Answers
[tex]\boxed{{\text{ClO,ClO}}_{\text{2}}^ - {\text{,ClO}}_{\text{3}}^ - {\text{,ClO}}_{\text{4}}^ - }[/tex] does not follow the octet rule.
Further Explanation:
Octet rule: states that for the stability of any element it must have a valence shell of eight electrons (octet means a group of eight). This rule is given by Kossel and Lewis.
Atoms of same or different elements with an incomplete electronic configuration that is having electrons less than 8 are unstable and they combine together to form stable molecules with a complete octet. They can combine by sharing of electrons or loss or gain of electrons.
In [tex]{\mathbf{ClO}}[/tex], Chlorine (Cl) has 7 electrons in its outer shell and oxygen has 6 electrons in its outer shell. So chlorine and oxygen share 2 electrons each to complete their octet and attain stability. Chlorine atom in the compound has ten electrons in its outer shell so it doesn’t follow the octet rule.
In [tex]{\mathbf{Cl}}{{\mathbf{O}}^ - }[/tex], Chlorine(Cl) has 7 electrons in its outer shell and oxygen has a unit negative charge on it [tex]\left( {{{\text{O}}^ - }} \right)[/tex] so, it has 7 electrons also in its outer shell. So chlorine and oxygen share one electron each to complete their octet and attain stability. All the atoms have eight electrons in their outer shell so they follow the octet rule.
In [tex]{\mathbf{ClO}}_2^ -[/tex], Chlorine (Cl) has 7 electrons in its outer shell and there is 2 oxygen attached to Cl. One has a unit negative charge on it [tex]\left( {{{\text{O}}^ - }} \right)[/tex] and the other is neutral. So, chlorine and neutral oxygen share 2 electrons each to get a stable octet and form a double bond. Similarly, chlorine shares 1 electron with oxygen ion with a negative charge on it forming a single bond with it. A chlorine atom has ten electrons in its outer shell so the compound doesn’t follow the octet rule.
In [tex]{\mathbf{ClO}}_3^ -[/tex], Chlorine (Cl) has 7 electrons in its outer shell and there is 3 oxygen attached to Cl. One has a unit negative charge on it [tex]\left( {{{\text{O}}^ - }} \right)[/tex] and the other two are neutral. So, chlorine and both neutral oxygen share 2 electrons to get a stable octet and form a double bond. Similarly, chlorine shares 1 electron with oxygen ion with a negative charge on it, forming a single bond with it. Here, Cl has ten electrons in its outer shell, so it doesn’t follow the octet rule.
In [tex]{\mathbf{ClO}}_4^ -[/tex] , Chlorine (Cl) has 7 electrons in its outer shell and there is 4 oxygen attached to Cl. One has a unit negative charge [tex]\left( {{{\text{O}}^ - }} \right)[/tex]on it and the other three are neutral. So, chlorine and both neutral oxygen share 2 electrons to get a stable octet and form a double bond. Similarly, chlorine shares 1 electron with oxygen ion with a negative charge on it, forming a single bond with it.
Here, Cl has fourteen electrons in its outer shell, so it doesn’t follow the octet rule.
Learn more:
1. The octet rule states that:https://brainly.com/question/1601832
2. How many covalent bond nitrogen forms: https://brainly.com/question/5974553
Answer details:
Grade: Secondary School
Subject: Chemistry
Chapter: Chemical Bonding
Keywords: octet rule, stability, oxygen, chlorine, covalent, sharing, ClO2-, ClO3-, ClO4-, ClO-and ClO.
The species that do not obey the octet rule are; ClO, ClO2^-, ClO3^-, ClO4^-.
The octet rule states that atoms must have eight electrons in their outermost shell in order to attain stability. Hence, stable molecules, ions and atoms are expected to contain atoms that obey the octet rule.
However, in some chemical species, atoms of elements do not obey the octet rule. For instance, in ClO, chlorine has seven valence electrons and oxygen has six electrons. The octet rule is clearly violated in this case.
Again, chlorine is able to expand its octet (contain more than eight valence electrons). This is seen in the ions; ClO2^- and ClO3^- in which chlorine contains ten valence electrons and ClO4^- in which chlorine contains fourteen valence electrons.
Learn more: brainly.com/question/1601832
Related Questions
Which statement about ionic bonds is true?
A. In ionic bonds atoms share electrons to achieve a stable outer shell.
B. Ionic bonds occur between non-metals.
C. Ionic bonds occur between two metals.
D. In ionic bonds one atom accepts electrons from another atom to achieve a stable outer shell.
Answers
In ionic bonds one atom accepts electrons from another atom to achieve a stable outer shell.
Answer:
In ionic bonds one atom accepts electrons from another atom to achieve a stable outer shell.
Explanation:
Answer via Educere/ Founder's Education
hi i'm having trouble solving this question
" if 7.54 grams of Al reacts with excess of H2SO4, and makes 0.33g of H2 . What is the % yield of H2gas produced?
Answers
m(Al) = 7.54 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 7.54 g ÷ 27 g7mol.
n(Al) = 0.28 mol; amount of substance.
From chemical reaction: n(Al) : n(H₂) = 1 : 1.
n(H₂) = 0.28 mol.
m(H₂) = 0.28 mol · 2 g/mol.
m(H₂) = 0.56 g.
