows.
[tex]\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}[/tex]
[tex]\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}[/tex]
[tex]\lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}[/tex]
= [tex]303.7 \times 10^{-10} m[/tex]
or, = [tex]303.7^{o}A[/tex]
Therefore, wavelength of light given off from the [tex]Be^{3+}[/tex] ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels [tex]303.7^{o}A[/tex].
The ionization energy of the Be³+ ion in its first excited state is approximately 5250 kJ/mol. The wavelength of light emitted when an electron transitions from n=4 to n=2 in a Be³+ ion is approximately 75.9 nm.
Ionization Energy and Wavelength Calculation for Be³+
The problem involves calculating the ionization energy of a Be³+ ion in its first excited state and the wavelength of light emitted during an electron transition.
a) Ionization Energy
For a one-electron species like Be³+, the energy in the nth state is given by:
E = –BZ²/n², where Z is the charge on the nucleus and B = 2.180 × 10⁻¹⁸ J.
In the first excited state (n = 2) for Be³+ (Z = 4), we get:
En = –B(4)²/2² = –2.180 × 10⁻¹⁸ J × 16/4 = –8.72 × 10⁻¹⁸ J.
To find the ionization energy, we need the energy to liberate 1 mole of Be³+ ions:
Ionization Energy = |En| × Avogadro's Number = 8.72 × 10⁻¹⁸ J × 6.022 × 10²³ mol⁻¹ = 5.25 × 10⁶ J/mol = 5250 kJ/mol.
b) Wavelength of Light
The energy difference between the fourth (n=4) and second (n=2) energy levels is calculated using:
ΔE = E₄ - E₂ = -BZ² / 4² - (-BZ² / 2²) = BZ² (1/4² - 1/2²)
ΔE = 2.180 × 10⁻¹⁸ J × 16 × (1/16 - 1/4) = 2.180 × 10⁻¹⁸ J × 16 × (-¾) = -2.616 × 10⁻¹⁷ J
The emitted photon's wavelength is given by:
λ = hc /ΔE = (6.626 × 10⁻³⁴ Js × 3.00 × 10⁸ m/s) / 2.616 × 10⁻¹⁷ J ≈ 7.59 × 10⁻⁸ m = 75.9 nm
Chlorophenols impart unpleasant taste and odor to drinking water at concentrations as low as 5 mg/m3. They are formed when the chlorine dis- infection process is applied to phenol-containing waters. What is the threshold for unpleasant taste and odor in units of (a) mg/L, (b) mg/L, (c) ppmm, and (d) ppbm
Answer:
See explaination
Explanation:
Chlorine’s disinfection properties have helped improve the lives of billions of people around the world. Chlorine also is an essential chemical building block, used to make many products that contribute to public health and safety, advanced technology, nutrition, security and transportation.
Please kindly check attachment for the step by step solution of the given problem.
The threshold for unpleasant taste and order in the respective units are;
A) 0.005 mg/L
A) 0.005 mg/LB) 5 μg/L
A) 0.005 mg/LB) 5 μg/LC) 0.005 ppm
A) 0.005 mg/LB) 5 μg/LC) 0.005 ppmD) 5 ppb
We are given the concentration of chlorophenols as; M = 5 mg/m³
A) Converting mg/m³ to mg/L means that;
1 mg/m³ = 0.001 mg/L. Thus;
5 mg/m³ = (5 × 0.001)/1
>> 0.005 mg/L
B) Converting mg/m³ to μg/L means that;
1 mg/m³ = 1 μg/L
Thus; 5 mg/m³ = 5 × 1/1 μg/L
>> 5 μg/L
C) Converting mg/m³ to ppm gives;
1 mg/m³ = 0.001 ppm
Thus;
5 mg/m³ = (5 × 0.001)/1 ppm
>> 0.005 ppm
D) Converting mg/m³ to ppb gives;
1 mg/m³ = 1 ppm
Thus;
5 mg/m³ = 5 × 1/1 ppb
>> 5 ppb
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Despite the destruction from Hurricane Katrina in September 2005, the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered an atmospheric pressure of 88.2 kPa on October 19, 2005, that was 2.0 kPa lower than Hurricane Katrina. What was the difference in pressure between the two hurricanes?
