Answer:
(A). C6H5Br + Mg(in ether) -----------> C6H5MgBr.
(B). C6H5MgBr + O = C = O -----------> C6H5-COO^- Mg^+ Br.
(C). C6H5-COO^- Mg^+ Br + HCl --------> C6H5-COOH + Mg^+Br(OH).
PRODUCTS=> C6H5-COOH and Mg^+Br(OH).
Explanation:
A Grignard reagent is a reagent that/which is an organometallic compound that is R -Mg- X. The R = alkyl, vinyl or allyl and the X = halogens.
It must be noted that an important reaction of Grignard reagent is its reaction with compounds containing the Carbonyl that is -CO functional group and this kind of Reaction is known as a Grignard Reaction.
So, in this question we are told that;
=> "1-bromo-benzene andits subsequent reaction with solid carbon dioxide (CO2) followed by acidic workup (using HCl asthe acid). "
Thus;
(A). C6H5Br + Mg(in ether) -----------> C6H5MgBr.
(B). C6H5MgBr + O = C = O -----------> C6H5-COO^- Mg^+ Br.
(C). C6H5-COO^- Mg^+ Br + HCl --------> C6H5-COOH + Mg^+Br(OH).
If you have 101 g of hydrogen gas (H2) and excess amount of nitrogen gas (N2), how many grams of ammonia gas (NH3) can you make?
Answer:
572. 3 g of NH3
Explanation:
Equation of the reaction: 3H2 + N2 ----> 2NH3
From the equation of reaction, 3 moles of H2 reacts with 1 mole of N2 to produce 2 moles of NH3.
Since N2 is in excess in the given reaction, H2 is the limiting reactant.
Molar mass of H2 = 2 g/mol
Molar mass of NH3 = 17 g/mol
Therefore 3 * 2 g of H2 reacts to produce 2 * 17 g of NH3
6 g of H2 produces 34 g of NH3
101 g of H2 will produce (34 * 101)/6 g of NH3 = 572.3 g of NH3
Therefore, 572.3 g of NH3 are produced
Answer:
572.33g of NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of the H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar Mass of H2 = 2x1 = 2g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6g
Molar Mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34g.
From the balanced equation above,
6g of H2 reacted to produce 34g of NH3.
Finally, we can determine the mass of ammonia (NH3) produced by reacting 101g of H2 as follow:
From the balanced equation above,
6g of H2 reacted to produce 34g of NH3.
Therefore, 101g of H2 will react to produce = ( 101 x 34) / 6 = 572.33g of NH3.
Therefore, 572.33g of NH3 is produced from the reaction.
The boiling point of diethyl ether, CH3CH2OCH2CH3, is 34.500 °C at 1 atmosphere. Kb(diethyl ether) = 2.02 °C/m In a laboratory experiment, students synthesized a new compound and found that when 14.94 grams of the compound were dissolved in 279.5 grams of diethyl ether, the solution began to boil at 35.100 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?
Answer:
The correct answer is 179.94 g/mol.
Explanation:
Based on the given question, the boiling point of diethyl ether us 34.500 degree C at 1 atm pressure. The boiling point of the solution given is 35.100 degree C. The Kb of diethyl ether given is 2.02 degree C/m. The weight of the compound given is 14.94 grams, the weight of the solvent (diethyl ether) is 279.5 grams.
The molecular weight of the compound can be determined by using the formula,
deltaTb = Kb * molality
Tb-To = Kb * molality
Tb-To = Kb*wt/mol.wt*1000/w (solvent)
35.100 - 34.500 = 2.02 * 14.94 / mol. wt * 1000 g / 279.5 g
0.6 = 2.02 * 53.45/ mol.wt
mol. wt = 2.02*53.45/0.6
mol. wt = 179.94 g/mol
Hence, the molecular weight of the compound is 179.94 gram per mol.