%yield = 0.33 g ÷ 0.56 g · 100%.
%yield = 58.9%.
The ka of hypochlorous acid (hclo) is 3.0 x 10-8 at . what is the % ionization of hypochlorous acid in a 0.015 m aqueous solution of hclo at ?
Answers
Answer is: the % ionization of hypochlorous acid is 0.14.
Balanced chemical reaction (dissociation) of an aqueous solution of hypochlorous acid:
HClO(aq) ⇄ H⁺(aq) + ClO⁻(aq).
Ka = [H⁺] · [ClO⁻] / [HClO].
[H⁺] is equilibrium concentration of hydrogen cations or protons.
[ClO⁻] is equilibrium concentration of hypochlorite anions.
[HClO] is equilibrium concentration of hypochlorous acid.
Ka is the acid dissociation constant.
Ka(HClO) = 3.0·10⁻⁸.
c(HClO) = 0.015 M.
Ka(HClO) = α² · c(HClO).
α = √(3.0·10⁻⁸ ÷ 0.015).
α = 0.0014 · 100% = 0.14%.
The percent ionization of the solution is 0.14%.
The equation of the reaction is;HClO(aq) ⇄ H^+(aq) + ClO^-(aq)
I 0.015 0 0
C -x +x +x
E 0.015 - x x x
Now the Ka of the solution is obtained from;Ka = [ H^+] [ClO^-]/[HClO]
3.0 x 10^-8= x^2/ 0.015 - x
3.0 x 10^-8(0.015 - x) = x^2
4.5 x 10^-10 - 3.0 x 10^-8x = x^2
x^2 + 3.0 x 10^-8x - 4.5 x 10^-10 = 0
x = 0.000021 M
Percent ionization = 0.000021 M/0.015 M × 100/1Percent ionization = 0.14%
Hence, the percent ionization of the solution is 0.14%.
Learn more about percent ionization: https://brainly.com/question/9173942
An apple pudding is prepared by combining apples, sugar, butter, and lemon juice. To which category does it belong? compound mixture element molecule atom
Answers
Each individual ingredient is an element.
Compound means they are chemically bound together which is not true in this case.
Select the single best answer. why is the boiling point of propanamide, ch3ch2conh2, considerably higher than the boiling point of n,n−dimethylformamide, hcon(ch3)2 (213°c vs. 153°c), even though both compounds are isomeric amides?
Answers
As shown below, propanamide contains two hydrogen atoms directly attached to Nitrogen atom. In this case the hydrogen atoms become partially positive and can form Hydrogen Bond Interactions with the Nitrogen atoms of neighbor propanamide molecule.
Due to the involvement of Hydrogen bonding which is strongest among intermolecular interactions increases the boiling point of propanamide which are absent in N,N−dimethylformamide.
The higher boiling point of propanamide compared to N,N-dimethylformamide is due to the ability of propanamide to form stronger intermolecular hydrogen bonds, requiring more energy to break and hence, raising its boiling point.
Explanation:The boiling point of propanamide, CH3CH2CONH2, is considerably higher than that of N,N-dimethylformamide, HCON(CH3)2 because propanamide can form robust intermolecular hydrogen bonds within its molecules. These require a significant amount of energy to break, hence increasing its boiling point. However, in N,N-dimethylformamide, hydrogen bonding is inhibited because the hydrogen atoms are bonded to carbon rather than a more electronegative element like nitrogen or oxygen, which would contribute to stronger hydrogen bonding. Therefore, the boiling point of N,N-dimethylformamide is lower.
Learn more about Boiling Point of Compounds here:https://brainly.com/question/36036751
How many grams of NAOH are contained within 0.785 moles of NAOH?
Answers
Answer: The mass of NaOH for given number of moles is 31.4 grams.
Explanation:
To calculate the mass for given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Molar mass of NaOH = 40 g/mol
Moles of NaOH = 0.785 moles
Putting values in above equation, we get:
[tex]0.785mol=\frac{\text{Mass of NaOH}}{40g/mol}\\\\\text{Mass of NaOH}=(0.785mol\times 40g/mol)=31.4g[/tex]
Hence, the mass of NaOH for given number of moles is 31.4 grams.
A sample of seawater contains 1.3g of calcium ions in 3,100kg of solution. what is the calcium ion concentration of this solution in units of ppm?
Answers
since 1 kg is 10⁻⁶ mg, 1 kg is equivalent to million mg.
therefore mg/kg is also ppm.
there are 1.3 g of Ca ions in 3100 kg
if 3100 kg contains - 1.3 g of Ca
then 1 kg contains - 1.3 g / 3100 kg
then Ca ions - 0.42 x 10⁻³ g/kg
Ca ion concentration - 0.42 mg/kg
therefore Ca ion concentration is 0.42 ppm
Sodium is produced by electrolysis of molten sodium chloride. what are the products at the anode and cathode, respectively?
Answers
NaCl ------> Na⁺ +Cl⁻
Anod (+) | Cathode (-)
2Cl⁻-2e⁻ ---> Cl2(gas) | Na⁺ + e⁻ -----> Na⁰(s)
Which symbol can be used to indicate a chemical reaction that takes place in the presence of a catalyst?
A: 350° C
B: Pt
C: 0.5 atm
D: the triangle
Answers
Following are the reason for identifying it as a correct answer.
