Answer:
1. Strong electrolytes: HCl and NaOH.
Weak electrolytes: CH₃COOH and NH₃.
Nonelectrolytes: (NH₂)₂CO (urea) and CH₃OH (methanol).
2. C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]
3. The spectator ions are Na⁺ and C₂H₃O₂⁻.
4. Written in explanation section.
Explanation:
An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.A characteristic of strong electrolytes is that the solute is assumed to be 100 percent dissociated into ions in solution, therefore they are great conductors of electricity. Eg: HCl and NaOH.
On the other hand, weak electrolytes are not completely dissociated in solution, therefore they are poor conductors of electricity. Eg: CH₃COOH and NH₃.
(By dissociation we mean the breaking up of the compound into cations and anions).
A nonelectrolyte does not conduct electricity when dissolved in water. Eg: (NH₂)₂CO (urea) and CH₃OH (methanol).
2. The products formed on each reaction are (always remember to balance the equations):
A) [tex]2 LiOH (aq) + Na₂S (aq) → Li₂S (aq) + 2NaOH (aq)[/tex]
B) [tex](NH₄)₂SO4 (aq) +2 LiCl (aq) → 2NH₄Cl (aq) + Li₂SO₄ (aq)[/tex]
C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]
D) [tex]D) KNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + KOH (aq) [/tex]
The reaction C will produce SrSO₄, a white color precipitate.
3. When ionic compounds dissolve in water, they break apart into their component cations and anions. To be more realistic, the equations should show the dissociation of dissolved ionic compounds into ions. This is called a ionic equation which shows dissolved species as free ions. To see whether a precipitate might form from this solution, we first combine the cation and anion from different compounds, and refer to the solubility rules. The spectator ions are ions that are not involved in the overall reaction.
Therefore, for the equation chosen above:
[tex]Sr²⁺ + 2C₂H₃O₂⁻ + 2Na⁺ + SO₄²⁻ → SrSO₄ + 2Na⁺ + 2C₂H₃O₂⁻[/tex]
Because spectator ions appear on both sides of an equation, they can be eliminated from the ionic equation.
[tex]Sr²⁺ + SO₄²⁻ → SrSO₄[/tex]
Finally, we end up with the net ionic equation, which shows only the species that actually take part in the reaction.
In this reaction, the spectator ions are Na⁺ and C₂H₃O₂⁻.
4. Molecular equations:
A) [tex]2 LiOH (aq) + Na₂S (aq) → Li₂S (aq) + 2NaOH (aq)[/tex]
B) [tex](NH₄)₂SO4 (aq) +2 LiCl (aq) → 2NH₄Cl (aq) + Li₂SO₄ (aq)[/tex]
C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]
D) [tex]D) KNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + KOH (aq) [/tex]
Complete ionic equations:
A) [tex]2Li⁺ + OH⁻ + 2Na⁺ + S²⁻ → 2Li⁺ + S²⁻ + 2Na⁺ +OH⁻[/tex]
B) [tex]2NH₄⁺ + SO₄²⁻ + 2Li⁺ + 2Cl⁻ → 2Li⁺ +S²⁻ +2NH₄⁺ +2Cl⁻ + 2Li⁺ + SO₄²⁻[/tex]
C) [tex]Sr²⁺ + 2C₂H₃O₂⁻ + 2Na⁺ + SO₄²⁻ → SrSO₄ + 2Na⁺ + 2C₂H₃O₂⁻[/tex]
D) [tex]K⁺ + NO₃⁻ + Na⁺ +OH⁻ → Na⁺ + NO₃⁻ + K⁺ + OH⁻[/tex]
If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, no precipitation reaction occurs. The only net equation can be written for reaction C):
C) [tex]Sr²⁺ + SO₄²⁻ → SrSO₄[/tex]
Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What is the theoretical yield of carbon dioxide formed from the reaction of 1.28 g of methane and 10.1 g of oxygen gas? Be sure your answer has the correct number of significant digits in it. 02
Answer:
Theoretical yield = 3.51 g
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
For [tex]CH_4[/tex] :-
Mass of [tex]CH_4[/tex] = 1.28 g
Molar mass of [tex]CH_4[/tex] = 16.04 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{1.28\ g}{16.04\ g/mol}[/tex]
[tex]Moles_{CH_4}= 0.0798\ mol[/tex]
For [tex]O_2[/tex] :-
Mass of [tex]O_2[/tex] = 10.1 g
Molar mass of [tex]O_2[/tex] = 31.998 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{10.1\ g}{31.998\ g/mol}[/tex]
[tex]Moles_{O_2}= 0.3156\ mol[/tex]
According to the given reaction:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
1 mole of methane gas reacts with 2 moles of oxygen gas
0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas
Moles of oxygen gas = 0.1596 moles
Available moles of oxygen gas = 0.3156 moles
Limiting reagent is the one which is present in small amount. Thus, [tex]CH_4[/tex] is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of methane gas on reaction produces 1 mole of carbon dioxide.
0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.
Mole of carbon dioxide = 0.0798 mole
Molar mass of carbon dioxide = 44.01 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]0.0798\ moles= \frac{Mass}{44.01\ g/mol}[/tex]
Mass of [tex]CO_2[/tex] = 3.51 g
Theoretical yield = 3.51 g
The theoretical yield of carbon dioxide formed from the reaction of 1.28 g of methane and 10.1 g of oxygen gas is 3.51 g.
The balanced equation for the reaction between methane (CH4) and oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O) is:
CH4 + 2O2 → CO2 + 2H2O
To find the theoretical yield of carbon dioxide, we need to calculate the moles of methane and oxygen and use the stoichiometry of the balanced equation.
Given: Mass of methane (CH4) = 1.28 g
Mass of oxygen gas (O2) = 10.1 g
Step 1: Calculate the moles of methane and oxygen using their molar masses:
Moles of CH4 = mass / molar mass = 1.28 g / 16.04 g/mol = 0.0798 mol
Moles of O2 = mass / molar mass = 10.1 g / 32.00 g/mol = 0.3156 mol
Step 2: Determine the limiting reactant (the reactant that is completely consumed first) by comparing the mole ratios of CH4 and O2 in the balanced equation. From the equation, for every 1 mol of CH4 there are 2 moles of O2 needed. Therefore, the mole ratio of CH4 to O2 is 1:2.
Since the mole ratio of CH4 to O2 is 1:2, and there are fewer moles of CH4 (0.0798 mol) compared to the moles of O2 (0.3156 mol), CH4 is the limiting reactant.