Dry chemical hand warmers utilize the oxidation of iron to form iron oxide according to the following reaction: 4Fe(s)+3O2(g)→2Fe2O3(s) Standard thermodynamic quantities for selected substances at 25 ∘C Reactant or product ΔH∘f(kJ/mol) Fe(s) 0.0 O2(g) 0.0 Fe2O3(s) −824.2 Calculate ΔH∘rxn for this reaction.
Answer:
-1648.4 kJ/mol
Explanation:
Based on Hess's law:
ΔHr = ∑n×ΔH°f(products) - ∑n×ΔH°f(reactants)
In the reaction:
4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)
ΔHr = 2 ΔH°f {Fe₂O₃} - (4ΔH°f {Fe(s)} + 3ΔH°f{O₂(g)}
As:
ΔH°f {Fe₂O₃} = -824.2kJ/mol
ΔH°f {Fe(s)} = 0.0kJ/mol
ΔH°f{O₂(g)} = 0.0kJ/mol.
Thus,
ΔHr = 2 ₓ -824.2kJ/mol =
-1648.4 kJ/molAnswer:
-1648.4 kJ
Explanation:
The product has the only nonzero heat of formation, so it is the only value needed to calculate the enthalpy of this reaction. Normally, you would want to express the enthalpy of a reaction with respect to one mole of a chemical species, whether it is a reactant or product. However, since the balanced chemical equation contains only coefficients greater than 1, you should consider how the enthalpy relates to one mole of each substance according to the coefficients. In other words, − 1648.4 kJ of heat is released when 4 mol of Fe reacts with 3 mol of O2 to produce 2 mol of Fe2O3 .
When pressure is increased on the following equilibrium, where will the shift be? 3H2 + N2 2NH3
Answer:
Explanation:
it is based on le chatliers principles
the left side of reaction you have 4 moles , where as at the right hand side you have 2 moles,,,,
so when you increase the pressure the reaction will shift towards the lower moles producing reaction that is reaction move towards forward in you case.
Consider each pair of compounds listed below and determine whether a fractional distillation would be necessary to separate them or if a simple distillation would be sufficient.
a. Ethyl acetate and hexane
b. Diethyl Ether and 1-butanol
c. Bromobenzene and 1,2-dibromobenzene
You are trying to recrystallize compound X. You consider using ethyl acetate as your recrystallizing solvent and test a small amount of compound X with ethyl acetate. You find that compound X is soluble in ethyl acetate at room temperature and at boiling. Is ethyl acetate a good recrystallization solvent? No, the sample needs to be insoluble or sparingly soluble at room temperature so that the maximum amount of purified crystals form at room temperature and in the ice bath. Yes, you want the sample to fully dissolve at room temperature and boiling so that it will crystallize in the ice bath. Yes, you can only be sure that all the impurities dissolved if the sample is soluble at room temperature
Answer:
No, the sample needs to be insoluble or sparingly soluble at room temperature so that the maximum amount of purified crystals form at room temperature and in the ice bath.
Explanation:
For a solvent to be adequate it must completely dissolve the substance to be purified when it is hot, that is, at boiling temperature only. It should be practically insoluble when the solvent is cold or at room temperature. This must occur in this way since impurities must be removed by hot filtering or dissolved in the mother liquor.
A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Litre volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3 1-
Answer:
[tex]M=0.213M[/tex]
Explanation:
Hello,
In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:
[tex]n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-[/tex]
[tex]n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-[/tex]
[tex]n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-[/tex]
We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:
[tex]n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-[/tex]
Finally, we compute the molarity:
[tex]M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M[/tex]
Regards.
Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 liter container forms an equilibrium mixture containing 0.198 mole of H2O and corresponding amounts of CO, H2, and CH4.