Option A reflects the temperature at which reaction has been performed. It doesn't signify use of catalyst.
Option B signifies that, Pt is employed as a foreign material in the reaction system. The role of this foreign material is to accelerate the reaction rate. Such material is referred as catalyst. With reference to catalyst, it must be noted that, Pt is a heterogeneous catalyst.
Option C signifies the pressure of reactant gas, at which reaction is carried out. atm is the unit of pressure of reactant gas.
Option D is the symbol that is very often used to depict that reaction is performed at higher temperature.
Thus, among all the available options, option B i.e. Pt is a correct answer
The normal boiling point of ethanol (c2h5oh) is 78.3 °c and its molar enthalpy of vaporization is 38.56 kj/mol. what is the change in entropy in the system in j/k when 42.2 grams of ethanol at 1 atm condenses to a liquid at the normal boiling point?
Answers
= enthalpy of vaporization/boiling point temperature
when we have the enthalpy of vaporization = 38560 J/mol
and the boiling point temperature in Kelvin = 78.3 + 273 = 351.3 K
by substitution:
∴the entropy change of vaporization = 38560J/mol/351.3K
= 109.76 J/K/mol
and when the liquid has lesser entropy than the gas and we here convert from
gas to liquid so, the change in entropy = -109.76 J/K/mol
now, we need the moles of C2H5OH = mass/molar mass when the molar
mass of C2H5OH = 46 g/mol and mass = 42.2 g
∴ moles of C2H5OH = 42.2 g / 46 g/mol = 0.92 moles
when 1 mol of C2H5OH turns in liquid entropy change →-109.76 J/K/mol
∴ 0.92 mol of C2H5OH → X
∴ X entropy change when 0.92 mol = -109.76 *0.92 mol / 1 mol
= 84.64 J/K
The change in the randomness of the system is the entropy change. The entropy change after condensation at the standard boiling point is 84.64 J/K.
What is the entropy change?When a system undergoes the addition or deletion of the reactant and the products, then the disorder of the system is known as entropy change.
Given,
Enthalpy of vaporization = 38560 J/mol
Boiling point temperature = 351.3 K
[tex]\begin{aligned}\text{Entropy change of vaporization} &= \dfrac{\text{enthalpy of vaporization}}{\text{boiling point temperature}}\\\\&= \dfrac{38560}{351.3}\\\\&=109.76 \;\rm J/K/mol\end{aligned}[/tex]
Here, liquid has less entropy than gas hence the change in entropy is -109.76 J/K/mol.
Moles of ethanol is calculated as:
[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar\; mass}\\\\&= \dfrac{42.2}{46}\\\\&= 0.92 \;\rm moles\end{aligned}[/tex]
If 1 mole of ethanol has an entropy change of -109.76 J/K/mol. Then, 0.92 moles will have,
[tex]\dfrac{-109.76 \times 0.92}{1} = 84.64\;\rm J/K[/tex]
Therefore, 84.64 J/K is the entropy change.
Learn more about entropy change here:
https://brainly.com/question/9864321
Write the balanced nuclear equation for β− decay of sodium−26. include both the mass numbers and the atomic numbers with each nuclear symbol. use the sup-subscript button in the answer palette to enter these numbers correctly. greek letters can be accessed in the drop-down menu that says -select−.
Answers
After doing β- decay, the formed daughter nucleus has the same number of mass number as the parent atom but the atomic number has increased by 1 and neutron number has reduced by 1.
If Na²⁶₁₁ (number of neutrons = 26 - 11 = 15) undergoes a β- decay, then produced daughter atom must have 12 protons and 14 neutrons. Hence, the daughter atom should be Mg.
Na²⁶₁₁ → Mg²⁶₁₂ + ₋₁⁰β + energy
The balanced nuclear equation for beta decay of sodium-26 is represented as Na²⁶₁₁ → Mg²⁶₁₂ + β₋₁⁰ + energy.
What is beta decay?Beta decay is one of the type of radioactive nuclear decay reaction in which emission of a beta particle takes place from the atomic nuclear.
Beta particle define by the symbol β₋₁⁰ i.e. this particles has no mass but having a negative charge on it. In this reaction atomic mass of the parent atom is equal to the new formed daughter atom. But in this reaction number of neutron is reduced by 1 and number of proton is increased by one of parent atom. So, balanced nuclear equation for β− decay of sodium−26 is represnted as:
Na²⁶₁₁ → Mg²⁶₁₂ + β₋₁⁰ + energy
Hence balanced nuclear equation for the beta decay is represented above.
To know more about beta decay, visit the below link:
https://brainly.com/question/13174032
If you ix 20.0 ml of a 3.00m sugar solution with 30.0 ml of a 5.69 m sugar solution, you will end up with a sugar solution of?
Answers
20 mL / 1000 mL x 3 M = .06 moles sugar
30 mL / 1000 mL x 5.69 M = .1707 moles sugar
Total = .2307 moles
Find how many mL of solution you have. 20 mL + 30 mL = 50 mL
1000 mL / 50 mL x .2307 moles = 4.614 M solution
If a person mixes 20.0 ml of a 3.00m sugar solution with 30.0 ml of a 5.69m sugar solution, you will end up with a sugar solution of:
4.614m solutionAccording to the given question, we can see that there is a mixture of 20 milliliters of 3 moles of sugar solution with 30 milliliters of 5.69 moles of another sugar solution, then we need to find the total sugar solution from the mixtures.