Step 3: Calculate the moles of CO2 produced using the mole ratio from the balanced equation:
Moles of CO2 = moles of CH4 x (1 mol of CO2 / 1 mol of CH4) = 0.0798 mol x (1 mol / 1 mol) = 0.0798 mol
Step 4: Convert the moles of CO2 to grams using the molar mass of CO2:
Mass of CO2 = moles x molar mass = 0.0798 mol x 44.01 g/mol = 3.51 g
Therefore, the theoretical yield of carbon dioxide formed from the reaction of 1.28 g of methane and 10.1 g of oxygen gas is 3.51 g (to the correct number of significant digits).
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Metals with ________ electron configurations characteristically form diamagnetic, square planar complexes.
a. d6
b. d8
c. d10
d. d0
e. d9
Answer:
B
Explanation:
Diamagnetism , paramagnetism and ferromagnetism are terms which are used to describe the magnetic activity of transition metals. These terms helps to know the response the metal will have when placed in a magnetic field. These activities can be discerned from the d-block electronic configuration. If the number of electrons are even, this means they all form a pair and the metal is diamagnetic. If otherwise, there is an unpaired electron which causes the paramagnetic activity of the metal in magnetic field.
To the question, options A-D are diamagnetic but configurations with d8 are the ones that form square planar complexes
Which of the following correctly represents the transmutation in which a curium-242 nucleus is bombarded with an alpha particle to produce a californium-245 nucleus?^242_96 Cm(^1_0 n, ^4_2 He)^245_98 Cf^242_96 Cm(^4_2 He, ^1_1 p)^245_98 Cf^242_96 Cm(^4_2 He, 2^1_1 p)^245_98 Cf^242_96 Cm(^4_2 He, ^1_0 n)^245_98 Cf^242_96 Cm(^4_2 He, ^1_-1 e)^245_98 Cf
Answer: The chemical equation is written below.
Explanation:
Transmutation is defined as the process in which one chemical isotope gets converted to another chemical isotope. The number of protons or neutrons in the isotope gets changed.
The chemical equation for the reaction of curium-242 nucleus with alpha particle (helium nucleus) follows:
[tex]_{96}^{242}\textrm{Cm}+_4^2\textrm{He}\rightarrow _{98}^{245}\textrm{Cf}+_0^1\textrm{n}[/tex]
The product formed in the nuclear reaction are californium-245 nucleus and a neutron particle.
Write balanced half-reactions for the following redox reaction: 5Cl2(g)+2Mn+2(aq)+8H2O(l)→ 10Cl−(aq)+2MnO−4(aq)+16H+(aq)a. reduction: __.b. oxidation: ___.
Answer:
a. 5Cl₂ + 10e⁻ → 10Cl⁻
b. 4H₂O + Mn²⁺ → MnO₄⁻ + 5e⁻ + 8H⁺
Explanation:
5Cl2(g)+2Mn+2(aq)+8H2O(l)→ 10Cl−(aq)+2MnO4-(aq)+16H+(aq)
First of all, think in all the oxidation numbers for each compound. That's the way, you can notice the half reaction.
When the oxidation number, decrease, you have reduction. Compound is wining electrons.
When the oxidation number increase, you have oxidation. Compound is losing electrons.
Cl2(g) - Atoms in ground state, has 0 as oxidation number. In products side, you have the anion chloride which act with -1, so chlorine has been reduced.
Mn2+ - The manganese ion is already telling you with 2, that is its oxidation number. On the side products, the element was transformed into the permanganate anion; how oxygen acts with -2 and there are 4 atoms, it has -8 as oxidation state but since the general charge is -1 the Mn is acting with +7. From + 2 it went to +7, which means that it increased, so it has oxidized.
5Cl₂ + 10e⁻ → 10Cl⁻ - REDUCTION
The chlorine had to gain 1 electron to go from 0 to -1, but being a diatomic molecule were 2 electrons, but finally so that the charges are balanced and because there are 5 moles, it ends up gaining 10 electrons.
Mn²⁺ → MnO₄⁻ - OXIDATION
Look that in main reaction we have H⁺, that is the clue to notice us, that we are in acidic medium. So if we have 4 O, in MnO₄⁻, we have to complete with 4H₂O in the other side of O. And to ballance the protons, we have to add 8H⁺ in product side
4H₂O + Mn²⁺ → MnO₄⁻ + 5e⁻ + 8H⁺ OXIDATION
How do u finish this?. You have to multipply .2, the oxidation one to balance the e⁻ so, they can be cancelled.
5Cl₂ + 10e⁻ → 10Cl⁻
(4H₂O + Mn²⁺ → MnO₄⁻ + 5e⁻ + 8H⁺ ).2
8H₂O + 2Mn²⁺ → 2MnO₄⁻ + 10e⁻ + 16H⁺
5Cl₂ + 10e⁻ + 8H₂O + 2Mn²⁺ → 10Cl⁻ + 2MnO₄⁻ + 10e⁻ + 16H⁺
5Cl₂ + 8H₂O + 2Mn²⁺ → 10Cl⁻ + 2MnO₄⁻ + 16H⁺
The half-reactions for the given redox reaction includes the reduction of Mn2+(aq) to MnO4-(aq) with the addition of 5 electrons and 8 hydrogen ions and the oxidation of Cl2(g) to 10 Cl-(aq), losing 10 electrons in the process.
Explanation:The redox reaction is related to the process of reduction and oxidation which often takes place in electrochemical cells. This process includes half-reactions, which separately represent the oxidation and reduction in the redox reaction. For the given equation, the oxidation and reduction half-reactions are as follows:
a. Reduction: Mn2+(aq) + 5e- + 8H+(aq) → MnO4-(aq) + 4H2O(l).
b. Oxidation: Cl2(g) → 10e- + 10Cl-(aq).
These equations represent the transformation of Mn2+ and Cl2 in the given redox reaction where Mn2+ is being reduced and Cl2 is being oxidized.