Answer:
4.41
Explanation:
Step 1: Write the balanced equation
CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)
Step 2: Calculate the respective concentrations
[tex][CO]_i = \frac{0.500mol}{5.00L} = 0.100M[/tex]
[tex][H_2]_i = \frac{1.500mol}{5.00L} = 0.300M[/tex]
[tex][H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M[/tex]
Step 3: Make an ICE chart
CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)
I 0.100 0.300 0 0
C -x -3x +x +x
E 0.100-x 0.300-3x x x
Step 4: Find the value of x
Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396
Step 5: Find the concentrations at equilibrium
[CO] = 0.100-x = 0.100-0.0396 = 0.060 M
[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M
[CH₄] = x = 0.0396 M
[H₂O] = x = 0.0396 M
Step 6: Calculate the equilibrium constant (Kc)
[tex]Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41[/tex]
How could the government enforce ethical standards of scientific
experiments?
A. The government could encourage scientists to make up their own
minds about ethics.
B. The government could take away research funds if ethical
standards are not met.
C. The government could let scientists monitor each other to
encourage ethical behavior.
D. The government could encourage the public to take a stand
against unethical scientists.
Answer: D. The Government could take away research funds if ethical standards are not met
The government enforce ethical standards of scientific experiments
B. The government could take away research funds if ethical
standards are not met.
Ethical standards are a set of principles established by the founders of the organization to communicate its underlying moral values. This code provides a framework that can be used as a reference for decision making processes.
How does the government control scientific research?Politicians and bureaucrats control scientific research and research outcomes by selectively funding projects that look for potential disasters, ideally global disasters.
What are the 8 ethical standards?This analysis focuses on whether and how the statements in these eight codes specify core moral norms (Autonomy, Beneficence, Non-Maleficence, and Justice), core behavioral norms (Veracity, Privacy, Confidentiality, and Fidelity), and other norms that are empirically derived from the code statements.
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A 75 gram solid cube of mercury (II) oxide has a density of 2.4 x 103 kg/m3 .
What is the length of one side of the cube in cm?
The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?
The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?
Answer:
0.031 m
HgO(s) ⇒ Hg(l) + 1/2 O₂(g)
Chemical change
Element
Explanation:
A 75 gram solid cube of mercury (II) oxide has a density of 2.4 × 10³ kg/m³. What is the length of one side of the cube in cm?
Step 1: Convert the mass to kilograms
We will use the relationship 1 kg = 1,000 g.
[tex]75g \times \frac{1kg}{1,000g} = 0.075kg[/tex]
Step 2: Calculate the volume (V) of the cube
[tex]0.075kg \times \frac{1m^{3} }{2.4 \times 10^{3} kg} = 3.1 \times 10^{-5} m^{3}[/tex]
Step 3: Calculate the length (l) of one side of the cube
We will use the following expression.
[tex]V = l^{3} \\l = \sqrt[3]{V} = \sqrt[3]{3.1 \times 10^{-5} m^{3} }=0.031m[/tex]
The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?
The balanced chemical equation is:
HgO(s) ⇒ Hg(l) + 1/2 O₂(g)
This is a chemical change because new substances are formed.
The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?
The liquid gray substance is Hg(l), which is an element because it is formed by just one kind of atoms.
A compound is known to be Na2CO3, Na2SO4, NaOH, NaCl, NaC2H3O2, or NaNO3. When a barium nitrate solution is added to a solution containing the unknown a white precipitate forms. No precipitate is observed when a magnesium nitrate solution is added to a solution containing the unknown. What is the identity of the unknown compound
Answer:
Na₂SO₄
Explanation:
Barium nitrate, Ba(NO₃)₂ produce precipitate with SO₄²⁻, CO₃²⁻. That means the precipitate could be obtained from Na₂SO₄ and Na₂CO₃.
Also, magnesium nitrate, Mg(NO₃)₂, produce precipitate just with CO₃²⁻. As the unknown solution produce no precipitate, the unknown compound is:
Na₂SO₄
RUIGA GIRLS
CHEMISTRY FORM 3. 23/06/2020
MR. GICHURU
IZ
1
Narne the elements present in
Common salt
(2 miks)
Hydrated copper (11) Sulphate.