Complete Question
Despite the destruction from Hurricane Katrina in Sept 2005,the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered at atmospheric pressure of 88.2 kPa on Oct 19, 2005,about 2 kPa lower than Hurricane Katrinia. What was the difference between the two hurricanes in:
millimeters of Hg? ___ mm Hg
atmospheres? ___ atm
millibars? ___ mb
Answer:
The difference in pressure in mm of Hg is [tex]Pg = 15.05 \ mm\ Hg[/tex]
The difference in pressure in atm is [tex]P_{atm} = 0.0198 \ atm[/tex]
The difference in pressure in millibar is [tex]P_{bar} = 20 \ mbar[/tex]
Explanation:
From the question we are told that
The pressure of Hurricane Wilma is [tex]P = 88 kPa[/tex]
The difference is pressure is [tex]\Delta P =2\ k Pa[/tex]
Generally
[tex]1\ atm = 1.01 *10^5 \ Pa[/tex]
[tex]1 \ atm = 760\ mmHg[/tex]
The pressure difference in mm Hg is mathematically evaluated as
[tex]Pg = \Delta P * \frac{1000 Pa}{1kPa} * \frac{1\ atm}{1.01 *10^5} * \frac{760 mm Hg}{1 \ atm}[/tex]
Substituting the value
[tex]Pg = 2 kPa * \frac{1000 Pa}{1kPa} * \frac{1\ atm}{1.01 *10^5} * \frac{760 mm Hg}{1 \ atm}[/tex]
[tex]Pg = 15.05 \ mm\ Hg[/tex]
The pressure difference in atm is
[tex]P_{atm } = 2kPa * \frac{1000}{1\ kPa} *\frac{1 \ atm}{1.01 *10^5 Pa}[/tex]
[tex]= 0.0198 \ atm[/tex]
The pressure in millibars is
[tex]= 2kPa * \frac{1000Pa}{1 kPa} * \frac{1\ bar }{1*10^5 Pa} * \frac{1000 \ milli bar}{1 bar}[/tex]
[tex]= 20 \ m bar[/tex]
Final answer:
The pressure difference between Hurricane Wilma and Hurricane Katrina was 2.0 kPa, with Wilma's pressure being 88.2 kPa and Katrina's being 90.2 kPa. The pressure drop during a hurricane on the northeast U.S. coast in torr is 50.8 torr. A nonvolatile liquid is used in barometers and manometers because volatility would affect the accuracy of pressure measurements.
Explanation:
The difference in atmospheric pressure between Hurricane Wilma and Hurricane Katrina can be calculated using the information provided. If Wilma's pressure was 88.2 kPa and that was 2.0 kPa lower than Katrina's, then Katrina's pressure was 88.2 kPa + 2.0 kPa = 90.2 kPa. Therefore, the difference in pressure between the two hurricanes was 2.0 kPa.
To calculate the drop in pressure in torr during a hurricane on the northeastern United States coast: the pressure usually at 30.0 in. Hg can drop to 28.0 in. Hg. Since 1 in. Hg is equivalent to 25.4 mm Hg, and 1 mm Hg is the same as 1 torr, the drop in pressure is (30.0 in. Hg - 28.0 in. Hg) x 25.4 mm Hg/in. Hg = 50.8 torr.
It is necessary to use a nonvolatile liquid in a barometer or manometer because a volatile liquid would evaporate and form a vapor that would affect the pressure measurement, leading to inaccurate readings.
Which will not appear in the equilibrium constant expression for the reaction below?
H2O(g) + C(s) +
H2() + CO(g)
[H20]
[C]
[H2]
[CO]
Concentration of C, [C] will not appear in the equilibrium constant expression for the reaction.
What is meant by equilibrium constant ?Ratio of product of concentration of the products to the product of concentration of reactants.
Here,
H₂O(g) + C(s) ⇆ H₂ + CO
The expression for equilibrium constant for the reaction is given by,
K = [H₂] [CO]/ [H₂O] [C]
where K is the equilibrium constant of the reaction.
We know, the concentration of solids does not change in a reaction and therefore the concentration is taken as unity.
In the given reaction, since the reactant C is in solid form, its concentration is taken as 1.
So, K = [H₂] [CO]/ [H₂O]
As a result, the concentration of C, [C] will not be included in the expression for equilibrium constant.
Hence,
Concentration of C, [C] will not appear in the equilibrium constant expression for the reaction.
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In the reaction H2O(g) + C(s) → H2(g) + CO(g), the component that does not appear in the equilibrium constant expression is C(s), carbon in its solid form. This is because pure solids do not change concentration and do not affect the state of equilibrium.
