Answer:
D. exosphere is the outer layer of the thermosphere
Question 2
Bromine will react very fast (almost instantly) with which compound?
O 1-Pentene
Cyclohexane
Heptane
Benzene
Answer:
1-Pentene
Explanation:
If we look at all the options listed, we will notice that the rate of reaction of bromine with each one differs significantly.
For 1-pentene, addition of bromine across the double bond is a relatively fast process. It is usually used as a test for unsaturation. Bromine water is easily decolorized by alkenes.
Cyclohexane, heptane are alkanes. They can only react with chlorine in the presence of sunlight. This is a substitution reaction. It does not occur easily. A certain quantum of light is required for the reaction to occur.
For benzene, bromine can only react with it by electrophilic substitution in which the benzene ring is retained. A Lewis acid is often required for the reaction to occur and it doesn't occur easily.
A box sits on a table. A short arrow labeled F subscript N = 100 N points up. A short arrow labeled F subscript g = 100 N points down. A long arrow labeled F subscript P = 75 N points right. A short arrow labeled F subscript f = 10 N points left. What is the net force acting on the box? 285 N 185 N 85 N 65 N
Answer:65N
Explanation:75-10=65
100-100=0
Therefore=65n
Answer:
the answer is 65N
At STP, what is the volume in milliliters of 0.0395 mol of fluorine gas, F2?
Answer:
884.8 mL
Explanation:
If we have STP condition (Standard temperature and pressure), we will have the following relationship:
[tex]1~mol=22.4~L[/tex] or [tex]1~mol=22400~mL[/tex]. With this in mind we can do the conversion:
[tex]0.0395~mol\frac{22400~mL}{1~mol}=884.8~mL[/tex]
In the 0.0395 mol of fluorine gas, we will have 884.8 mL of gas at STP conditions.
An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.323 M HI, 4.34E-2 M H2 and 4.34E-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.228 mol of HI(g) is added to the flask?
Answer:
[HI] = 0.5239M
[H₂] = 7.05x10⁻²
[I₂] = 7.05x10⁻²
Explanation:
The reaction of HI to produce H₂ and I₂ is:
2HI → H₂ + I₂
Where K of reaction is defined as:
K = [H₂] [I₂] / [HI]²
Replacing with the concentrations of the gases in equilibrium:
K = [4.34x10⁻²] [4.34x10⁻²] / [0.323] ²
K = 0.0181
If you add 0.228 mol = 0.228M (Because volume of the flask is 1.0L), the concentration when the system reaches the equilibrium are:
[HI] = 0.228M + 0.323M - X = 0.551M - X
[H₂] = 4.34x10⁻² + X
[I₂] = 4.34x10⁻² + X
Where X is reaction coordinate.
Replacing in K formula:
K = 0.0181 = [4.34x10⁻²+ X] [4.34x10⁻²+ X] / [0.551 - X] ²
0.0181 = 0.00188356 + 0.0868 X + X² / 0.303601 - 1.102 X + X²
0.005495 - 0.01995 + 0.0181X² = 0.00188356 + 0.0868 X + X²
0 = -0.003611 + 0.10675X + 0.9819X²
Solving for X:
X = -0.136 → False solution, there is no negative concentrations.
X = 0.0271 → Right solution.
Replacing for concentrations of each species:
[HI] = 0.5239M[H₂] = 7.05x10⁻²[I₂] = 7.05x10⁻²PLEASE ANWSER I WILL GOVE YOU POINTS ! And a like Many statues are made out of calcium (marble).Acid rain contains sulfuric acid,which reacts with the calcium carbonate.how many grams of calcium carbonate are consumed if 500.g of sulfuric acid fall onto the statue ? 2CaCO3+1H2SO4->1co2+1H2O+CaSo4
Answer:
1.02 × 10³ g
Explanation:
Step 1: Write the balanced equation
2 CaCO₃ + H₂SO₄ ⇒ CO₂ + H₂O + CaSO₄
Step 2: Calculate the moles corresponding to 500 g of sulfuric acid
The molar mass of sulfuric acid is 98.08 g/mol.