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In Ontario, some electricity comes from coal-burning generators. Coal is a natural form of carbon that has a large amount of sulphur mixed in with it. Answer the following questions based on the burning of coal to produce energy.
a) Write the word equations and balanced chemical equations for the burning of carbon and the burning of sulphur. (4 marks)
b) Which of these products is harmful to the environment? How is it harmful? (2 marks)
c) Write the word equation and balanced chemical equation for the reaction that produces this harmful environmental effect. (2 marks)
d)Explain why it is important to make sure your furnace is tuned up and in proper working order before the winter
Answer:
a)
[tex]C(s)+O_{2}(g) \rightarrow CO_{2}(g)[/tex]
[tex]S(s)+O_{2}(g) \rightarrow SO_{2}(g)[/tex]
b)
Sulphurdioxide
c)
[tex]SO_{2}(g) + H_{2}O(l) \rightarrow H_{2}SO_{3}(l)[/tex]
[tex]SO_{2}(g) + \frac{1}{2}O_{2} \rightarrow SO_{3}(g)[/tex]
d)
Incomplete combustion produce harmful gases.
Explanation:
a)
Carbon reacts with atmospheric oxygen to form carbondioxide as well as sulphur dioxide.
The chemical equations are as follows.
[tex]C(s)+O_{2}(g) \rightarrow CO_{2}(g)[/tex]
[tex]S(s)+O_{2}(g) \rightarrow SO_{2}(g)[/tex]
b)
Sulfur dioxide is very harmful to the environment to cause acid rains.
This harmful gas mix with rain water to form sulphuric acid.
c)
The balanced chemical equation for the reaction that produces harmful environmental effects is as follows.
[tex]SO_{2}(g) + H_{2}O(l) \rightarrow H_{2}SO_{3}(l)[/tex]
[tex]SO_{2}(g) + \frac{1}{2}O_{2} \rightarrow SO_{3}(g)[/tex]
[tex]SO_{3}(g) +H_{2}O(l) \rightarrow H_{2}SO_{4}(l)[/tex]
d)
In the absence of proper amount of oxygen required for combustion, incomplete combustion will take place which will result in formation of more carbondioxide an other harmful gases.
Calculate the change internal energy (ΔE) for a system that is giving off 65.0 kJ of heat and is performing 855 J of work on the surroundings. Calculate the change internal energy (ΔE) for a system that is giving off 65.0 kJ of heat and is performing 855 J of work on the surroundings. 9.00 x 102 kJ 64.1 kJ -9.00 x 102 kJ -64.1 kJ -65.9 kJ
A system that is giving off 65.0 kJ of heat and is performing 855 J of work on the surroundings has a change in the internal energy of -65.9 kJ.
The system is giving off 65.0 kJ of heat (Q). By convention, when a system is releasing heat, Q < 0. Then, Q = -65.0 kJ.The system is also performing 855 J of work (W) on the surroundings. By convention, when a system performs work on the surroundings, W < 0. Then, W = -855 J (-0.855 kJ)According to the First Law of Thermodynamics, we can calculate the change in the internal energy (ΔE) of the system using the following expression.
[tex]\Delta E = Q + W = (-65.0 kJ) + (-0.855 kJ) = -65.9 kJ[/tex]
A system that is giving off 65.0 kJ of heat and is performing 855 J of work on the surroundings has a change in the internal energy of -65.9 kJ.
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An unknown quantity of sugar was completely dissolved in water at 75 degrees Celsius to form a clear and colorless solution. The temperature of the solution was then lowered to 25 degrees Celsius while being mixed.
Question 1: Was the solution at 75 degrees Celsius saturated, unsaturated or supersaturated?
Question 2: Explain your reason for choosing the answer in question 1.
Answer:
67
Explanation:
Question 1: The solution at 75 degrees Celsius was likely saturated.
Question 2: The reason for choosing "saturated" is that the sugar was completely dissolved in water at 75 degrees Celsius, forming a clear and colorless solution.
What happens in such solutionsIn a saturated solution, the maximum amount of solute (sugar, in this case) that can dissolve at that temperature has already dissolved. When the solution is clear and colorless at that temperature, it suggests that the solution contains as much sugar as it can hold under those conditions.
If it were unsaturated, there would still be room for more sugar to dissolve, and if it were supersaturated, it would become unstable and potentially precipitate out excess solute when the temperature was lowered.
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A buffer is prepared by mixing hypochlorous acid ( HClO ) and sodium hypochlorite ( NaClO ) . If a strong base, such as NaOH , is added to this buffer, which buffer component neutralizes the additional hydroxide ions ( OH − ) ? ClO − HClO Write a balanced chemical equation for the reaction of the selected buffer component and the hydroxide ion ( OH − ) . Do not include physical states.
The chemical equation for the neutralization of hydroxide ion by HClO is:
HClO + OH ---> H2O + ClO-What a buffer?A buffer is a solution which resists changes to its pH when a small quantity of strong acid or base is added to it.
A solution of weak acid such as hypochlorous acid (HClO) and its basic salt that is sodium hypochlorite (NaClO) forms a buffer solution
When a strong base is added to the buffer, the excess hydroxide ion will be neutralized by hydrogen ions from the acid, HClO.
The chemical equation for the neutralization of hydroxide ion with acid follows:
HClO + OH ---> H2O + ClO-Therefore, the balanced chemical equation is such that the excess OH- is neutralized.
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Consider the following data for air trapped in a flask: Pressure = 0.988 atm Room Temperature = 23.5°C Volume of the flask = 1.042 L For this calculation, assume air is 78.5% nitrogen and 21.5% oxygen (by number of moles). R = 0.0821 L•atm•mol-1•K-1 N = 14.01 g/mol O = 16.00 g/mol What is the mass of air in the flask
Answer:
total mass will be = = 1.207g
Explanation:
First what is given
Pressure P= 0.988 atm Room TemperatureT = 23.5°C= 296.5 K
Volume V= 1.042 L
Nitrogen in air is 80 % (moles number) = 0.8
Ideal gas constant R = 0.0821 L atmmol-1K-1
Given mass of N = 14.01 g/mol
Given mass of oxygen O = 16.00 g/mol
Total number of moles = ?
