The compounds of interest are Na2S and H2SO4. Th.pdf
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The compounds of interest are Na2S and H2SO4. The molecular equation then is: Na2S + H2SO4 ---> Na2SO4 + H2S. Just exchange one anion for the other. Now, we need to know a little bit about the properties of these compounds. Almost all sodium salts are soluble, so both Na2S and Na2SO4 are fully dissociated: 2 Na+ + S2- + 2 H+ + SO42- ---> 2 Na+ + SO42- + H2S. Crossing out terms that appear on both sides, we obtain the net ionic equation: 2 H+ + S2- ----> H2S. Conveniently, H2S is a gas, and escapes from solution. Thus, it selectively removes H+ and S2- ions from the equilibrium, driving the reaction to completion (favoring Na2SO4) Solution The compounds of interest are Na2S and H2SO4. The molecular equation then is: Na2S + H2SO4 ---> Na2SO4 + H2S. Just exchange one anion for the other. Now, we need to know a little bit about the properties of these compounds. Almost all sodium salts are soluble, so both Na2S and Na2SO4 are fully dissociated: 2 Na+ + S2- + 2 H+ + SO42- ---> 2 Na+ + SO42- + H2S. Crossing out terms that appear on both sides, we obtain the net ionic equation: 2 H+ + S2- ----> H2S. Conveniently, H2S is a gas, and escapes from solution. Thus, it selectively removes H+ and S2- ions from the equilibrium, driving the reaction to completion (favoring Na2SO4).
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(NH4)2SO4 is a soluble salt and is fully ionized in solution(NH4).pdf
(NH4)2SO4 is a soluble salt and is fully ionized in solution:
(NH4)2SO4 => 2 NH4+ + SO42-
NH4+ is a weak acid and ionizes further: NH4+<=> NH3 + H+
The H+ formed can react with C2O42- to form the weak acid oxalic acid (H2C2O4):
C2O42- + 2 H+ <=> H2C2O4
Removal of C2O42- causes the equilibrium: ZnC2O4 <=> Zn2+ + C2O42-
to shift to the right according to Le Chatelier\'s principle and more ZnC2O4 dissolves.
Thus the solubility of ZnC2O4 increases in the presence of (NH4)2SO4.
Solution
(NH4)2SO4 is a soluble salt and is fully ionized in solution:
(NH4)2SO4 => 2 NH4+ + SO42-
NH4+ is a weak acid and ionizes further: NH4+<=> NH3 + H+
The H+ formed can react with C2O42- to form the weak acid oxalic acid (H2C2O4):
C2O42- + 2 H+ <=> H2C2O4
Removal of C2O42- causes the equilibrium: ZnC2O4 <=> Zn2+ + C2O42-
to shift to the right according to Le Chatelier\'s principle and more ZnC2O4 dissolves.
Thus the solubility of ZnC2O4 increases in the presence of (NH4)2SO4..
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(NH4)2SO4 is a soluble salt and is fully ionized in solution(NH4).pdf
(NH4)2SO4 is a soluble salt and is fully ionized in solution:
(NH4)2SO4 => 2 NH4+ + SO42-
NH4+ is a weak acid and ionizes further: NH4+<=> NH3 + H+
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Thus the solubility of ZnC2O4 increases in the presence of (NH4)2SO4.
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(NH4)2SO4 is a soluble salt and is fully ionized in solution:
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generally, compounds with Na+ or NO3- are soluble in water and will .pdf
generally, compounds with Na+ or NO3- are soluble in water and will be ions in solution.
Therefore you\'re left with Ba2+ and SO42- to react together and form a salt. BaSO4 is not a
very soluble salt, so it\'s going to precipitate out of the solution. (Ksp is about 1.1x10-10).
NaNO3 is going to just stay in solution as the ions, so Na+ and NO3- are the spectator ions.