As a result of this, we need to first convert the milliliters to moles.
20 mL/1000 mL x 3 moles = 0.06 moles of sugar
30 mL/ 1000 mL x 5.69 moles = 0.1707 moles of sugar
Please note that we are dividing by 1000 mL to convert to liters.
Now, we add up the two values, 0.06 + 0.1707 = 0.2307 moles
Next, we add the total millimeters we have so far which would be
20 mL + 30 mL = 50 mL
Finally, we would convert the 50 mL to liters before we get our final answer.
1000 mL / 50 mL = 20 L
Now, we multiply 20 L by 0.2307 moles = 4.614 M solution
Read more here:
https://brainly.com/question/24247427
A 36.41-g sample of calcium carbonate (CaC O 3 )
contains 14.58 g of calcium and 4.36 g of carbon.
What is the mass of oxygen contained in the sam-
ple? What is the percent by mass of each element in
this compound?
Answers
m(O) = 36.41 g - 14.58 g - 4.36 g.
m(O) = 17.47 g.
ω(Ca) = m(Ca) ÷ m(CaCO₃) · 100%.
ω(Ca) = 14.58 g ÷ 36.41 g · 100%.
ω(Ca) = 40 %; mass percent of calcium.
ω(C) = m(C) ÷ m(CaCO₃) · 100%.
ω(C) = 4.36 g ÷ 36.41 g · 100%
ω(C) = 12%; mass percent of carbon.
ω(O) = 100% - ω(Ca) - ω(C).
ω(O) = 100% - 40% - 12%.
ω(O) = 48%; mass percent of oxygen.
Na has an atomic mass of 23.0g and O has an atomic mass of 16.0g. How many grams of Na are needed to completely react with 40.0g of O2?
Answers
find the moles of O2 used = mass/molar mass
= 40 g/32g/mol = 1.25 moles
write the reacting equation
4Na+ O2 = 2Na2O
by use of mole ratio between Na to O2 which is 4 :1
the moles of Na = 1.25 x 4 = 5 moles
mass of Na = mass x molar mass
= 5 moles x 23 g /mol= 115 moles
Final answer:
To completely react with 40.0g of O2, 115.0g of Na are required. This calculation uses stoichiometry and the balanced chemical equation 4Na + O2 → 2Na2O, considering the atomic masses of Na and O.
Explanation:
The question relates to how many grams of sodium (Na) are needed to react completely with 40.0 grams of oxygen (O2). This is a stoichiometry problem that involves understanding the molar mass of the reactants and the chemical reaction equation. The first step is to understand the reaction between sodium and oxygen, which typically forms sodium oxide (Na2O). The balanced chemical equation for this reaction is:
4Na + O2 → 2Na2O
From this equation, we can see that 4 moles of Na react with 1 mole of O2 to produce 2 moles of Na2O. Using the given atomic mass of O (16.0 g/mol) and knowing that oxygen is diatomic in its normal state (O2), the molar mass of O2 is 32.0 g/mol. Therefore, 40.0 g of O2 corresponds to 40.0 g / 32.0 g/mol = 1.25 moles of O2.
Since 4 moles of Na are required for every 1 mole of O2, the moles of Na needed is 4 * 1.25 moles = 5 moles of Na. Given the atomic mass of Na as 23.0 g/mol, the mass of Na required is 5 moles * 23.0 g/mol = 115.0 grams. Therefore, to completely react with 40.0 grams of O2, 115.0 grams of Na are needed.
Suppose a student titrates a 10.00-ml aliquot of saturated ca(oh)2 solution to the equivalence point with 16.08 ml of 0.0199 m hcl. what was the initial [oh − ]?
Answers
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O
Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O
Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L
∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
= 0.0199 X 16.08 X 10^(-3)
= 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = [tex] \frac{number of moles of [OH-]}{volume of solution (L)} [/tex] = [tex] \frac{3.1999 X 10^(^-^4^)}{10 X 10^(^-^3^)} [/tex] = 0.03199 M
The reaction 4 a + c + h → d has the mechanism below. what is the rate law? 1) 2 a → b fast 2) 2 b → b2 fast 3) b2 + c → g slow 4) g + h → d fast
a. rate = k[a]2[c]
b. rate = k[a]4[h][c]
c. rate = k[a]2
d. rate = k[b]2[c]
e. rate = k[a]4[c]
Answers
Final answer:
To determine the rate law, the rate-determining step must be identified, which is the third step in the given reaction mechanism. The rate law correlates with the concentration of reactants in this slow step, resulting in a rate law of rate [tex]= k[A]^2[C][/tex] correct answer to the question is a. rate = [tex]k[A]^2[C].[/tex]
Explanation:
To determine the rate law for a reaction with a given mechanism, it is essential to identify the rate-determining step, as the rate law is based on this slowest step. In the provided mechanism, the slow step is the third one, where B2 reacts with C to form G. The overall reaction is 4 A + C + H → D, and the steps are:
2 A → B (fast)
2 B → B2 (fast)
B2 + C → G (slow)
G + H → D (fast)
Because step 3 is the slow, rate-determining step, the rate law will be based on the concentrations of the reactants in this step. Since B is formed from A in a fast step and B2 is formed from 2 B, the rate of formation of B2 is dependent on the concentration of A. However, as B2 forms immediately before the slow step, we look at the concentration of A instead of B2 when writing the rate law.