So first we have to find the total number of moles by using formula
Total number of moles
n = PV/ RT
adding the values
moles n= 0.988atm x 1.042L / (0.0821L-atm/mole-K x 296.6K)
= 1.0294 / 24.35
= 0.042 moles (total number of moles)
So by using Nitrogen percentage
Moles of nitrogen = total moles x 80/100
= 0.042moles x 0.8
So moles of O2= Total moles – moles of N2
= 0.042moles - 0.034moles
Moles of O2 = 0.008moles
Now for finding the mass of the N2 and oxygen
Mass of Nitrogen N2 = no of moles x molar mass
= 0.034 x 28
= 0.952 g
Mass of oxygen O2 = no of moles x molar mass
= 0.008 x 32
= 0.256 g
total mass will be = Mass of Nitrogen N2 + Mass of oxygen O2
=0.952 g + 0.256 g
=1.207g
The mass of air in the flask = 1.207g
Given:
Pressure, P= 0.988 atm
Room Temperature, T = 23.5°C= 296.5 K
Volume, V= 1.042 L
Nitrogen in air is 80 % (moles number) = 0.8
Ideal gas constant R = 0.0821 L atm [tex]mol^{-1}K^{-1}[/tex]
Molar mass of N = 14.01 g/mol
Molar mass of oxygen O = 16.00 g/mol
To find:
Total number of moles = ?
Calculation for number of moles:From ideal gas law:
n = PV/ RT
n= 0.988atm * 1.042L / (0.0821 L atm [tex]mol^{-1}K^{-1}[/tex] * 296.6K)
n = 1.0294 / 24.35
n= 0.042 moles
Using mol fraction we will calculate moles for nitrogen and oxygen:
Moles of nitrogen = total moles * 80/100
Moles of nitrogen = 0.042moles * 0.8 = 0.034 moles
So, Moles of O₂ = Total moles – moles of N₂
Moles of O₂ = 0.042 moles - 0.034 moles
Moles of O₂ = 0.008 moles
Calculation for mass:
Mass of Nitrogen N₂ = no of moles x molar mass
= 0.034 x 28
= 0.952 g
Mass of oxygen O₂ = no of moles x molar mass
= 0.008 * 32
= 0.256 g
Total mass will be = Mass of Nitrogen N₂ + Mass of oxygen O₂
=0.952 g + 0.256 g
Total mass = 1.207g
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GenAlex Medical, a leading manufacturer of medical laboratory equipment, is designing a new automated system that can detect therapeutic levels of dissolved acetaminophen (10. to 30./μgmL), using a blood sample that is as small as 2.0mL. Calculate the minimum mass in milligrams of acetaminophen that the new system must be able to detect. Round your answer to 2 significant digits.
Answer:
The minimum mass of acetaminophen that the new system must be able to detect is 0.02 milligrams.
Explanation:
Range of the equipment to detect acetaminophen in blood = 10.0 to 30.0 μg/mL
Minimum amount of acetaminophen that can be detected= 10.0 μg/mL
Volume of blood sample = 2 mL
Amount of acetaminophen in 2 ml sample of blood =
[tex]2 mL\times 10.0 \mu g/mL=20.0 \mu g[/tex]
20.0 μg = 0.02 mg (1 μg = 0.001 mg)
The minimum mass of acetaminophen that the new system must be able to detect is 0.02 milligrams.
Draw all of the constitutional isomeric monochlorination products resulting from the reaction of 2,3−dimethylbutane under radical substitution conditions.
(a) primary alkyl halide
(b) tertiary alkyl halide
Answer:
a) 2-chloro-2,3-dimethylbutane
b) 1-chloro-2,3-dimethylbutane
Explanation:
The monochlorination is a reaction in which we have to add only 1 Cl to the molecule. In this case, we will have to add a Cl to a primary carbon (a) and to a tertiary carbon (b).
In the monochlorination of the primary carbon, we can choose any methyl carbon. For the monochlorination of the terciary carbon we have to choose an CH carbon.
(See the figure)
Final answer:
When 2,3-dimethylbutane undergoes monochlorination under radical substitution conditions, multiple constitutional isomers can be formed due to the presence of primary and tertiary carbon atoms.
Explanation:
When 2,3-dimethylbutane undergoes monochlorination under radical substitution conditions, multiple constitutional isomers can be formed. The monochlorination reaction involves the substitution of a hydrogen atom on the carbon chain with a chlorine atom. Since 2,3-dimethylbutane has two different types of carbon atoms (primary and tertiary), the resulting constitutional isomers will also differ.
(a) Primary alkyl halide: The primary carbon atom is bonded to one other carbon atom and one hydrogen atom. So, a constitutional isomer could be formed by substituting the hydrogen atom with a chlorine atom.
(b) Tertiary alkyl halide: The tertiary carbon atom is bonded to three other carbon atoms. So, a constitutional isomer could be formed by substituting one of the carbon atoms with a chlorine atom.
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for H2O, and on which side of the balanced equation should it appear?
MnO4–(aq)+Br–(aq)→Mn2+(aq)+Br2(l)
Select one:
a. 1, reactant side
b. 2, product side
c. 8, product side
d. 16, reactant side
e. 4, product side
Answer:
c. 8, product side
Explanation:
In order to balance a redox reaction we use the ion-electron method, which has the following steps:
Step 1: identify oxidation and reduction half-reaction.
Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)
Reduction: Br⁻(aq) → Br₂(l)
Step 2: perform the mass balance adding H⁺ and H₂O where necessary
8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)
2 Br⁻(aq) → Br₂(l)
Step 3: perform the electrical balance adding electrons where necessary.
8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)
2 Br⁻(aq) → Br₂(l) + 2 e⁻
Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.
2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))
5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)
Step 5: add both half-reactions side to side.
16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻
16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)
In Part B you calculated ΔG∘ for the following redox reaction:Cu2+(aq)+Co(s)→Cu(s)+Co2+(aq), ΔG∘=−1.18×105JBased on the value of ΔG∘ for the given redox reaction, identify the spontaneity of the reaction.a) nonspontaneousb) at equilibriumc) spontaneous
Answer:
c) spontaneous
Explanation:
According the equation of Gibb's free energy -
∆G = ∆H -T∆S
∆G = is the change in gibb's free energy
∆H = is the change in enthalpy
T = temperature
∆S = is the change in entropy .
And , the sign of the ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,
i.e. ,
if
ΔG < 0 , the reaction is Spontaneous ΔG > 0 , the reaction is non Spontaneous ΔG = 0 , the reaction is at equilibriumFrom the question ,
The value for ΔG is negative ,
Hence ,
ΔG < 0 , the reaction is Spontaneous
The standard cell potential (E°cell) for the reaction below is +1.10V. The cell potential for this reaction is ________ V when the concentration of [Cu2+]=1.0⋅10−5M and [Zn2+]=2.5M. Zn (s) + Cu2+ (aq) → Cu (s) + Zn2+ (aq) The standard cell potential () for the reaction below is . The cell potential for this reaction is ________ when the concentration of and (s) + (aq) (s) + (aq) 0.78 1.10 0.94 1.26 1.42
Answer: 0.94 V
Explanation:
For the given chemical reaction :
[tex]Zn(s)+Cu^{2+}(aq)\rightarrow Cu(s)+Zn^{2+}[/tex]
Using Nernst equation :
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}[/tex]
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = [tex]298K[/tex]
n = number of electrons in oxidation-reduction reaction = 2
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.10 V
[tex]E_{cell}[/tex] = emf of the cell = ?