(2 ks)
Sulphuric (VI) acid,
2 Why is a reaction between zinc metal and Nitric acid not suitable for preparing
hydrogen gae in the laboratory
(2 mi)
(1 m)
3.
What is relative atomic mass?
b)
Define 'isotopes
c)Determine the relative atomic mass of element K whose isotople misure occur in
the proportione:
(2 marks)
Which of the following best describes isotopes?
An element with the same number of neutrons, but a different number of protons.
An element with the same number of protons, but a different number of electrons.
An element with the same number of electrons, but a different number of neutrons
An element with the same number of protons, but a different number of neutrons
Answer: An element with the same number of protons, but a different number of neutrons
Explanation:
The # of protons in an atom is what determines what atom it is (hydrogen has 1 proton, helium has 2 protons, etc ...). You cannot change the number of protons in an atom without changing what element the atom is.
The number of electrons in atoms varies greatly because electrons are constantly gained, lost, and shared during chemical reactions.
An isotope is a variation of the same element (so they must have the same # of protons) that have different masses (and therefore a different number of neutrons).
The answer is the fourth choice, "An element with the same number of protons, but a different number of neutrons"
The isotopes refer to an element that consists of a similar number of protons but have a distinct no of neutrons.
What are isotopes:It is considered to be the members of the family with respect to the elements that consist of a similar number of protons but have a distinct no of neutrons. The no of protons in the nucleus measured the atomic number of elements based on the periodic table.
Therefore, the fourth option is correct.
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What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together
Answer:
B. Mixing a solute and a solvent
Explanation:
Hello,
In this case, solutions are defined as liquid homogeneous mixtures formed when two substances having affinity are mixed. It is important to notice that the two substances are known as solute, which is added to other substance that is the solvent. Therefore, answer is B. Mixing a solute and a solvent.
Notice that when two insoluble substances are mixed no solution is formed. Furthermore, if two solutes together or a solute and a precipitate are mixed, no liquid homogeneous solution is formed, as commonly solutes are solid, nevertheless, when liquid, one should have to act as the solvent.
Best regards.
Answer:
B. Mixing a solute and a solvent
Explanation:
ap3x
Calculate Keq for these reactions and predict if the equilibrium will lie to the right or to the left as written. (You may enter your answer in scientific notation, e.g. 1.0*10^-9. Enter your answer to two significant figures.) Reaction 1: + + pKa = 9 pKa = 38 Keq = Equilibrium position = _______ Reaction 2: + + pKa = 35 pKa = 25 Keq = Equilibrium position = _______
Complete Question
The complete question is shown on the first uploaded image
Answer:
For reaction 1
[tex]K_{eq} = 10^{29}[/tex]
The equilibrium position is to the right
For reaction 2
[tex]K_{eq} = 10^{-6.66}[/tex]The equilibrium position is to the left
Explanation:
Generally [tex]pKa[/tex] is mathematically evaluated as
[tex]pKa = pKa _ \ {left }} - pKa _ \ {right }}[/tex]
And equilibrium position [tex]K_a[/tex] is mathematically evaluated as [tex]K_{eq} = 10^\ {-pK_a}[/tex]
From the question we are told that
For reaction 1
[tex]pKa_\ {left}} \ = 9[/tex]
[tex]pKa_\ {right }} \ = 38[/tex]
So
[tex]pKa = 9-38[/tex]
[tex]pKa =-29[/tex]
So [tex]K_{eq} = 10^{-(-29)}[/tex]
[tex]K_a = 10^{29}[/tex]
This implies that the equilibrium position is to the right
For reaction 2
[tex]pKa_\ {left}} \ = 15.9[/tex]
[tex]pKa_\ {right }} \ = 9.24[/tex]
So
[tex]pKa = 15.9-9.24[/tex]
[tex]pKa = 6.66[/tex]
So [tex]K_{eq} = 10^{-(6.66)}[/tex]
[tex]K_{eq} = 10^{-6.66}[/tex]
This implies that the equilibrium position is to the left
Calculate the pH of this solution 0.0043 M of H2SO4=
Answer:
pH = - log [concentration]
pH = - log (0.0043M)
pH = 2.37
How many moles of solute are contained in the following solution: 15.25 mL of a 2.10 M CaCl₂
Answer:
0.032moles
Explanation:
2.10moles in 1000ml what about 15.25ml
(15.25×2.10)÷1000
0.032moles
Precision can be defined as the?