Explanation:To answer which component will not appear in the equilibrium constant expression for the reaction H2O(g) + C(s) → H2(g) + CO(g), we need to understand how equilibrium expressions are formed. The equilibrium constant expression for a reaction is based on the balanced chemical equation and includes the concentrations of the gaseous and aqueous species, raised to the power of their coefficients in the balanced equation. Pure solids and pure liquids are not included in the expression because their concentrations don't change during the reaction.
In the given reaction, C(s), which is carbon in its solid form, does not appear in the equilibrium constant expression because as a pure solid, its concentration is constant and does not affect the state of equilibrium. Therefore, the equilibrium constant expression for this reaction would be K = [CO][H2], excluding both the water (H2O) because it is in gaseous form and included in the expression, and the carbon (C) because it is a solid.
I need help please answer
Answer:
A
Explanation:
Just thinking logically.
5. When doing an experiment, aside from doing calculations, what can you do to
determine which reactant is the limiting reactant and which is the excess
reactant? *
Answer:
check which reactant is totally consumed and which one remains in the mixture
Explanation:
Apart from doing calculations during an experiment, one can determine which reactant is limiting and which one is in excess by checking the resulting mixture for the presence of reactants.
A limiting reactant is one that determines the amount of product formed during a reaction. It is usually a reactant that is lower than stoichiometry amount.
On the other hand, an excess reactant is one that is present in more than the stoichiometrically required amount during a reaction.
Limiting reactants will be totally consumed in a reaction while excess reactant would still be seen present in mixture after the reaction has stopped.
Hence, apart from using stoichiometric calculation to determine which reactant is limiting or in excess during an experiment, one can just check the final mixture of the reaction for the presence of any of the reactants. The reactant that is detected is the excess reactant while the one without traces in the final mixture is the limiting reactant.
The complete combustion of methane is: CH4 + 2O2 ! 2H2O + CO2 a. Calculate the standard Gibbs free energy change for the reaction at 298 K (i.e. ). b. Calculate the energetic (ΔH) and entropic contributions (TΔS) to the favorable standard Gibbs free energy change at 298 K and determine which is the dominant contribution,? c. Estimate the equilibrium constant at 298 K.
Answer:
a
The standard Gibbs free energy change for the reaction at 298 K is
[tex]\Delta G^o_{re} = -800.99 kJ/moles[/tex]
b
The energetic (ΔH) is [tex]\Delta H^o _{re} = -802.112 \ kJ/mole[/tex]
The entropic contributions is [tex]T \Delta S = -1.126 \ kJ/mole[/tex]
Energetic is the dominant contribution
c
The equilibrium constant at 298 K is [tex]K = 2.53[/tex]
Explanation:
From the question we are told that
The chemical reaction is
[tex]CH_4 + 2 O_2 ----> 2 H_2 O + CO_2[/tex]
Generally ,
The free energy of formation of [tex]CH_4[/tex] is a constant with a value
[tex]\Delta G^o_f __{CH_4}} = -50.794 \ kJ / moles[/tex]
The free energy of formation of [tex]O_2[/tex] is a constant with a value
[tex]\Delta G^o_f __{O_2}} = 0 \ kJ / moles[/tex]
The free energy of formation of [tex]H_2O[/tex] is a constant with a value
[tex]\Delta G^o_f __{H_2O}} = -228.59 \ kJ / moles[/tex]
The free energy of formation of [tex]CO_2[/tex] is a constant with a value
[tex]\Delta G^o_f __{H_2O}} = -394.6 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]CH_4[/tex] at standard condition i is a constant with a value
[tex]\Delta H^o_f __{CH_4}} = -74.848 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]CO_2[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{CO_2}} = -393.3 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]O_2[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{O_2}} = 0 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]H_2O[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{H_2O}} = -241.83 \ kJ / moles[/tex]