Now put all the given values in the above equation, we get:
[tex]E_{cell}=+1.10-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{2.5}{1.0\times 10^{-5}}[/tex]
[tex]E_{cell}=+1.10-0.16V=0.94V[/tex]
The cell potential for this reaction is 0.94 V
The cell potential for the given reaction depends on the concentration of the reactants. We can use the Nernst equation to calculate the cell potential when the concentration of [Cu2+]=1.0⋅10−5M and [Zn2+]=2.5M.
Explanation:The standard cell potential (E°cell) for the given reaction is +1.10V. To find the cell potential for this reaction when the concentrations of [Cu2+]=1.0⋅10−5M and [Zn2+] = 2.5M, we need to use the Nernst equation:
Ecell = E°cell - (0.0592/n) x log(Q)
Here, n is the number of electrons involved in the reaction, and Q is the reaction quotient, which is equal to [Zn2+]/[Cu2+]. In this case, n = 2. Plugging in the values, we get:
Ecell = 1.10V - (0.0592/2) x log((2.5M)/ (1.0⋅10−5M))
Solving this equation will give us the cell potential for the given concentrations.
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Using a 300 MHz NMR instrument:
a. How many Hz downfield from TMS is a signal at 2.5 ppm?
b. If a signal comes at 1200 Hz downfield from TMS, at what ppm does it occur?
c. If two peaks are separated by 2 ppm, how many I-AZ does this correspond to?
Answer:
a. 750Hz, b. 4.0ppm, c. 600Hz
Explanation:
The Downfield Shift (Hz) is given by the formula
Downfield Shift (Hz) = Chemical Shift (ppm) x Spectrometer Frequency (Hz)
Using the above formula we can solve all three parts easily
a. fspec = 300 MHz, Chem. Shift = 2.5ppm, 1MHz = 10⁶ Hz, 1ppm (parts per million) = 10⁻⁶
Downfield Shift (Hz) = 2.5ppm x 300MHz x (1Hz/10⁶MHz) x (10⁻⁶/1ppm)
Downfield Shift = 750 Hz
The signal is at 750Hz Downfield from TMS
b. Downfield Shift = 1200 Hz, Chemical Shift = ?
Chemical Shift = Downfield shift/Spectrometer Frequency
Chemical Shift = (1200Hz/300MHz) x (1ppm/10⁻⁶) = 4.0 ppm
The signal comes at 4.0 ppm
c. Separation of 2ppm, Downfield Shift = ?
Downfield Shift (Hz) = 2(ppm) x 300 (MHz) x (1Hz/10⁶MHz) x (10⁻⁶/1ppm) = 600 Hz
The two peaks are separated by 600Hz
A phosphate buffer solution (25.00 mL sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The components of the buffer were sodium monohydrogenphosphate and sodium dihydrogenphosphate. The first endpoint occurred at a volume of 10.32 mL, and the second occurred after an additional 18.62 mL was added, for a total volume of 28.94 mL. What was the total concentration of phosphate (in any form) in the buffer?
Answer:
0,07448M of phosphate buffer
Explanation:
sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:
Na₂HP + HCl ⇄ NaH₂P + NaCl (1)
NaH₂P + HCl ⇄ H₃P + NaCl (2)
The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:
0,01862L of HCl×[tex]\frac{0,1000mol}{L}[/tex]= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.
The concentration is:
[tex]\frac{1,862x10^{-3}moles}{0,02500L}[/tex] = 0,07448M of phosphate buffer
I hope it helps!
During the combustion of a peanut that weighed 0.341 g, the temperature of the 100 mL of water in the calorimeter rose from 23.4ºC to 37.9ºC. The peanut didn't burn completely. If the serving size is 28.0 grams of peanuts, what is the Cal from fat per serving (Cal/serving)?
Select one:
46.2 Cal/serving 1
19 Cal/serving
28.5 Cal/serving
138 Cal/serving
Answer:
119 kCal per serving.
Explanation:
The heat energy necessary to elevates water's temperature from 23.4°C to 37.9°C can be calculated by the equation below:
Q = mcΔT
Q: heat energy
m: mass in g
c: specific heat capacity in cal/g°C
ΔT = temperature variation in °C
m is the mass of water, considering the density of water to be 1g/mL, 100 mL of water weights 100g. Therefore:
Q = 100 g x 1.00 cal/g°C x (37.9 - 23.4)°C
Q = 1450 cal
1450 cal ____ 0.341 g peanuts
x ____ 28 g peanuts
x = 119061.58 cal
This means that the cal from fat per serving of peanuts is at least 119 kCal.
The temperature change of the water indicated 1.45 Calories for 0.341 grams of peanut. Therefore, a serving size of 28.0 grams contains option 2) 119 Calories from fat.
To determine the Calories from fat per serving, we must first calculate the energy released by the peanut during combustion and then scale it to a full serving size.
Determine the temperature change in the water:Thus, the Calories from fat per serving is option 2) 119 Cal/serving.
Calculate the standard potential for the following galvanic cell: Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) which has the overall balanced equation: Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s) Express your answer to three significant figures and include the appropriate units
Answer:
The standard potential for this galvanic cell is 1.25 V
Explanation:
Step 1
Oxidation: Fe(s) → Fe^2+ (aq) + 2e- E° = 0.447 V
Reduction: Ag+(aq) + e- → Ag(s) E° = 0.7996 V
The electron number for the oxidation is 2; the electron number for the reduction is 1.