[tex]500g \times \frac{1mol}{98.08g} = 5.10mol[/tex]
Step 3: Calculate the moles of calcium carbonate that react with 5.10 moles of sulfuric acid
The molar ratio of CaCO₃ to H₂SO₄ is 2:1. The moles that react of calcium carbonate are (2/1) × 5.10 mol = 10.2 mol
Step 4: Calculate the mass corresponding to 10.2 moles of calcium carbonate
The molar mass of calcium carbonate is 100.09 g/mol.
[tex]10.2mol \times \frac{100.09 g}{mol} =1.02 \times 10^{3} g[/tex]
A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric acid will: (Assume that the volume does not change upon the addition of hydrochloric acid.)
a. Raise the pH slightly
b. Lower the pH slightly
c. Raise the pH by several units
d. Lower the pH by several units
e. Not change the pH
f. Exceed the buffer capacity
Answer:
Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.
Explanation:
The pH of the buffer solution before the addition of HCl is:
[tex]pH = pKa + log(\frac{[KF]}{[HF]})[/tex]
[tex]pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04[/tex]
The hydrochloric acid added will react with the potassium fluoride as follows:
H₃O⁺(aq) + F⁻(aq) ⇄ HF(aq) + H₂O(l)
The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:
[tex] \eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol [/tex]
[tex] \eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles [/tex]
The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:
[tex] \eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles [/tex]
Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.
We can calculate the pH after the addition of HCl:
HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq) (1)
The number of moles of HF after the reaction of KF with HCl is:
[tex] \eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles [/tex]
And the concentration of HF after the reaction of KF with HCl is is:
[tex] C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L [/tex]
Now, from the equilibrium of equation (1) we have:
[tex] Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]} [/tex]
[tex] Ka = \frac{x^{2}}{0.531 - x} [/tex] (2)
By solving equation (2) for x we have:
x = 0.0187
Finally, the pH after the addition of HCl is:
[tex] pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73 [/tex]
Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.
I hope it helps you!
Given the reaction: 2Na(s) + 2H20(1) 2Na+(aq) + 2OH(aq) + H2(g)
This reaction goes to completion because one of the products formed is
1.
an insoluble base
2.
a soluble base
3
a precipitate
4.
a gas
A soluble base is formed when sodium reacts with water.
What happens when sodium reacts with water?
When sodium reacts with water, it produces strongly alkalic sodium hydroxide which is also called caustic soda and hydrogen gas. In this chemical reaction, energy is absorbed which means it is an exothermic reaction so we can conclude that a soluble base is formed when sodium reacts with water.
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Reactions that undergo shifts in their equilibrium must be...
A.reversible reactions
B.chemical reactions
C.physical reactions
D.nuclear reactions
Answer:
A. Reversible reactions.
Explanation:
Reactions that undergo shifts in their equilibrium must be reversible reactions.
A reversible reaction is a chemical reaction that is capable of forming it's reactants back from the resulting formation of products. This simply means that, reversible reactions are chemical reactions that are in equilibrium because the forward and reverse path happens at the same rate.
For example, the reaction of hydrogen gas [tex]H_{2}[/tex] and iodine gas [tex]I_{2}[/tex] to form a chemical compound called hydrogen Iodide [tex]HI[/tex] is a reversible chemical reaction.
Forward reaction:
[tex]H_{2} + I_{2} ----> 2HI[/tex]
Reverse reaction:
[tex]2HI----> H_{2} + I_{2}[/tex]
A reversible reaction is denoted with a double arrow.
Also, reversible chemical reactions are controlled by the Le Chatelier's principle.
You are required to prepare 500 ml of a 6.00 M solution of HNO3 from a stock solution of 12.0 M. Describe in detail how you would go about preparing this solution. Clearly state the volume of stock solution used, the glassware's used and the procedure.
Answer: 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.