With the stoichiometry of 2 A forming B and then 2 B forming B2, we can note that the concentration of B (and hence B2) is proportional to [tex][A]^2[/tex]efore, for the rate-determining step, the rate law is:
[tex]rate = k[B2][C] → rate = k[A]^2[C][/tex]
So the correct rate law for the overall reaction is:
[tex]rate = k[A]^2[C][/tex]
[tex]rate = k[A]^2[C][/tex]
Nitrogen-13 has a half-life of 10 minutes. how much of a 128 mg sample would remain after 20 minutes?
Answers
Nitrogen-13, because it has a half-life of ten minutes, will experience two half-lives over the course of the twenty minute period. This means that 25% of the isotope will remain after this.
0.25 x 128mg = 32mg
32mg of Nitrogen-13 will remain after 20 minutes.
After 20 minutes, approximately 32 mg of the 128 mg sample of Nitrogen-13 would remain.
Explanation:The half-life of a radioactive isotope is the time required for half of the atoms in a sample to decay.
In this case, we are given that the half-life of Nitrogen-13 is 10 minutes. This means that after every 10 minutes, half of the sample will decay.
Since 20 minutes have passed, we need to determine how many half-lives have occurred. There have been 2 half-lives because 20 divided by 10 equals 2.
Therefore, after 2 half-lives, one-fourth of the sample will remain (since half of the original sample will decay after each half-life).
To find out how much of a 128 mg sample would remain after 20 minutes, we multiply the original amount by one-fourth:
128 mg x 1/4 = 32 mg
After 20 minutes, approximately 32 mg of the 128 mg sample of Nitrogen-13 would remain.
Learn more about Half-life here:https://brainly.com/question/24710827
#SPJ3
What is the [H+] if the pH of a solution is 7.90
Answers
pH is calculated using the hydrogen ion concentration
pH = -log[H⁺]
if we know the pH we can calculate the H⁺ concentration
[H⁺] = antilog(-pH)
[H⁺] = 1.26 x 10⁻⁸ M
therefore [H⁺] is 1.26 x 10⁻⁸ M
Answer:
1.26 x 10^-8
Explanation
Brainly cant blur this out
What are the oxidation numbers of potassium, sulfur, and oxygen, respectively, in k2so4?
Answers
In the compound K2SO4, the oxidation numbers for potassium (K), sulfur (S) and oxygen (O) are +1, +6 and -2, respectively.
Explanation:In the compound K2SO4, the oxidation numbers for potassium (K), sulfur (S) and oxygen (O) can be determined based on established guidelines for assigning oxidation states.
In every stable (neutral) atom, the oxidation number is always zero. Therefore, for potassium, in its ionic form, it has an oxidation number of +1.
In general, the oxidation number of oxygen in its compounds is -2. The compound K2SO4 contains 4 oxygen atoms. Therefore, the total oxidation state contributed by oxygen is -8.
To ensure that the compound is electrically neutral, the total oxidation number should be zero. Hence, for sulfur, you would calculate its oxidation state as follows: total oxidation state of compound = (+1 x 2 for potassium) + oxidation state of sulfur + (-2 x 4 for oxygen) = 0. Solving this equation gives the oxidation state for sulfur as +6.
So, the oxidation numbers for potassium, sulfur and oxygen in K2SO4 are +1, +6 and -2, respectively.
Learn more about Oxidation Numbers here:https://brainly.com/question/29100691
#SPJ12
Lakes that have been acidified by acid rain can be neutralized by limiting the addition of limestone how much limestone is required to completely neutralize a 4.3 billion liter lake with a ph of 5.5
Answers
[H+] = 10^-pH = 10^-5.5
We then use the volume to solve for the number of moles of H+:
moles H+ = 10^-5.5M * 4.3x10^9 L = 13598 moles
From the balanced equation of the neutralization of hydrogen ion by limestone written as
CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2CO3(aq)
we use the mole ratio of limestone CaCO3 and H+ from their coefficients, which is 1 mole of CaCO3 is to react with 2 moles of H+, to compute for the mass of the limestone:
mass CaCO3 = 13598mol H+(1mol CaCO3/2mol H+)
(100.0869g CaCO3/1mol CaCO3)(1kg/1000g)
= 680 kg
The amount of limestone required to neutralize the lake is approximately [tex]\( 2.653 \times 10^4 \, \text{kg} \).[/tex]
To determine how much limestone ([tex]CaCO_3[/tex]) is required to neutralize a lake acidified by sulfuric acid ([tex]H_2SO_4[/tex]), we need to follow these steps:
1. Calculate the total mass of [tex]H_2SO_4[/tex] in the lake.
2. Determine the moles of [tex]H_2SO_4[/tex] present.
3. Use stoichiometry to find the moles of [tex]CaCO}_3[/tex] required to neutralize the [tex]H_2SO_4[/tex]
4. Convert the moles of [tex]CaCO}_3[/tex] to mass in kilograms.
Step 1: Calculate the Total Mass of [tex]H_2SO_4[/tex] in the Lake
The lake volume is [tex]\( 5.2 \times 10^9 \) liters.[/tex]
The concentration of [tex]H_2SO_4[/tex] is [tex]\( 5.0 \times 10^{-3} \) g/L.[/tex]
Total mass of [tex]H_2SO_4[/tex]
[tex]\[ \text{Total mass of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} \][/tex]
[tex]\[ \text{Total mass of H}_2\text{SO}_4 = 5.0 \times 10^{-3} \, \text{g/L} \times 5.2 \times 10^9 \, \text{L} \][/tex]
[tex]\[ \text{Total mass of H}_2\text{SO}_4 = 2.6 \times 10^7 \, \text{g} \][/tex]
Step 2: Determine the Moles of [tex]H_2SO_4[/tex]
Molar mass of [tex]H_2SO_4[/tex]
[tex]\[ \text{H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol} \][/tex]
Moles of [tex]H_2SO_4[/tex]
[tex]\[ \text{Moles of H}_2\text{SO}_4 = \frac{\text{Total mass of H}_2\text{SO}_4}{\text{Molar mass of H}_2\text{SO}_4} \][/tex]
[tex]\[ \text{Moles of H}_2\text{SO}_4 = \frac{2.6 \times 10^7 \, \text{g}}{98 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of H}_2\text{SO}_4 = 2.653 \times 10^5 \, \text{mol} \][/tex]
Step 3: Use Stoichiometry to Find the Moles of [tex]CaCO}_3[/tex] Required
The neutralization reaction between [tex]H_2SO_4[/tex] and [tex]CaCO}_3[/tex] is:
[tex]\[ \text{H}_2\text{SO}_4 + \text{CaCO}_3 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2\text{O} \][/tex]
From the balanced equation, 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]CaCO}_3[/tex]
Therefore, moles of [tex]CaCO}_3[/tex] required:
[tex]\[ \text{Moles of CaCO}_3 = \text{Moles of H}_2\text{SO}_4 \][/tex]
[tex]\[ \text{Moles of CaCO}_3 = 2.653 \times 10^5 \, \text{mol} \][/tex]
Step 4: Convert the Moles of [tex]CaCO}_3[/tex] to Mass in Kilograms
Molar mass of [tex]CaCO}_3[/tex]
[tex]\[ \text{CaCO}_3 = 40 + 12 + 3 \times 16 = 100 \, \text{g/mol} \][/tex]
Mass of [tex]CaCO}_3[/tex]
[tex]\[ \text{Mass of CaCO}_3 = \text{Moles of CaCO}_3 \times \text{Molar mass of CaCO}_3 \][/tex]
[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^5 \, \text{mol} \times 100 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^7 \, \text{g} \][/tex]
Convert grams to kilograms:
[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^7 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} \][/tex]
[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^4 \, \text{kg} \][/tex]
The complete Question is
Lakes that have been acidified by acid rain can be neutralized by the addition of limestone (CaCO3). How much limestone in kg would be required to completely neutralize a 5.2 x 10^9 L lake containing 5.0 x 10-3 g of H2SO4 per liter?
The normal boiling point of water is 100.0 °c and its molar enthalpy of vaporization is 40.67 kj/mol. what is the change in entropy in the system in j/k when 51.1 grams of steam at 1 atm condenses to a liquid at the normal boiling point?
Answers
Δ H = T * ΔS
when Δ H is the change in enthalpy
and T is the temperature in Kelvin = 100+273= 373 K
and ΔS is the change in entropy
but first, we need to get the value of ΔH:
ΔH = mass * molar mass per mole * enthalpy of vaporization
by substitution:
ΔH = 51.1 g * (1 mole/18g) * 40.67KJ/mole
= -115.5 KJ
we use negative sign because this is an exothermic reaction.
by substitution on the first formula, we will get the change of entropyΔS:
ΔS = ΔH / T
= -115.5 KJ / 373K
= 310 J/K
The change in entropy in the system is 308.3 J/K.
Explanation:The change in entropy in the system can be calculated using the formula ∆S = ∆Hvap/T. Here, ∆Hvap is the molar enthalpy of vaporization of water, which is given as 40.67 kJ/mol. T is the temperature in Kelvin, which can be calculated by adding 273.15 to the boiling point of water in Celsius. So, T = 100.0 + 273.15 = 373.15 K. Plugging in these values in the formula, we get:
∆S = (40.67 kJ/mol)/(373.15 K) = 0.1089 kJ/(mol·K)
Now, we need to convert grams of steam to moles of steam. The molar mass of water is 18.015 g/mol. So, 51.1 g of steam is equal to (51.1 g)/(18.015 g/mol) = 2.835 mol. Multipling this with the change in entropy, we get:
∆S = (0.1089 kJ/(mol·K)) · (2.835 mol) = 0.3083 kJ/K
Finally, to convert kJ/K to J/K, we multiply by 1000:
∆S = (0.3083 kJ/K) · (1000 J/1 kJ) = 308.3 J/K
Learn more about entropy here:https://brainly.com/question/38613023
#SPJ11
Which of the following is not a group found in a protein?
Carboxyl Group
L Group
R Group
Amino Group
Answers
Answer:
L Group
Explanation:
Proteins are made up of amino acids, these amino acids determine the protein's functional groups, in the image I annexed you can observe that amino acids have an amino group (NH2), an acid group(COOH) and a R group, a lateral chain that varies depending on the amino acid, in the protein these amino acids are linked through a peptide bond leaving one side of the chain with an acid group and the other side with an amide chain, keeping the R groups as well.