Answer:Precision can be defined as the. reproducibility of a measured value. Precision is how close the measured values are to each others. In contrast with accuracy, accuracy is the agreement between a measured value and an accepted value.
Explanation:
Please help! (:
question above — how much money would you need to buy 7.0 lb of arugula? If 27lb of arugula cost $16
Answer:
$11.81
Explanation:
27 lb cost $16
27/16=$1.69 per pound
$1.69*7=$11.81 for 7 lbs
uses of sodium chloride in daily life
Answer:
sodium chloride can be used as salt
extraction sodium metal by electrolysis
a common chemical in laboratory experiments
Answer:
sodium chloride can be used as preservatives,
in preserving foods.
The temperature program for a separation starts at a temperature of 50 °C and ramps the temperature up to 270 °C at a rate of 10 °C/minute. Which statement is NOT true for this separation?
A) At 10 °C/minute, a total of 22 minutes is needed to reach 270 oC.
B) Strongly retained solutes will remain at the head of the column while the temperature is low.
C) Weakly retained solutes will separate and elute early in the separation.
D) The vapor pressure of strongly retained solutes will increase as temperature increases.
E) Strongly retained analytes will give broad peaks.
Answer:
The correct answer to the question is Option E (Strongly retained analytes will give broad peaks).
Explanation:
The other options are true because:
A. Initial temp = 50 °C
Final temp = 270 °C
Differences in temp = 270 - 50 = 220°C
Rate = 10 °C/minute.
So, at 10 °C/minute,
total of 220°C /10 °C = number of minutes required to reach the final temp.
220/10 = 22 minutes
B. A column has a minimum and maximum use temperature. Solutes that are already retained would remain stationary while temperatures are low. This would only change if there is an increase in temperature. Heat transfers more energy to the liquid which would make the solute interact with the column phase.
C. Weakly retained solutes may contain larger molecules, will separate by absorbing into the solvent early in separation making the mobile phase separates out into its components on the stationary phase.
D. Retained solute's vapor pressure is higher at higher temperatures making it possible for particle to escape more from the solute when the temperature is high than when it is low.
Calculate the osmotic pressure of a solution prepared by dissolving 65.0 g of Na2SO4 in enough water to make 500 mL of solution at 20°C. (Assume no ion pairing – in other words, assume that the electrolyte completely dissociates into its constituent ions.)
Answer:
66.0 atm
Explanation:
We can calculate the osmotic pressure (π) using the following expression.
[tex]\pi = i \times M \times R \times T[/tex]
where,
i: van 't Hoff indexM: molarityR: ideal gas constantT: absolute temperatureStep 1: Calculate i
Sodium sulfate completely dissociates according to the following equation.
Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻
Since it produces 3 ions, i = 3.
Step 2: Calculate M
We can calculate the molarity of Na₂SO₄ using the following expression.
[tex]M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M[/tex]
Step 3: Calculate T
We will use the following expression.
K = °C + 273.15
K = 20°C + 273.15 = 293 K
Step 4: Calculate π
[tex]\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm[/tex]
How would I find the quantity of heat absorbed or released when 2.0g of LiOH is dissolved in 100g of H₂0 when the enthalpy of the solution is -23.6KJ/mol?
Answer:
1.97kJ of energy are released.