Explanation:
According to the dilution law,
[tex]C_1V_1=C_2V_2[/tex]
where,
[tex]C_1[/tex] = concentration of stock solution = 12.0 M
[tex]V_1[/tex] = volume of stock solution = ?
[tex]C_2[/tex] = concentration of diluted solution= 6.00 M
[tex]V_2[/tex] = volume of diluted acid solution = 500 ml
Putting in the values we get:
[tex]12.0\times V_1=6.00\times 500[/tex]
[tex]V_1=250ml[/tex]
Thus 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.
1. the purpose of the aqueous solutions in a galvanic cell is to?
a. provide ions to be oxidized and reduced
b. provide a path for the free flow of electrons
c. maintain charge balance in the cell
d. correct any volume changes in the cell
2. given these half-reactions,
B2 + 2e- -> 2B- Ecell= 0.662 V
A+ Ie- -> A Ecell= -1.305V
what is the standard potential for the overall reaction. 2A+B2 -> 2AB
a. -1.97 V
b. -0.643 V
c. +3.272 V
d. +1.967 V
Answer:
1. The correct option is;
c. maintains charge balance in the cell
2. The correct option is;
c. +3.272 V
Explanation:
The aqueous solution in a galvanic cell is the electrolyte which is a ionic solution containing that permits the transfer of ions between the separated compartment of the galvanic cell such that the overall system is electrically neutral
Therefore, the aqueous solution maintains the charge balance in the cell
2. Here we have;
B₂ + 2e⁻ → 2B⁻ Ecell = 0.662 V
A⁺ + 1e⁻ → A Ecell = -1.305 V
Hence for the overall reaction, we have;
2A + B₂ → 2AB gives;
(0.662) - 2×(-1.305) = +3.272 V.
What is kinetic energy?
a. The energy of change
b. The energy of distance or volume
c. The energy of motion
d. The energy of position or composition
Answer:
C. The energy of motion
Explanation:
Kinetic energy is the energy associated with the movement of objects.
The kinetic energy of an object depends on both its mass and velocity, with its velocity playing a much greater role.
Example of Kinetic Energy:
1. An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.
// have a great day //
Kinetic energy is the energy of motion.
So kinetic energy is present when objects move.
For example, a snowball rolling down a mountain.
A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction: →+2IO−3aq+12H+aq5Cos+I2s+6H2Ol5Co+2aq Suppose the cell is prepared with 6.64 M IO−3 and 1.54 M H+ in one half-cell and 7.82 M Co+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
Answer:
E = 1.47
Explanation:
To do this, you need to apply the Nerst equation which is the following:
E = E° - RT/nF lnQ (1)
Where:
E: cell voltage
E°: Standard potential reduction
R: universal constant
T: temperature of the system
n: number of electrons transfered during the reaction
F: Faraday constant.
Q: Equilibrium constant
However, as the reaction is taking place at 25 °C, and R and F have constant values, we can reduce the above expression to the following:
E = E° - 0.05916/n lnQ (2)
We can get the value of Q because it has to do with the reaction which is the following:
2IO₃⁻(aq) + 12H⁺(aq) + 5Co(s) ----------> I₂(s) + 5Co²⁺(aq) + 6H₂O(l)
Now, using only the aqueous state the expression of Q will be:
Q = [Co²⁺]⁵ / [H⁺]¹² [IO₃⁻]²
Replacing the values we have:
Q = (7.82)⁵ / (1.54)¹² * (6.64)²
Q = 3.728
Knowing this, all we need to know now is the standard potential reduction of the reaction. To do so, we need to write the two semi equations of reduction and oxidation:
2IO₃⁻ + 12H⁺ + 10e⁻ ---------> I₂ + 6H₂O E₁° = 1.20 V
5Co ---------> 5Co²⁺ + 10e⁻ E₂° = 0.28 V
E° = 1.2 + 0.28 = 1.48 V
Now that we have all the values (n = 10) we can write now the nernst equation to calculate the cell voltage:
E = 1.48 - 0.05916/10 ln (3.728)
E = 1.48 - 0.005916 (1.315872)
E = 1.47 V
This will be the cell voltage
Which of following changes that affect the composition of our atmosphere involve physical changes and which involve chemical reactions? Drag the appropriate items to their respective bins. View Available Hint(s) Reset Help Human activities such as the combustion of fossil fuels, has increased the levels of greenhouse At dew point, the water vapor begins to condense out of air. gases. Human activities, such as the combustion of fossil fuels, generate the aerosols. You observe hail when the temperatures are below the freezing point. Carbon dioxide is produced during respiration of plants, animals, fungi, and microorganisms. Chemical reaction Physical change
Answer:
1) Human activities such as the combustion of fossil fuels, has increased the levels of greenhouse gases---- Chemical reaction.