The standard Gibbs free energy change for the reaction at 298 K is mathematically represented as
[tex]\Delta G^o_{re} = (\Delta G^o_f __{H_2O}} + (2 * \Delta G^o_f __{H_2O}} )) - ((\Delta G^o_f __{CH_4}} + (2 * \Delta G^o_f __{O_2}}))[/tex]
Substituting values
[tex]\Delta G^o_{re} =\Delta G= ( (-394.6 ) + (2 * (-228.59)) ) - ((-50.794) +(2* 0))[/tex]
[tex]\Delta G^o_{re} = -800.99 kJ/moles[/tex]
The Enthalpy of formation of the reaction is
[tex]\Delta H^o _{re} =( \Delta H^o_f __{CH_4}} + (2 * (\Delta H^o_f __{H_2O}} ))) - ( \Delta H^o_f __{CH_4}} + (2 * \Delta H^o_f __{O_2}}))[/tex]
Substituting values
[tex]\Delta H^o _{re} = \Delta H = ((-393.3) + 2 * ( -241.83)) - ( -74.848 + (2 * 0))[/tex]
[tex]\Delta H^o _{re} = -802.112 \ kJ/mole[/tex]
The entropic contributions is mathematically represented as
[tex]T \Delta S = \Delta H -\Delta G[/tex]
Substituting values
[tex]T \Delta S =-802 .112-(-800.986)[/tex]
[tex]T \Delta S = -1.126 \ kJ/mole[/tex]
Comparing the values of [tex]T \Delta S \ and \ \Delta G[/tex] we see that energetic is the dominant contribution
The standard Gibbs free energy change for the reaction at 298 K can also be represented mathematically as
[tex]\Delta G = -RT lnK[/tex]
Where R is the gas constant with as value of [tex]R = 8.314 *10^{-3} kJ/mole[/tex]
K is the equilibrium constant
T is the temperature with a given value of [tex]T = 298K[/tex]
Making K the subject we have
[tex]K = e ^{- \frac{\Delta G }{RT} }[/tex]
Substituting values
[tex]K = e ^{- \frac{-800.99 }{(8.314 *10^{-3} ) * (298)} }[/tex]
[tex]K = 2.53[/tex]
Final answer:
To calculate the standard Gibbs free energy change for the combustion of methane, use the formula ΔG = ΔH - TΔS, where ΔG is the standard Gibbs free energy change, ΔH is the standard enthalpy change, T is the temperature in Kelvin, and ΔS is the standard entropy change.
Explanation:
The standard Gibbs free energy change for a reaction can be calculated using the formula:
ΔG = ΔH - TΔS
where ΔG is the standard Gibbs free energy change, ΔH is the standard enthalpy change, T is the temperature in Kelvin, and ΔS is the standard entropy change.
In this case, we are given the thermochemical equation for the combustion of methane as: CH4 + 2O2 → 2H2O + CO2
To calculate the standard Gibbs free energy change, we need the values of ΔH and ΔS for this reaction at 298 K. Once we have these values, we can plug them into the formula to get the answer.
ball has a volume of 5.27 liters and is at a temperature of 27.0°C. A pressure gauge attached to the ball reads 0.25 atmosphere. The atmospheric pressure is 1.00 atmosphere.
Answer:
The absolute pressure inside the ball is 1.25atm.
Explanation:
Given-
Volume, V = 5..27L
Temperature, T = 27°C
Gauge Pressure, Pg = 0.25 atm
Atmospheric pressure, Patm = 1 atm
Absolute pressure, Pabs = ?
We know,
Therefore, absolute pressure inside the ball is 1.25atm.
Answer:
The absolute pressure inside the ball is 1.25 atmospheres.
The ball contains 0.267 moles of air
Explanation:
The fuel value of hamburger is approximately 3.3 kcal/g. If a man eats 0.75 pounds of hamburger for lunch and none of the energy is stored in his body, estimate the amount of water that would have to be lost in perspiration to keep his body temperature constant. The heat of vaporization of water may be taken as 2.41 kJ/g.
Answer:
[tex]m_{w} = 1755.323\,g[/tex]
Explanation:
The energy stored in the hamburger is:
[tex]Q = (0.75\,pd)\cdot \left(\frac{453\,g}{1\,pd} \right)\cdot \left(3.3\,\frac{kcal}{g} \right)\cdot \left(4.186\,\frac{kJ}{kcal} \right)[/tex]
[tex]Q = 4693.239\,kJ[/tex]
The amount of water perspired is derived from the following formula:
[tex]Q = m_{w}\cdot (c_{p}\cdot \Delta T + L_{v})[/tex]
[tex]m_{w} = \frac{Q}{c_{p}\cdot \Delta T + L_{v}}[/tex]
[tex]m_{w} = \frac{4693.239\,kJ}{\left(4.186\times 10^{-3}\,\frac{kJ}{g\cdot ^{\circ}C} \right)\cdot (100^{\circ}C-37^{\circ}C)+2.41\,\frac{kJ}{g} }[/tex]
[tex]m_{w} = 1755.323\,g[/tex]