By multiplying the reduction reaction by 2, we make the electron number the same between the two half-equations
2 Ag+(aq) + 2e- → 2Ag(s) E° = 0.7996 V
Step 2: The overall balanced equation
Fe(s) + 2Ag+(aq) → Fe^2+(aq) + 2ag(s) E° = 0.447 V + 0.7996 V = 1.25V
The standard potential for this galvanic cell is 1.25 V
Calculate the percentage by mass of water in magnesium sulfate heptahydrate, MgSO4•7H2O
Enter your answer with 3 significant figures
Answer:
The percentage by mass of water in magnesium sulfate heptahydrate is 51.2 %
Explanation:
Step 1: Data given
Molar mass of MgSO4*7H2O = 246.5 g/mol
Molar mass of H2O = 18.02 g/mol
Molar massof MgSO4 = 120.37 g/mol
Step 2: Calculate % water in magnesium sulfate heptahydrate
Since we have 7 molecules of water in the heptahydrate, we will divide the molar mass of 7 molecules water by the molar mass of the heptahydrate.
m%(H2O) = (7*18.02)/ 246.5
m%(H2O) = (126.14 /246.5)*100%
m%(H2O) = 51.2 %
To controle this we will calculate the mass % of MgSO4
m%(MgSO4) = (120.37/ 246.50)*100%
m%(MgSO4) = 48.8%
51.2 + 48.8 = 100%
The percentage by mass of water in magnesium sulfate heptahydrate is 51.2 %
Answer:from reffered dfn of %by mass
%by M=mass of component particular/total mass
But mass directly proportional to Molar mass
% by M=molar mass of H2O cpn/total M
%by M=7((1×2)+16)/(7((1×2)+16)+(24+(16×4)+32))
=0.573
Part A Describe the electrodes in this nickel-copper galvanic cell. Drag the appropriate items to their respective bins. View Available Hint(s) ResetHelp Nickel Copper Standard reduction potentials for nickel(II) and copper(II) The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following: Ni2+(aq)+2e−→Ni(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.230 V E∘red=+0.337 V Part B What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.
Answer:
Part-A:
Anode is Nickel and Cathode is copper.
Part -B:
[tex]E^{0}_{cell}[/tex] of the reaction is 0.567V.
Explanation:
Nickel-Copper cell electrodes are "Ni" and "Cu" rod.
Nickel electrode is dipped in [tex]Ni^{+2}[/tex] solution.
Copper electrode dipped in [tex]Cu^{+2}[/tex] solution.
Part -A:
Anode:
At anode oxidation takes place
[tex]Ni(s) \rightarrow Ni^{+2}(aq)+2e^{-}[/tex]
Hence, anode is Nickel.
Cathode:
At cathode reduction takes place.
[tex]Cu^{+2}+2e^{-} \rightarrow Cu(s)[/tex]
Hence, Cathode is copper.
Part-B:
[tex]E^{0}_{cell}=E^{0}_{cathode}-E^{0}_{anode}[/tex]
[tex]=0.337-(-0.230)=0.567V[/tex]
Hence, [tex]E^{0}_{cell}[/tex] of the reaction is 0.567V.
Anodes and cathodes are the electrodes where oxidation and reduction take place.
What are the electrodes in galvanic cell?Electrochemical cells have two electrodes which is called anode and cathode. The anode is the electrode where oxidation occurs while the other hand, cathode is the electrode where reduction takes place so we can conclude that anodes and cathodes are the electrodes where oxidation and reduction take place.
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what is the molarity of the solution formed when 7.88 grams of NaCl is mixed with enough water to make 350. mL of solution?
Answer:
The molarity of the solution is 0.386 M
Explanation:
Step 1: Data given
Mass of NaCl = 7.88 grams
Volume of the solution = 350 mL
Molar mass of NaCl = 58.5 g/mol
Step 2: Calculate number of moles
Number of moles NaCl = mass of NaCl / molar mass of NaCl
Number of moles NaCl = 7.88 grams / 58.5 g/mol
Number of moles = 0.135 moles
Step 3: Calculate molarity of the solution
Molarity = Number of moles NaCl / volume
Molarity = 0.135 moles / 0.350 L
Molarity = 0.386 M
The molarity of the solution is 0.386M
Two solutions, initially at 24.60 °C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°C). When a 100.0 mL volume of 0.100 M AgNO3 solution is mixed with a 100.0 mL sample of 0.200 M NaCl solution, the temperature in the calorimeter rises to 25.30 °C. Determine the ∆H°rxn in kJ/mol AgCl for the reaction as written below. The density of the final solution is 1.00 g/mL and heat capacity of the final solution is 4.18 J/goC.
Answer:
∆H°rxn in kJ/mol AgCl = - 59.61 kJ/mol
Explanation:
To solve this question we need to calculate the heat absorbed by the cup calorimeter and by the water in the solutions. We will also need to calculate the amount in moles produced by the reaction since we want to know the ∆H°rxn in kJ/mol AgCl .
mol AgNO₃ = 100 mL x 1L/1000 mL x 0.100 mol/L = 0.01 mol
mol NaCl = 100 mL x 1L/1000 mL x 0.200 mol/L = 0.02 mol
Therefore our limiting reagent is the 0.01 mol AgNO₃ and 0.01 mol AgCl will be produced according to the stoichiometry of the reaction:
AgNO₃ + NaCl ⇒ AgCl + NaNO₃
Heat absorbed by the water:
qw = m(H₂O) x c x ΔT where m (H₂O) = 200 g ( the density of final solution is 1 g/ml)
c = specific heat of water = 4.18 J/gºC
ΔT = change in temperature = (25.30 - 24.60 ) ºC = 0.7ºC
qw = 200 g x 4.18 J/gºC x 0.7 ºC = 585.20 J
Heat Absorbed by the calorimeter :
q cal = C cal x ΔT = 15.5 J/ºC x 0.7ºC = 10.85 J
Total Heat released by the combustion = qw + qcal = 585.20 J +10.85 J
= 596.05 J
We have to change the sign to this quantity since it is an exotermic reaction ( ΔT is positive ) and have the ∆Hrxn
∆H rxn = -596.05 J
but this is not what we are being asked since this heat was released by the formation of 0.0100 mol of AgCl so finally
∆H°rxn = -596.05 J /0.01 mol = -59,605 J/mol x 1 kJ/1000J = -59.61 kJ/mol
You want to make 100 mL of a 2.5 M stock solution of calcium chloride (molecular weight 110.98 g/mol). How many grams of calcium chloride do you need to weigh out in order to make this solution? Round your answer to the nearest tenth of a gram.
Answer:
27.7 g
Explanation:
100 mL = 0.100 LTo solve this problem we use the definition of molarity:
M = mol / LFrom the problem we're given M = 2.5 M and volume = 0.100 L. We use the above formula and calculate the required moles of calcium chloride (CaCl₂):
2.5 M = moles / 0.100 Lmoles = 0.25 moles CaCl₂Finally we use the molecular weight of calcium chloride to calculate the required mass:
0.25 moles CaCl₂ * 110.98 g/mol = 27.745 g CaCl₂Rounding to the nearest tenth of a gram the answer is 27.7 g.