2) At dew point, the water vapor begins to condense out of air---- Physical changes.
3) Human activities, such as the combustion of fossil fuels, generate the aerosols---- Chemical reaction.
4) You observe hail when the temperatures are below the freezing point---- Physical changes.
5) Carbon dioxide is produced during respiration of plants, animals, fungi, and microorganisms----- Chemical changes.
The total pressure of a mixture of H2, He, and Ar is 99.3 kPa. The partial pressure of the He is 42.7 kPa, and the partial pressure of Ar is 54.7 kPa. What is the partial pressure of hydrogen
Answer:
[tex]P_{H_2}=1.9kPa[/tex]
Explanation:
Hello,
Here, by using the Dalton's law, we can quantify the total pressure of a gaseous mixture by knowing the partial pressure of each gas, in case, hydrogen, helium and argon:
[tex]P_T=P_{H_2}+P_{He}+P_{Ar}[/tex]
In such a way, since we actually know the partial pressure of helium and argon, and the total pressure, we can compute the partial pressure of hydrogen as shown below:
[tex]P_{H_2}=P_T-P_{He}+P_{Ar}=99.3kPa-42.7kPa-54.7kPa\\\\P_{H_2}=1.9kPa[/tex]
Best regards.
A 60 g piece of aluminum at 20°C is cooled to -196°C by placing it in a large container of liquid nitrogen at that temperature. How much nitrogen is vaporized? (Assume that the specific heat of aluminum is constant and is equal to 0.90 kJ/kg·K and that the vaporized nitrogen's temperature does not change.)
Answer:
0.0586 kg
Explanation:
From the question,
Heat lost by the aluminium = heat gain by nitrogen.
CM(t₁-t₂) = cm................... Equation 1
Where C = Specific heat capacity of aluminum, M = mass of aluminum, t₂ = Final temperature, t₁ = initial temperature, c = latent heat of vaporization of nitrogen, m = mass of nitrogen.
make m the subject of the equation
m = CM(t₁-t₂)/c................ Equation 2
Given: C = 900 J/kg.K, M = 60 g = 0.06 kg, t₁ = 20 °C, t₂ = -196 °C
Constant: c = 199200 J/kg
Substitute these values into equation 2
m = 900×0.06×[20-(-196)]/199200
m = 900×0.06×216/199200
m = 0.0586 kg.
Answer:
[tex]m_{N_2}=58.6gN_2[/tex]
Explanation:
Hello,
In this case, for an average temperature of -176 °C, the vaporization enthalpy of liquid nitrogen is 199.2 J/g, thus, we first compute the heat lost by the aluminium by considering it cooled mass, specific heat and temperature change:
[tex]Q=mCp(T_2-T_1)=60g*0.90\frac{J}{g\°C}*(-196-20)\°C\\ \\Q=-11664J[/tex]
Next, heat lost by the aluminium is gained by the nitrogen:
[tex]-Q_{Al}=Q_{N_2}=11664J[/tex]
Therefore, the vaporized nitrogen is:
[tex]m_{N_2}=\frac{Q_{N_2}}{\Delta H_v}=\frac{11664J}{199.2J/g}\\\\m_{N_2}=58.6gN_2[/tex]
Best regards.
why larger amount of H2SO4 is used in the assay of H2O2?