Ba2+(aq) + NO3-(aq) + Na+(aq) + SO42-(aq) ---> BaSO4 (s) + NO3-(aq) + Na+(aq)
(if looking for only 2 products, write the Na+ and NO3- ions as NaNO3 (aq) )
The balanced net ionic equation is Ba2+ (aq) + SO42- (aq) --> BaSO4 (s)
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generally, compounds with Na+ or NO3- are soluble in water and will be ions in solution.
Therefore you\'re left with Ba2+ and SO42- to react together and form a salt. BaSO4 is not a
very soluble salt, so it\'s going to precipitate out of the solution. (Ksp is about 1.1x10-10).
NaNO3 is going to just stay in solution as the ions, so Na+ and NO3- are the spectator ions.
Ba2+(aq) + NO3-(aq) + Na+(aq) + SO42-(aq) ---> BaSO4 (s) + NO3-(aq) + Na+(aq)
(if looking for only 2 products, write the Na+ and NO3- ions as NaNO3 (aq) )
The balanced net ionic equation is Ba2+ (aq) + SO42- (aq) --> BaSO4 (s).
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1. Copper is above silver in the activity series. Thus Cu metal will.pdf
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Okay, here are the steps to balance this reaction:
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MnO4- is reduced, so it is the oxidizing agent.
MnO4- + 5e- → Mn2+
SO2 is oxidized, so it is the reducing agent.
SO2 → SO42- + 4e-
Step 2) Balance other elements: No need here.
Step 3) Balance O by adding H2O:
MnO4- + 5e- → Mn2+ + 4H2O
SO2 → SO42- + 4e-
Step 4) Balance H by adding H+:
M
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Reduction: 1/2O2 + H+ → H2O
Balanced half-reactions:
Ag → Ag+
1/2O2 + H+ → H2O
Overall balanced reaction:
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The correct coefficients are:
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Balancing ANY chemical equation is done exactly t.pdf
Balancing ANY chemical equation is done exactly the same. Easiest way is to start
with a primary element such as Carbon, then just work your way back and forth between
products and reactants until everything is balanced. Be sure to check your equation once you\'re
done to make sure everything is balanced completely. Types of Chemical reactions: 1.)
Combustion: A combustion reaction is when oxygen combines with another compound to form
water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An
example of this kind of reaction is the burning of propane: C3H8 + 5O2 --> 3CO2 + 4H2O
*Hint: Combustion reactions always have O2 in the reactants and CO2 and H2O somewhere in
the products. 2.) Synthesis: A synthesis reaction is when two or more simple compounds
combine to form a more complicated one. These reactions come in the general form of: A + B --
-> AB One example of a synthesis reaction is the combination of iron and sulfur to form iron
(II) sulfide: 8 Fe + S8 ---> 8 FeS 3.) Decomposition: A decomposition reaction is the opposite
of a synthesis reaction - a complex molecule breaks down to make simpler ones. These reactions
come in the general form: AB ---> A + B One example of a decomposition reaction is the
electrolysis of water to make oxygen and hydrogen gas: 2 H2O ---> 2 H2 + O2 4.) Single
displacement: This is when one element trades places with another element in a compound.
These reactions come in the general form of: A + BC ---> AC + B One example of a single
displacement reaction is when magnesium replaces hydrogen in water to make magnesium
hydroxide and hydrogen gas: Mg + 2 H2O ---> Mg(OH)2 + H2 5) Double displacement: This is
when the anions and cations of two different molecules switch places, forming two entirely
different compounds. These reactions are in the general form: AB + CD ---> AD + CB One
example of a double displacement reaction is the reaction of lead (II) nitrate with potassium
iodide to form lead (II) iodide and potassium nitrate: Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3 6)
Acid-base: This is a special kind of double displacement reaction that takes place when an acid
and base react with each other. The H+ ion in the acid reacts with the OH- ion in the base,
causing the formation of water. Generally, the product of this reaction is some ionic salt and
water: HA + BOH ---> H2O + BA One example of an acid-base reaction is the reaction of
hydrobromic acid (HBr) with sodium hydroxide: HBr + NaOH ---> NaBr + H2O *Hint: An acid
plus a base always produces a salt and H2O as a product.