To make a 2.5 M stock solution of calcium chloride, you would need to weigh out approximately 276.2 grams of calcium chloride.
Explanation:To make a 2.5 M stock solution of calcium chloride, you need to calculate the number of grams of calcium chloride required. The molecular weight of calcium chloride is 110.98 g/mol. The formula to calculate the number of grams is:
Grams = Moles x Molecular Weight
Since you want to make 100 mL of a 2.5 M solution, you first need to convert mL to moles using the formula:
Moles = Volume (L) x Molarity (M)
Finally, you can substitute the calculated moles into the first formula to find the grams, and round to the nearest tenth of a gram.
Using these calculations, you would need to weigh out approximately 276.2 grams of calcium chloride to make the 2.5 M stock solution.
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The overall reaction in a commercial heat pack can be representedas shown below.
4 Fe(s) + 3 O2(g) 2Fe2O3(s) ΔH = -1652 kJ
(a) How much heat (kJ) is released when4.48 mol iron is reacted with excessO2?
(b) How much heat is released when 1.76 mol Fe2O3 isproduced?
(c) How much heat is released when 2.66 g iron is reacted with excessO2?
(d) How much heat is released when 12.8 g Fe and 1.49 gO2 are reacted?
Explanation:
[tex]4 Fe(s) + 3O_2(g)\rightarrow 2Fe_2O_3(s) ,\Delta H = -1652 kJ[/tex]
a) Heat released when 4.48 moles of iron is reacted with excessive oxygen:
According to reaction, when 4 moles of an iron reacts with 3 moles of oxygen it gives 1625 kilo Joules of heat.
Then 4.48 moles of iron will give:
[tex]\frac{1652 kJ}{4}\times 4.48=1850.24 kJ[/tex]
1850.24 kJ of heat is released when 4.48 moles of an iron is reacted with excess oxygen.
b) Heat released when 1.76 mol [tex]Fe_2O_3[/tex] is produced.
According to reaction, when 2 moles of an [tex]Fe_2O_3[/tex] are produced 1625 kilo Joules of heat is released
Then heat released on production of 1.76 mol [tex]Fe_2O_3[/tex] :
[tex]\frac{1652 kJ}{2}\times 1.76=1453.76 kJ[/tex]
1453.76 kJ heat is released when 1.76 mol [tex]Fe_2O_3[/tex] is produced.
c) Heat released when 2.66 grams of iron is reacted with excessive oxygen:
Moles of iron = [tex]\frac{2.66 g}{56 g/mol}=0.0475 mol[/tex]
According to reaction, when 4 moles of an iron reacts with 3 moles of oxygen it gives 1625 kilo Joules of heat.
Then 0.0475 moles of iron will give:
[tex]\frac{1652 kJ}{4}\times 0.0475=19.62 kJ[/tex]
19.62 kJ of heat is released when 2.66 grams of iron is reacted with excessive oxygen.
d) Heat released when 12.8 g Fe and 1.49 g oxygen gas are reacted:
Moles of iron = [tex]\frac{12.8 g}{56 g/mol}=0.2286 mol[/tex]
According to reaction, 4 moles of iron reacts with 3 moles of oxygen.Then 0.2286 mol will react with:
[tex]\frac{3}{4}\times 0.2286 mol=0.17145 mol[/tex] of oxygen
Moles of oxygen = [tex]\frac{1.49 g}{56 g/mol}=0.04656 mol[/tex]
According to reaction, 3 moles of oxygen gas reacts with 4 moles of iron .Then 0.04656 mol will react with:
[tex]\frac{4}{3}\times 0.04656 mol=0.06208 mol[/tex] of iron.
From this we can conclude that oxygen is in limiting amount. So, the amount of energy release will depend upon moles of oxygen gas.
According to reaction, when 4 moles of an iron reacts with 3 moles of oxygen it gives 1625 kilo Joules of heat.
Then 0.04656 moles of oxygen gas will give:
[tex]\frac{1652 kJ}{3}\times 0.04656 =25.7011 kJ[/tex]
25.7011 kJ of Heat is released when 12.8 g Fe and 1.49 g oxygen gas are reacted
The heat released in an exothermic reaction can be calculated using stoichiometry based on the ΔH value. The released heat varies with the amount of reactant or product involved. By calculating the molar ratios, the heat released is computed for each sub-question accordingly.
Explanation:The reaction you provided loosens heat, hence it is an exothermic reaction. The heat released can be calculated using stoichiometry if we know the amount of reactants or products. ΔH (-1652 kJ) is the amount of heat released for the reaction of 4 moles of Fe with 3 moles of O2 to form 2 moles of Fe2O3.
When 4.48 moles of iron (Fe) reacts with an excess of O2, the heat released would be in direct proportion. So, heat released = (4.48 / 4) * -1652 = -1848 kJ.When 1.76 mole of Fe2O3 is produced, heat released = (1.76 / 2) * -1652 = -1456 kJ.When 2.66 g of iron is reacted, convert it to moles (2.66g / 55.85g/mol = 0.048 moles). Hence, the heat released in this case = (0.048 / 4) * -1652 = -19.92 kJ.When 12.8 g of Fe and 1.49 g O2 are reacted, convert both into moles and use the reactant which is in lesser amount to calculate the heat. In this case, it's O2 (1.49g / 32g/mol = 0.0466 moles). The heat released = (0.0466 /(3/2)) * -1652 = -51.40 kJ.Learn more about Heat Release in Exothermic Reaction here:https://brainly.com/question/1047086
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When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydrofluoric acid results. Write the net ionic equation for the reaction. (Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)
Answer:
F⁻(aq) + H⁺(aq) ⇄ HF(aq)
Explanation:
When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydrofluoric acid results. The corresponding molecular equation is:
KF(aq) + HCl(aq) ⇄ KCl(aq) + HF(aq)
The full ionic equation includes all the ions and the molecular species. HF is a weak acid so it exists mainly in the molecular form.
K⁺(aq) + F⁻(aq) + H⁺(aq) + Cl⁻(aq) ⇄ K⁺(aq) + Cl⁻(aq) + HF(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.