Answer:
This is because Hydrogen peroxide must be acidic, thereby making hydrogen ions taking part in the reaction which act as reactant. It is important also to use H2SO4 because permanganate is a strong oxidizing agent and it cannot oxidize sulphuric acid if enough is added but it can oxidize chloride ions to chlorine if hydrochloric acid is used.
Explanation:
Hydrogen peroxide is used as a chemical agents in cleaning processes for chemical industries and semiconductor plants. In the hydrogen peroxide assay, sulphuric acid is used and the reaction is an exothermic reaction. This produce strong peroxysulphuric acid
Convert 32 K to degrees Celsius.
Answer:
32 K is approx. -241.15° C
3. Use the balanced chemical equation from the last question to solve this situation: You combine 0.5 grams of Na2CO3 with excess CaCl2. How many grams of NaCl would you expect this reaction to produce? Show all work below. g
Answer:
0.27 g
Explanation:
The reaction equation:
[tex]Na_{2} CO_{3} + CaCl_{2}[/tex] → [tex]2NaCl + CaCO_{3}[/tex]
106g of Na2CO3 - 1 mole
0.5g of Na2CO3 = 0.5 ÷ 106
= 0.0047 moles.
1 mole of NaCl - 58.5
⇒ 0.0047 moles = 0.0047 × 58.5
= 0.27g.
When 0.5 grams of Na₂CO₃ react with excess CaCl₂, 0.6 g of NaCl are formed.
We combine 0.5 grams of Na₂CO₃ with excess CaCl₂ and we want to know the mass of NaCl produced. This is a stoichiometry problem.
What is stoichiometry?Stoichiometry refers to the relationship between the quantities of reactants and products before, during, and following chemical reactions.
First, we will write the balanced chemical equation.
Na₂CO₃ + CaCl₂ ⇒ 2 NaCl + CaCO₃
We will consider the following relationships.
The molar mass of Na₂CO₃ is 105.99 g/mol.The molar ratio of Na₂CO₃ to NaCl is 1:2.The molar mass of NaCl is 58.44 g/mol.[tex]0.5 g Na_2CO_3 \times \frac{1molNa_2CO_3}{105.99gNa_2CO_3} \times \frac{2molNaCl}{1molNa_2CO_3} \times \frac{58.44gNaCl}{1molNaCl} = 0.6gNaCl[/tex]
When 0.5 grams of Na₂CO₃ react with excess CaCl₂, 0.6 g of NaCl are formed.
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Consider the balanced chemical reaction below. When the reaction was carried out, the calculated theoretical yield for carbon dioxide was 93.7 grams, but the measured yield was 88.3 grams. What is the percent yield?
Answer:
Percent Yield = 94.237%
Explanation:
CO = Carbon Dioxide = Molar Mass 28g/mol
C = Carbon = 12g/mol
O = Oxygen = 16g/mol
Theoretical yield = 93.7 grams
Actual yield = 88.3 grams
Percent yield = (actual yield /theoretical yield )x100
Percent Yield = (88.3/93.7)x100
Percent Yield = 94.237%
Before running other simulations, try to think of solutions to this global energy dilemma. Write a paragraph in response to the following questions: Should we invest more heavily in renewables at a high up-front financial cost, and/or should we invest in finding more fossil fuels, and/or should we ignore the CO2 limit
Answer:
Yes, we should invest in renewable energies that decrease the levels of carbon dioxide in the atmosphere, that is, decrease the partial pressure of carbon dioxide in the environment since it is lethal for human life, in turn I consider that the exploitation of sources Oil was widely used but I think they should have an end, since they pollute oceans, oxygen, and fuels that are not necessary for today's technology since automobiles could be a base of electric or solar energy
Explanation:
Environmental pollution is a very serious current problem, since we should stop investing in oil sources and be able to keep those profits or that economy in another way or replace them with another resource, since if this greenhouse effect is not perceived as a current serious problem, we could run the risk of suffering natural catastrophes, systematic diseases, affections in the same life.