Explanation:
The dissolution of LiOH in water is:
LiOH(s) → Li⁺(aq) + OH⁻(aq) ΔH = -23.6kJ
That means, when 1 mole of LiOH is dissolved, there are released (Because of the - in the enthalpy) 23.6kJ
2.0 g of LiOH (Molar mass: 23.95g/mol) are:
2.0g LiOH × (1 mol / 23.95g) =0.0835 moles of LiOH.
As 1 mole of LiOH release 23.6kJ, 0.0835moles release:
0.0835moles × (-23.6kJ / 1mole) = 1.97kJ of energy are released
A stock solution of HNO3 is prepared and found to contain 14.9 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, what is the concentration of the diluted solution
Answer:
[tex]0.745~M[/tex]
Explanation:
In this case, we have a dilution problem. So, we have to use the dilution equation:
[tex]C_1*V_1=C_2*V_2[/tex]
Now, we have to identify the variables:
[tex]C_1~=~14.9~M[/tex]
[tex]V_1~=~25~mL[/tex]
[tex]C_2~=~?[/tex]
[tex]V_2~=~0.5~L[/tex]
Now, we have different units for the volume, so we have to do the conversion:
[tex]0.5~L\frac{1000~mL}{1~L}=~500~mL[/tex]
Now we can plug the values into the equation:
[tex]C_2=\frac{14.9~M*25~mL}{500~mL}=0.745~M[/tex]
I hope it helps!
Given that S is the central atom, draw a Lewis structure of OSF4 in which the formal charges of all atoms are zero. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons.
Answer:
Here's what I get
Explanation:
A Lewis structure shows the valence electrons surrounding the atoms.
Your structure has two problems:
It shows too many valence electrons It violates the octet rule for O — there are 10 electrons around the O atom.Here's one way to draw a Lewis structure.
1. Draw a trial structure
Make F and O terminal atoms and give each one an octet (Fig. 1).
2. Count the valence electrons in the trial structure
5 BP + 15 LP = 10 + 30 = 40 electrons
3. Check the number of valence electrons available
1 S = 1 × 6 = 6 electrons
1 O = 1 × 6 = 6
4 F = 4 × 7 = 28
TOTAL = 40 electrons
The trial structure has the correct number of electrons.
4. Determine the formal charge on each atom.
To get the formal charges, we cut the covalent bonds in half.
Each atom gets the electrons on its side of the cut.
Formal charge = valence electrons in isolated atom - electrons on bonded atom
FC = VE - BE
(a) On S
VE = 6
BE = 5 bonding electrons = 5
FC = 6 - 5 = +1
(b) On O:
VE = 6
BE = 3 LP(six electrons) + 1 bonding electron = 7
FC = 6 - 7 = -1
(c) On F:
VE = 6
BE = 3 lone pairs(6 electrons) + 1 bonding electron = 6 + 1 =7
FC = 7 - 7 = 0
5. Minimize the formal charges
We must rearrange the valence electrons so that S gets one more and O gets one fewer.
Move a lone pair from the O to make an S=O double bond (Fig. 2).
6. Recalculate the formal charges
(a) On S
VE = 6
BE = (3 bonding electrons) = 6
FC = 6 - 6 = 0
(b) On O:
VE = 6
BE = 2 LP(four electrons) + 2 bonding electrons = 6
FC = 6 - 6 = 0
Fig. 2 shows the Lewis structure in which all atoms have a formal charge of zero.
The formal charge of the atoms can be concluded zero with the bond formation between the sulfur and oxygen atom.
The lewis structure can be defined as the dot structure of the valence bond with the bonded atoms. The formal charge can be calculated with the difference in the valence electrons and the bonding electrons.
The formal charge of an atom can be zero when the valence electrons and the bonding electrons are equal. In the structure of [tex]\rm OSF_4[/tex], the formal charge has been assigned zero with the bond formation resulting in the valence electrons and bonding electrons being equal.