F⁻(aq) + H⁺(aq) ⇄ HF(aq)
Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the number of unpaired electrons for each: (a) [Noble gas] ns^2 np^5 (b) [noble gas] ns^2 (n-1)d^2 (c) [noble gas] ns^2 (n-1)d^10 np^1 (d)[noble gas] ns^2 (n-2)f^6
Answer:
(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17 i.e. halogen group of the periodic table.
(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4 of the periodic table.
(c) [noble gas] ns² (n-1)d¹⁰ np¹: Number of unpaired electron is 1 and belongs to group 13 of the periodic table.
(d)[noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8 of the periodic table.
Explanation:
In the Periodic table, the chemical elements are arranged in 7 rows, called periods and 18 columns, called groups. They are organized in increasing order of atomic numbers.
(a) [Noble gas] ns² np⁵
As the total number of electrons in the p-orbital is 5. Therefore, the number of unpaired electron is 1.
This element has 2 electrons in ns orbital and 5 electrons in np orbital. So there are 7 valence electrons.
Therefore, this element belongs to the group 17 i.e. halogen group of the periodic table.
(b) [noble gas] ns² (n-1)d²
As the total number of electrons in the d-orbital is 2. Therefore, the number of unpaired electrons is 2.
This element has 2 electrons in ns orbital and 2 electrons in (n-1)d orbital. So there are 4 valence electrons.
Therefore, this element belongs to the group 4 of the periodic table.
(c) [noble gas] ns² (n-1)d¹⁰ np¹
As the total number of electrons in the p-orbital is 1. Therefore, the number of unpaired electron is 1.
This element has 2 electrons in ns orbital and 1 electron in np orbital. So there are 3 valence electrons.
Therefore, this element belongs to the group 13 of the periodic table.
(d)[noble gas] ns² (n-2)f⁶
As the total number of electrons in the f-orbital is 6. Therefore, the number of unpaired electron is 6.
This element has 2 electrons in ns orbital and 6 electrons in (n-2)f orbital. So there are 8 valence electrons.
Therefore, this element belongs to the group 8 of the periodic table.
The group of elements that corresponds to each of the following are:
(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17.
(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4.
(c) [noble gas] ns² (n-1)d¹⁰ np¹:Number of unpaired electron is 1 and belongs to group 13.
(d) [noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8.
Periodic Table:Periods are horizontal rows (across) the periodic table, while groups are vertical columns (down) the table. The elements are arranged in increasing order of their atomic number.
Group 17 is known as Halogen group, it has only on unpaired electron that means it needs one electron more to complete its octet or attain noble gas configuration. Group 4 is the second group of transition metals in the periodic table. Group 13, is also known as Boron group and it lies in p block elements. Group 8, is also known as Iron family.
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What is/are the product(s) of a neutralization reaction of a carboxylic acid? View Available Hint(s) What is/are the product(s) of a neutralization reaction of a carboxylic acid? a neutral compound and water a carboxylate salt and water a carboxylate salt a base and water g
Answer:
A carboxylate salt and water
Explanation:
A carboxylic acid is an organic compound that has general formula RCOOH, where R is a carbon chain. Because it's an acid, the neutralization will happen when it reacts with a base, such as NaOH.
When this reaction occurs, the base will dissociate in Na⁺ and OH⁻, and the acid will ionize in RCOO⁻ and H⁺, so the products will be RCOO⁻Na⁺ (a carboxylate salt) and H₂O (water).
The room temperature electrical conductivity of a semiconductor specimen is 2.8 x 104 (?-m)-1. The electron concentration is known to be 2.9 x 1022 m-3. Given that the electron and hole mobilities are 0.14 and 0.023 m2/V-s, respectively, calculate the hole concentration (in m-3)
Answer:
7.43 × 10²⁴ m⁻³
Explanation:
Data provided in the question:
Conductivity of a semiconductor specimen, σ = 2.8 × 10⁴ (Ω-m)⁻¹
Electron concentration, n = 2.9 × 10²² m⁻³
Electron mobility, [tex]\mu_n[/tex] = 0.14 m²/V-s
Hole mobility, [tex]\mu_p[/tex]= 0.023 m²/V-s
Now,
σ = [tex] nq\mu_n+pq\mu_p[/tex]
or
σ = [tex] q(n\mu_n+p\mu_p)[/tex]
here,
q is the charge on electron = 1.6 × 10⁻¹⁹ C
p is the hole density
thus,
2.8 × 10⁴ = 1.6 × 10⁻¹⁹( 2.9 × 10²² × 0.14 + p × 0.023 )
or
1.75 × 10²³ = 0.406 × 10²² + 0.023p
or
17.094 × 10²² = 0.023p
or
p = 743.217 × 10²²
or
p = 7.43 × 10²⁴ m⁻³
What is the mass of 3 L of water vapor at 120◦C and 388 torr?
Answer:
Explanation:
Use gas equation to calculate mol of H2O gas:
P = pressure = 389/760 = 0.512atm
V = volume = 4L
n=???
R = 0.082057
T = 121+273 = 394K
0.512 * 4 = n*0.082057*394
n = (0.512*4) / (0.082057*394)
n = 2.048 / 32.33
n = 0.063 mol H2O
Molar5 mass H2O = 18g/mol
Mass of water = 0.063*18
Mass of water valour = 1.14g
100 POINTS!!! WILL MARK BRAINIEST!!! What is the maximum number of grams of SO2 that can be fromed when 10.0 g of H2S reacts with 8.5 of oxygen? Given the equation 2H2S + 3O2 --> 2SO2 + 2H2O. PLS SHOW WORK!!
Answer:
Mass = 12.82 g
Explanation:
Given data:
Mass of oxygen = 8.5 g
Mass of H₂S = 10.0 g
Mass of SO₂ = ?
Solution:
Chemical equation;
2H₂S + 3O₂ → 2SO₂ + 2H₂O
Number of moles of H₂S:
Number of moles = mass/ molar mass
Number of moles = 10.0 g / 34 g/mol
Number of moles =0.3 mol
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 8.5 g / 32 g/mol
Number of moles = 0.3 mol
Now we will compare the moles of SO2 with oxygen and hydrogen sulfide.
O₂ : SO₂
3 : 2
0.3 : 2/3×0.3=0.2 mol
H₂S : SO₂
2 : 2
0.3 : 0.3
The number of moles of SO₂ produced by oxygen are less so it will limiting reactant.
Mass of SO₂:
Mass = number of moles × molar mass
Mass = 0.2 mol × 64.1 g/mol
Mass = 12.82 g