What is the molarity of a solution that has 4.4 moles BaCl2 in 1.5 Liters of solution?
Answer:
2.9 M
Explanation:
Step 1: Given data
Moles of barium chloride (solute): 4.4 moles
Volume of solution: 1.5 liters
Step 2: Calculate the molarity of barium chloride in the solution
The molarity is a way to quantitatively express the concentration of a solute in a solution. The molarity is equal to the moles of solute divided by the volume, in liters, of solution.
[tex]M=\frac{4.4mol}{1.5L} =2.9 M[/tex]
Select the atoms or ions drawn with valid Lewis dot structures. A) A carbon has a dot on top, right, bottom and to the left.a nitrogen has one dot on top, left and to the bottom and has a charge of minus 3.a nitrogen has a dot on top, right, bottom and to the left. B) An oxygen has two dots on top and bottom and one dot to the left and to the right. C) A carbon has two dots on top, right, bottom and to the left and a charge of plus four. D) An oxygen has two dots on top, left and to the bottom and a charge of minus 2.
Answer:
B, C
Explanation:
The atoms or ions with the valid Lewis dot structures are B and C.
In A;
The Lewis structure of the carbon is correct. Each of the four dots represent the four valence electrons.
The nitrogen with one dot on top, left and to the bottom and has a charge of minus 3 is wrong. For it to have a charge of -3 it must have 8 lewis dots ( two on the top, right, bottom and to the left)
The nitrogen with four dots (on top, right, bottom and to the left) is wrong.
In B;
An oxygen has two dots on top and bottom and one dot to the left and to the right. This is correct , the 6 dots represent the valence electrons of oxygen.
In C;
A carbon has two dots on top, right, bottom and to the left and a charge of plus four. This is correct because the charge indicates that it has gained four extra electrons so its valence electrons is now 8.
In D;
An oxygen has two dots on top, left and to the bottom and a charge of minus 2. This is wrong because the lewis dots are incomplete. Two dots are missing.
Calculate ΔHrxn for the following reaction: C(s) + H2O(g) --> CO(g) + H2(g) Use the following reactions and given ΔH values: C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ 2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ 2 H2 (g) + O2 (g) → 2 H2O (g), ΔH= -483.6 kJ Express your answer numerically, to four significant figures and in terms of kJ.
Answer:
ΔH = 130.5 kJ
Explanation:
Hello,
In this case, by using the Hess law, we compute the enthalpy of the required reaction:
C(s) + H2O(g) --> CO(g) + H2(g)
Thus, the first step is to keep the following reaction unchanged:
C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ
Next, we invert and halve this reaction:
2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ
So the enthalpy of reaction is inverted and halved:
CO2 (g) → CO (g) + 1/2 O2 (g) ΔH= 283 kJ
Then, we also invert and halve this reaction:
2 H2 (g) + O2 (g) → 2 H2O ΔH= -483.6 kJ
So the enthalpy of reaction is inverted and halved as well:
H2O → H2 (g) + 1/2 O2 (g) ΔH= 241.8 kJ
Finally, we add the three reactions to obtain the required reaction:
= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + 1/2 O2 (g) + CO (g) + 1/2 O2 (g) + CO2 (g)
= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + O2 (g) + CO (g) + CO2 (g)
= C (s) + H2O → H2 (g) CO (g)
So enthalpy is computed by:
ΔH = -393.5 kJ + 283 kJ + 241.8 kJ
ΔH = 130.5 kJ
Best regards.
Considering the Hess's Law, the enthalpy change for the reaction is 131.3 kJ.
Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.
In this case you want to calculate the enthalpy change of:
C(s) + H₂O(g) → CO(g) + H₂(g)
You know the following reactions, with their corresponding enthalpies:
Equation 1: C (s) + O₂(g) → CO₂ (g) ΔH = -393.5 kJ
Equation 2: 2 CO (g) + O₂ (g) → 2 CO₂ (g) ΔH= -566.0 kJ
Equation 3: 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ΔH= -483.6 kJ
First stepFirst, to obtain the enthalpy of the desired chemical reaction you need one mole of C(s) on reactant side and it is present in first equation so let's write this as such.