The lewis structure with the central S atom has been attached.
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How many grams of the salt CaF2 (g) are formed when 15.7 mL of 0.612 M KF reacts with an excess of aqueous calcium bicarbonate (Ca(HCO3)2) via a metathesis reaction?
Answer:
[tex]m_{CaF_2}0.375gCaF_2[/tex]
Explanation:
Hello,
In this case, for the studied reaction:
[tex]2KF+Ca(HCO_3)_2\rightarrow CaF_2+2KHCO_3[/tex]
Thus, the first step is to compute the reacting moles of potassium fluoride by using its volume and molarity:
[tex]n_{KF}=0.0157L*0.612\frac{mol}{L} =9.61x10^{-3}molKF[/tex]
Then, we apply the 2:1 molar ratio between potassium fluoride and calcium fluoride to compute the produced moles of calcium fluoride:
[tex]n_{CaF_2}=9.61x10^{-3}molKF*\frac{1molCaF_2}{2molKF} =4.80x10^{-3}molCaF_2[/tex]
Finally, by using the molar mass of calcium fluoride (78.07 g/mol) we can compute its produced grams:
[tex]m_{CaF_2}=4.80x10^{-3}molCaF_2*\frac{78.07gCaF_2}{1molCaF_2} \\\\m_{CaF_2}0.375gCaF_2[/tex]
Best regards.
consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.
Answer:
[tex]pH=10.5[/tex]
Explanation:
Hello,
In this case, the dissociation of the given weak acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Therefore, the law of mass action for it turns out:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
That in terms of the change [tex]x[/tex] due to the reaction extent is:
[tex]1x10^{-23}=\frac{x*x}{0.1-x}[/tex]
Thus, by solving with the quadratic equation or solver, we obtain:
[tex]x=31.6x10^{-12}M[/tex]
Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:
[tex]pH=-log(-31.6x10^{-12})\\pH=10.5[/tex]
Regards.
I WILL GIVE BRAINLIEST
Select the correct answer.
What effect does an increase in products have on the reaction rate of a mixture at equilibrium?
A.
The forward reaction rate increases.
B.
Both the forward and the reverse reaction rates decrease.
Both the forward and the reverse reaction rates increase.
D.
The reverse reaction rate increases.
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Answer:
At equilibrium the rate of the forward reaction is equal to the rate of the backward reaction.
When the product of a reaction at equilibrium is increased the equilibrium will shift left or to the reactant side. As a result the excess product will get converted to reactant. This is in accordance to Le Chatelier's principle.
Le Chatelier's principle states that when a system is subjected to stress the equilibrium will shift in a direction to minimize effect of the stress.
Thus the products added to the system at equilibrium will make the equilibrium shift to the reactant side, the rate of the reverse or backward reaction will increase.
Explanation:
Hope This Helps Amigo!
An ideal gaseous reaction occurs at a constant pressure of 35.0 atm and releases 66.8 kJ of heat. Before the reaction, the volume of the system was 8.20 L. After the reaction, the volume of the system was 2.21 L. Calculate the total change in internal energy for the system. Enter your answer numerically in units of kJ.
Answer:
U = -45.557kj
Explanation:
Before we can calculate the totally internal energy change in kilojoules firstly we need to calculate W
U=q + w .
We know that
w = PΔ V
where P is the pressure of
and V is the volume
then we can calculate the work
w = 35 atm * ( 8.20L - 2.21L)
W=35atm* 5.99L
W=209.65atmJ
But 1 atm = 101.325J
then ,
w = 209.65* 101.325 J = 21242.79 J
let us convert it to Kj
But we know that 1kJ = 10^3 J .
Then w = 21.243 kJ .
Then we can now calculate the internal energy as
U = 21.243- 66.8 kJ = -45.557kj
But we know that heat was released. Theeefore, the total internal energy change was -45.557kj