Solution
Balancing ANY chemical equation is done exactly the same. Easiest way is to start
with a primary element such as Carbon, then just work your way back and forth between
products and reactants until everything is balanced. Be sure to check your equation once you\'re
done to make sure everything is balanced completely. Types of Chemical reactions: 1.)
Combustion: A combustion reaction is when oxygen combines with an.
36.Kovacs reagent is used in Indole test. Kovacs reagent is 4 (p)-.pdf
Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and
therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular
weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =-
114.1/(2*1.14225) =40 kJ
Solution
Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and
therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular
weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =-
114.1/(2*1.14225) =40 kJ.
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generally, compounds with Na+ or NO3- are soluble in water and will .pdf
generally, compounds with Na+ or NO3- are soluble in water and will be ions in solution.
Therefore you\'re left with Ba2+ and SO42- to react together and form a salt. BaSO4 is not a
very soluble salt, so it\'s going to precipitate out of the solution. (Ksp is about 1.1x10-10).
NaNO3 is going to just stay in solution as the ions, so Na+ and NO3- are the spectator ions.
Ba2+(aq) + NO3-(aq) + Na+(aq) + SO42-(aq) ---> BaSO4 (s) + NO3-(aq) + Na+(aq)
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generally, compounds with Na+ or NO3- are soluble in water and will be ions in solution.
Therefore you\'re left with Ba2+ and SO42- to react together and form a salt. BaSO4 is not a
very soluble salt, so it\'s going to precipitate out of the solution. (Ksp is about 1.1x10-10).
NaNO3 is going to just stay in solution as the ions, so Na+ and NO3- are the spectator ions.
Ba2+(aq) + NO3-(aq) + Na+(aq) + SO42-(aq) ---> BaSO4 (s) + NO3-(aq) + Na+(aq)
(if looking for only 2 products, write the Na+ and NO3- ions as NaNO3 (aq) )
The balanced net ionic equation is Ba2+ (aq) + SO42- (aq) --> BaSO4 (s).
Writing net ionic equations
The document discusses the rules for writing net ionic equations: write weak electrolytes as molecules or atoms and strong electrolytes as ions, then cancel out any spectator ions that are common to the reactants and products. It provides examples of applying these rules to write balanced equations, identify electrolytes, write molecular and ionic forms, and derive the net ionic equation.
Chem anions
The document discusses qualitative inorganic analysis of anions, specifically focusing on carbonates/bicarbonates and sulfur-containing anions. It describes the general characteristics, solubility, and common reactions of these anion groups. Key points include that carbonates and bicarbonates react with acids to produce carbon dioxide gas, while sulfur-containing anions like sulfides, sulfites, and thiosulfates act as reducing agents and react with reagents to produce characteristic colors or precipitates. Common tests involve the use of acids, silver nitrate, barium chloride, and iron (III) chloride.
10) Ionic compounds are a combination of metals a.pdf
10) Ionic compounds are a combination of metals and non-metals. A covalent
compound is a combination of a non-metal and another non-metal An example of an ionic
compound would be Na2SO4. An example of a covalent compound would be CO2. 10b.) a.
Hydrochloric acid Anytime you have H-nameofanyhalogen and its aqueous (aq), you know it\'s
an acid. To name acids (of this type) you just say \"Hydro\" then the prefix name of halogen (ie.
chlor) and add \"ic\" to the end So HCL = Hydro chlor ic acid. If you had for instance HBr, it
would be Hydro brom ic acid... Hydrobromic Acid b. Hydrogen Chloride This is the same letters
as Hydrochloric acid but it\'s NOT (aq). It\'s a gas (g). That\'s how you know it\'s a regular
covalent compound and not an acid. And it\'s covalent because it\'s a mix of two non-metals. So
therefore it would just be the names Hydrogen, and then chloride. When you have a covalent
compound you always say the first element name, just spelled out normally, and then the prefix
of the next element with the suffix \"ide.\" So HCl would be Hydrogen Chlor ide. Hydrogen
Chloride. c. Methane Many compounds have everyday names that have been given to them.