Second stepNow, 1 mole of CO(g) must be a product and is present in the second equation. Since this equation has 2 moles of CO(g) on the reactant side, it is necessary to locate this component on the product side (invert it) and divide it by 2 to obtain 1 mole of CO(g).
When an equation is inverted, the sign of ΔH also changes.
And since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is divided by 2, the variation of enthalpy also is divided by 2.
Third step
Finally, 1 mole of H₂O(g) must be a reactant and is present in the third equation. Since this equation has 2 moles of CO(g) on the product side, it is necessary to locate this component on the product side (invert it) and divide it by 2 to obtain 1 mole of H₂O(g).
So, the sign of ΔH also changes and the variation of enthalpy is divided by 2.
In summary, you know that three equations with their corresponding enthalpies are:
Equation 1: C (s) + O₂(g) → CO₂ (g) ΔH = -393.5 kJ
Equation 2: CO₂ (g) → CO (g) + [tex]\frac{1}{2}[/tex] O₂ (g) ΔH= 283 kJ
Equation 3: H₂O (g) → H₂ (g) + [tex]\frac{1}{2}[/tex] O₂ (g) ΔH= 241.8 kJ
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
C(s) + H₂O(g) → CO(g) + H₂(g) ΔH= 131.3 kJ
Finally, the enthalpy change for the reaction is 131.3 kJ.
Learn more:
brainly.com/question/5976752?referrer=searchResults brainly.com/question/13707449?referrer=searchResults brainly.com/question/13707449?referrer=searchResults brainly.com/question/6263007?referrer=searchResults brainly.com/question/14641878?referrer=searchResults brainly.com/question/2912965?referrer=searchResultsA gas has a volume of 1140 ml at 37 ºC and 82.6 kPa pressure. Calculate its volume at STP.
Answer:
Volume, V2 at STP = 818.61ml.
Explanation:
Given the following parameters;
Volume, V1 = 1140ml
Volume, V2 =?
Temperature, T1 = 37ºC to Kelvin = 273+37 = 310K
Temperature, T2 = 273K is the standard temperature.
Pressure, P1 = 82.6kPa
Pressure, P2 = 101.3kPa is the standard pressure.
To solve for volume at standard temperature and pressure (STP), we would use the combined gas law;
The combined gas law is a combination of the other three gas laws, namely Gay Lusac's law, Boyle's law and Charles's law.
Combined gas law states that the ratio of the product or multiplication of volume and pressure to the temperature of a gas is equal to a constant.
Mathematically, [tex]PV/T = K[/tex]
P1V1/T1 = P2V2/T2
Let's make V2 to the subject formula;
Cross multiplying gives,
P1V1T2 = P2V2T1
Hence, [tex]V2 = (P1V1T2)/P2T1[/tex]
Substituting the parameters;
V2 = (82.6*1140*273)/101.3*310
V2 = 25706772/31403
V2 = 818.61ml.
Which of the following isotopes of chlorine has 16 neutrons in its nucleus?
33Cl
38Cl
42Cl
Answer:
33Cl
Explanation:
The atomic mass of an element is made up of the proton and neutron.
chlorine has a constant number of 17 protons.
33-17 = 16
Determine whether the following pairs of elements can form ionic compounds
Answer:
Oxygen and Magnesium.
Based on the difference in electronegativity, potassium and sulfur, chlorine and lithium, and oxygen and magnesium can form ionic compounds, whereas lithium and calcium, sulfur and bromine, and manganese and chlorine cannot.
Ionic compounds are formed when there is a high electronegativity difference between two elements.
To determine whether the following pairs of elements can form ionic compounds, we need to look at their electronegativity difference:
1. Lithium and calcium: The electronegativity difference between Lithium and calcium is only 0.9, which is less than 1.7. Therefore, they cannot form an ionic compound.