I hope you find this information useful and interesting! Good luck!
The mineral enargite is 48.41% cu, 19.02% as, and 32.57% s by mass. what is the empirical formula of enargite?
Answers
If we use 100 grams of enargite:
n(Cu) = 48,41 g ÷ 63.55 g/mol.
n(Cu) = 0.761 mol.
n(As) = 19.02 g ÷ 74.92 g/mol.
n(As) = 0.254 mol.
n(S) = 32.57 g ÷ 32.065 g/mol.
n(S) = 1.016 mol.
n(Cu) : n(As) : n(S) = 0.761 mol : 0.254 mol : 1.016 mol / 0.254.
n(Cu) : n(As) : n(S) = 3 mol : 1 mol : 4 mol.
A gas sample of argon, maintained at constant temperature, occupies a volume of 500. l at 4.00 atm. what is the new volume if the pressure were charged to 8.0 atm
Answers
It states that at constant temperature, pressure is inversely proportional to volume of gas.
PV = k
where P - pressure, V - volume and k - constant
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting these values in the equation
4.00 atm x 500 L = 8.0 atm x V
V = 250 L
new volume is 250 L
More people need to get a profile picture
Suppose that 26.89 ml of vinegar solution requires 33.23 ml of the 0.09892 m sodium hydroxide solution to reach the endpoint. calculate the molar concentration of the vinegar solution.
Answers
= 3.287
Thus, number of millimoles of vinegar present in sample = 3.287
Now, Molarity of vinegar solution = [tex] \frac{\text{number of millimoles of vinegar}}{\text{volume of solution (ml)}} [/tex]
= [tex] \frac{3.287}{26.89} [/tex]
= 0.1222 mol/dm3
Molar concentration of vinegar present in sample solution is 0.1222 M
How many compounds, of the ones listed below, have hydrogen bonding? ch3(ch2)2nh2 ch3(ch2)2nh(ch2)4ch3 (ch3ch2)2n(ch2)4ch3?
Answers
Following two compounds have Hydrogen Bond Interactions;
1) CH₃(CH₂)₂NH₂ (Propan-1-amine)
2) CH₃(CH₂)₂NH(CH₂)₄CH₃ (N-propylpentan-1-amine)
Explanation:
Hydrogen Bond Interactions are formed between those molecules which has hydrogen atoms covalently bonded to most electronegative atoms like Fluorine, Oxygen and Nitrogen. This direct attachment of Hydrogen to electronegative atom makes it partial positive resulting in hydrogen bonding with neighbor's partial negative most electronegative atom. So, in above selected compounds it can be seen that both compounds contain hydrogen atoms directly attached to Nitrogen atoms, Therefore, allowing them to form Hydrogen Bonding Interactions.
The correct answer is a. 2 compounds have hydrogen bonding among the listed below.
To determine how many of the given compounds have hydrogen bonding, we must identify if they have hydrogen atoms attached to highly electronegative atoms like nitrogen (N), oxygen (O), or fluorine (F).
The compounds listed are:
CH₃(CH₂)₂NH₂CH₃(CH₂)₃NH(CH₂)₂CH₃(CH₃CH₂)₂N(CH₂)₄CH₃Analysis
CH₃(CH₂)₂NH₂: Contains an N-H bond, so it can form hydrogen bonds.CH₃(CH₂)₃NH(CH₂)₂CH₃: Contains an N-H bond, so it can form hydrogen bonds.(CH₃CH₂)₂N(CH₂)₄CH₃: Does not contain an N-H bond, so it cannot form hydrogen bonds.Based on this analysis, 2 out of the 3 compounds can form hydrogen bonds. Therefore, the correct answer is a. 2.
Correct question is: How many compounds, of the ones listed below, have hydrogen bonding?
CH₃(CH₂)₂NH₂ , CH₃(CH₂)₃NH(CH₂)₂CH₃ , (CH₃CH₂)₂N(CH₂)₄CH₃ ?
a. 2
b. 1
c. 0
d. 3
What mass of solute is contained in 25.4 ml of a 1.56 m potassium bromide solution?
Answers
V(KBr) = 25.4 mL ÷ 1000 mL/L = 0.0254 L, volume of solution.
c(KBr) = 1.56 mol/L.
n(KBr) = c(KBr) · V(KBr).
n(KBr) = 1.56 mol/L 0.054 L.
n(KBr) = 0.0396 mol, amount of substance.
m(KBr) = n(KBr) · M(KBr).
m(KBr) = 0.0396 mol · 119 g/mol.
m(KBr) = 4.71 g.
M - molar mass.
Answer : The mass of solute is 4.72 grams.
Explanation :
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
[tex]\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
or,
[tex]\text{Molarity}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
In this question, the solute is potassium bromide.
Given:
Volume of solution = 25.4 mL
Molarity = 1.56 M
Molar mass of potassium bromide = 119 g/mole
Now put all the given values in this formula, we get:
[tex]1.56M=\frac{\text{Mass of solute}\times 1000}{119g/mole\times 25.4L}[/tex]
[tex]\text{Mass of solute}=4.72g[/tex]
Therefore, the mass of solute is 4.72 grams.