This is an example of that. You just have to memorize them there is no easy way of determining
an everyday name for these compounds. d. Methanol Yet another example, you just have to
memorize what these formulas are according to their name. 10c.) a. 2H2 + O2 ===> 2H2O b.
CH4 + 2O2 ===> CO2 + 2H2O As you can see, for both of these reactions they have the same
number of elements on both sides of the equation. Now, remember, you NEVER add subscripts
to any of the elements on either side of the equation when you\'re balancing. The subscripts stay
the same. What you DO add are coefficients in front of a compound. For instance the
unbalanced equation in your question (question a.) \"Hydrogen gas and oxygen gas react to form
liquid water.\" would be: H2 + O2 ===> H2O As you can see, it\'s not balanced..... the
hydrogens are balanced on both sides (2 on each side) but the oxygens are not. There are 2
oxygens on the left and only 1 on the right. So add a coefficion of 2 to the \"H2O\" so that you
can have 2 oxygens. Now our new reaction is: H2 + 02 ===> 2H2O However this is still not
balanced. As you can see we have 2 hydrogens on the left, but now we have 4 hydrogens on the
right. So, in order to fix this, add 2 in front of H2 on the left, so now we have: 2H2 + O2 ===>
2H2O Now we have a balanced equation with 4 hydrogens on the right, and 4 hydrogens on the
left. As well as 2 oxygens on the right and 2 oxygens on the left. This same type of procedure is
how you balance any reaction. 10d.) Al2(SO4)3 = Aluminum Sulfate To calculate percent
aluminum by mass you first find the mass of the compound. Molar Mass Al2(SO4)3: Al =
27.0(2) = 54g/mol S = 32.1(3) = 96.3g/mol O = 16.0(12) = 192g/mol Add them all together to
get the molar mass of Al2(SO4)3 = 342.3g/mol Now, to get the % by mass of Al in Al2(SO4).
Acids and bases dr.surendran prambadath
The document discusses acids and bases according to different theories including Arrhenius, Bronsted-Lowry, and Lewis concepts. It defines acids and bases, describes their properties, and explains neutralization reactions. Examples are provided of strong vs weak acids and bases as well as monoprotic, diprotic, and triprotic acids and bases based on their equivalent weights.
Anions 120108105251-phpapp02
1) The document discusses qualitative inorganic analysis of anions, focusing on the carbonate, bicarbonate, and sulfur-containing anion groups.
2) Key reactions described include the evolution of carbon dioxide gas from carbonates and bicarbonates with hydrochloric acid, and the evolution of sulfur dioxide gas from sulfites and thiosulfates with hydrochloric acid.
3) Precipitation reactions with reagents like barium chloride, calcium chloride, and silver nitrate are used to identify anions based on the insoluble compounds formed.
Anions
This document discusses qualitative inorganic analysis of anions, specifically focusing on carbonates/bicarbonates and sulfur-containing anions. It describes:
1) Carbonates and bicarbonates react with acids to produce carbon dioxide gas, which turns limewater milky. Sulfur-containing anions like sulfides, sulfites and thiosulfates are reducing and react with acids.
2) Common tests involve adding reagents like barium chloride, silver nitrate, and iron (III) chloride to detect anions via formation of insoluble precipitates.