2. Sulfur and bromine: The electronegativity difference between sulfur and bromine is 0.9, which is less than 1.7. Therefore, they cannot form an ionic compound.
3. Manganese and chlorine: The electronegativity difference between manganese and chlorine is 1.5, which is less than 1.7. Therefore, they cannot form an ionic compound.
4. Potassium and sulfur: The electronegativity difference between potassium and sulfur is 2.4, which is greater than 1.7. Therefore, they can form an ionic compound.
5. Chlorine and lithium: The electronegativity difference between chlorine and lithium is 2.8, which is greater than 1.7. Therefore, they can form an ionic compound.
6. Oxygen and magnesium: The electronegativity difference between oxygen and magnesium is 1.7, which is equal to 1.7. Therefore, they can form an ionic compound.
Therefore, based on the electronegativity difference between the elements, potassium and sulfur, chlorine and lithium, and oxygen and magnesium can form ionic compounds, while lithium and calcium, sulfur and bromine, and manganese and chlorine cannot form ionic compounds.
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Calculate the molar mass of a gaseous substance if 0.125 g of the gas occupies 93.3 mL at STP.
30.2 g/mol
30.4g/mol
30.6 g/mol
30.0 g/mol
None of the above
Answer:
30.0g/mol
Explanation:
Step 1: Given data
Mass of the gas: 0.125 gPressure (P): 1 atm (standard pressure)Temperature (T): 273.15 K (standard temperature)Volume (V): 93.3 mLStep 2: Calculate the moles of the gas
We will use the ideal gas equation.
[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1atm \times 0.0933L}{\frac{0.0821atm.L}{mol.K} \times 273.15K} = 4.16 \times 10^{-3} mol[/tex]
Step 3: Calculate the molar mass of the gas
4.16 × 10⁻³ moles correspond to a mass of 0.125 g. The molar mass of the gas is:
[tex]\frac{0.125g}{4.16 \times 10^{-3} mol} =30.0g/mol[/tex]
The molar mass of the 0.125 g of the gas occupies 93.3 mL at STP is 30.0 g/mol.
Number of moles of Gas at STP,
[tex]\bold{n =\dfrac {PV}{RT}}[/tex]
where,
P - pressure
V- volume
R- gas constant
T - temperature
Put the values in the formula,
[tex]\bold{n =\dfrac {1 \times 0.0933} {0.082 \times 273.15 }}\\\\\bold{n =4.16 \timesw 10^-^3}[/tex]
The molar mass of the gas can be calculated using formula,
[tex]\bold {m = \dfrac {w}{n}}\\\\\bold {m = \dfrac {0.125} {4.16 \times 10^-^3}}\\\\\bold {m = 30g/mol}[/tex]
The molar mass of the 0.125 g of the gas occupies 93.3 mL at STP is 30.0 g/mol.
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Trimix is a general name for a type of gas blend used by technical divers and contains nitrogen, oxygen and helium. In one Trimix blend, the partial pressures of each gas are 55.0 atm oxygen, 90.0 atm nitrogen, and 50.0 atm helium. What is the percent oxygen (by volume) in this trimex blend
Answer:
The correct answer is 28.2 %.
Explanation:
Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,
Ptotal = PO2 + PN2 + PHe
= 55 + 90 + 50
= 195 atm
The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100
Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,
%VO2 = (55/195) * 100 = 28.2%
%VN2 = (90/195) * 100 = 46%
%VHe = (50/195) * 100 = 25.64%
Hence, the percent oxygen by volume present in the blend is 28.2 %.
Ideally, how should a buffer be prepared?
A. in such a way that pH = POH
B. in such a way that pH = pka
c. in such a way that pH = Kg
D. in such a way that pH = 1
Answer:
I may not be correct but i think its pH = Kg
Explanation:
Answer:
B. in such a way that pH = pKa
Explanation:
Ideally, the pH of the desired solution should have the same pKa as the pH, making the ratio 1:1.
What are the differences between cholesterol and ergosterole?