What are the products obtained in the electrolysis of molten nai?
Answers
The electrolysis of molten sodium iodide (NaI) yields sodium metal and iodine gas. The process involves the migration of ions to respective electrodes, gaining or losing electrons, thus getting reduced or oxidized respectively.
Explanation:The products obtained from the electrolysis of molten sodium iodide (NaI) are sodium (Na) and iodine (I2). During electrolysis, sodium ions are reduced at the cathode to form sodium metal, shown by the half-equation 2Na+ + 2e- → 2Na. Meanwhile, iodide ions get oxidized at the anode to produce iodine gas, as depicted by 2I- → I2 + 2e-.
In a nutshell, the process involves electrolysis of molten sodium iodide using a Downs cell. Positively charged sodium ions migrate to the negatively charged cathode and gain electrons, reducing them to sodium metal. Conversely, the negatively charged iodide ions migrate to the positively charged anode and lose electrons, getting oxidized to iodine gas.
Learn more about Electrolysis of Molten Sodium Iodide here:https://brainly.com/question/34433122
#SPJ12
The electrolysis of molten sodium iodide (NaI) produces sodium (Na) and iodine (I2). In the process, sodium ions migrate to the cathode and are reduced to sodium metal, while iodide ions migrate to the anode and are oxidized to iodine gas.
Explanation:The process of electrolyzing molten sodium iodide, or NaI, is similar both in concept and execution to the electrolysis of molten sodium chloride, which is a better-known and more commonly discussed process. Electrolysis of molten sodium iodide will produce sodium (Na) and iodine (I) as products. Here's a description of how it works:
In the set-up, you would have a Downs cell. The Downs cell contains molten sodium iodide. The passage of a direct current through the cell causes the sodium ions to migrate to the negatively charged cathode and pick up electrons, reducing the ions to sodium metal. Iodide ions migrate to the positively charged anode, lose electrons, and undergo oxidation to iodine gas.
The overall cell reaction would be: 2NaI -> 2Na + I2
This is a simple explanation of the process and should suffice for your understanding. Do keep in mind that actual industrial processes could have additional steps and complexities not addressed here.
Learn more about Electrolysis here:https://brainly.com/question/33301115
#SPJ11
Calculate the k sp for zinc hydroxide if the solubility of zn (oh)2 in pure water is 2.1 × 10-4 g/l. calculate the k sp for zinc hydroxide if the solubility of zn (oh)2 in pure water is 2.1 × 10-4 g/l. 2.9 × 10-3 1.5 × 10-3 9.9 × 10-18 3.8 × 10-17
Answers
For such systems, k sp = 4S^3
where S = solubility in mol/l
Now for Zinc hydroxide, molecular weight = 99.4
so, solubility = 2.1 x 10-4 g/l = 2.11 X 10^-6 mol/l
Finally, k sp = 4S^3 = 4 X (2.11 X 10^-6)^3 = 3.77 X 10^-17
The [tex]\( K_{sp} \)[/tex] for zinc hydroxide is [tex]\[ 3.8 \times 10^{-17}}\][/tex]
1. Molar mass of [tex]\( \text{Zn(OH)}_2 \)[/tex]:
[tex]\[ \text{Zn} = 65.38 \, \text{g/mol}, \, \text{O} = 16.00 \, \text{g/mol}, \, \text{H} = 1.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass} = 65.38 + 2 \times (16.00 + 1.01) = 99.40 \, \text{g/mol} \][/tex]
2. Molar solubility [tex]\( s \)[/tex]:
[tex]\[ \text{Solubility} = 2.1 \times 10^{-4} \, \text{g/L} \][/tex]
[tex]\[ s = \frac{2.1 \times 10^{-4}}{99.40} \approx 2.11 \times 10^{-6} \, \text{mol/L} \][/tex]
3. Ksp expression:
[tex]\[ \text{Zn(OH)}_2 \rightleftharpoons \text{Zn}^{2+} + 2\text{OH}^- \][/tex]
[tex]\[ K_{sp} = [\text{Zn}^{2+}][\text{OH}^-]^2 = s \cdot (2s)^2 = 4s^3 \][/tex]
4. Calculate [tex]\( K_{sp} \):[/tex]
[tex]\[ K_{sp} = 4 \times (2.11 \times 10^{-6})^3 = 4 \times 9.39 \times 10^{-18} \approx 3.76 \times 10^{-17} \][/tex]
Recrystallizing an anhydride (such as the product of this reaction) from water or an alcohol is rarely a good idea. explain why. be specific.
Answers
Final answer:
Recrystallization of an anhydride in water or alcohol is not effective because anhydrides will react with these solvents to form carboxylic acids or esters, respectively, thereby altering the compound instead of purifying it.
Explanation:
Recrystallizing an anhydride from water or an alcohol is often not ideal due to the reactive nature of anhydrides. Anhydrides undergo hydrolysis in the presence of water to form carboxylic acids, a process which is energetically favorable and further driven by the stabilization through hydrogen-bonding interactions among water molecules and the carboxylic acids. Moreover, anhydrides react with alcohols to yield esters. This means that instead of recrystallizing the anhydride, you are likely to convert it to other products, which defeats the purpose of recrystallization since you want to purify the anhydride, not change its chemical structure.