3) The document provides details on solubility rules and characteristic reactions for each group of anions to allow their identification in mixtures
Oxidation reduction reactions honors
The document discusses oxidation-reduction (redox) reactions, where there is a transfer of electrons between reactants. It defines oxidation as the loss of electrons and reduction as the gain of electrons. An example redox reaction and its net ionic form are provided. The document explains how to determine the oxidation states of elements and identifies common oxidation states of nonmetals. It describes how to write half-reactions by separating a redox reaction into its oxidation and reduction components.
1. Copper is above silver in the activity series. Thus Cu metal will.pdf
1. Copper is above silver in the activity series. Thus Cu metal will displace Ag+ ions from
solution, forming Cu2+ ions and metallic Ag.
Observation: The reddish brown copper metal dissolves and silver metal precipitates out. The
colorless Ag+ solution turns blue due to Cu2+ ions formed.
2. Molecular equation:
Cu(s) + 2 AgNO3(aq) => 2 Ag(s) + Cu(NO3)2(aq)
3. Complete ionic equation (obtained by expanding soluble ions in molecular equation):
Cu(s) + 2 Ag+(aq) + 2 NO3-(aq) => 2 Ag(s) + Cu2+(aq) + 2 NO3-(aq)
4. Net ionic equation (obtained by cancelling common ions in complete ionic equation):
Cu(s) + 2 Ag+(aq) => 2 Ag(s) + Cu2+(aq)
Solution
1. Copper is above silver in the activity series. Thus Cu metal will displace Ag+ ions from
solution, forming Cu2+ ions and metallic Ag.
Observation: The reddish brown copper metal dissolves and silver metal precipitates out. The
colorless Ag+ solution turns blue due to Cu2+ ions formed.
2. Molecular equation:
Cu(s) + 2 AgNO3(aq) => 2 Ag(s) + Cu(NO3)2(aq)
3. Complete ionic equation (obtained by expanding soluble ions in molecular equation):
Cu(s) + 2 Ag+(aq) + 2 NO3-(aq) => 2 Ag(s) + Cu2+(aq) + 2 NO3-(aq)
4. Net ionic equation (obtained by cancelling common ions in complete ionic equation):
Cu(s) + 2 Ag+(aq) => 2 Ag(s) + Cu2+(aq).
Chemistry M2 Acids bases salts
Acids provide H+ ions in water, while bases provide OH- ions. Examples of acids include HCl and H2SO4, while examples of bases include NaOH and KOH. When acids and bases react, they undergo a neutralization reaction producing water and a salt. For example, HCl + NaOH produces H2O + NaCl. Strong acids and bases fully dissociate in water, while weak ones only partially dissociate.
Tutorial 2- Balancing Chemical Equations.ppt
Okay, here are the steps to balance this reaction:
Step 1) Identify oxidizing and reducing agents:
MnO4- is reduced, so it is the oxidizing agent.
MnO4- + 5e- → Mn2+
SO2 is oxidized, so it is the reducing agent.
SO2 → SO42- + 4e-
Step 2) Balance other elements: No need here.
Step 3) Balance O by adding H2O:
MnO4- + 5e- → Mn2+ + 4H2O
SO2 → SO42- + 4e-
Step 4) Balance H by adding H+:
M
Chapter 20 Lecture- Electrochemistry
Let's balance this step-by-step using the half-reaction method:
Oxidation: Ag → Ag+
Reduction: 1/2O2 + H+ → H2O
Balanced half-reactions:
Ag → Ag+
1/2O2 + H+ → H2O
Overall balanced reaction:
1Ag + 1/2O2 + 1H+ → 1Ag+ + 1H2O
The correct coefficients are:
1, 1/2, 1, 1, 1
The answer is 1.
Oxidation reduction reactions By MUHAMMAD FAHAD ANSARI 12 IEEM 14
The document summarizes various electrochemical reactions including oxidation-reduction reactions. It discusses reactions that produce electrical energy spontaneously like in batteries and those that require electrical energy like electrolysis. Key points covered include the definitions of oxidation, reduction, and oxidizing/reducing agents. Examples of electrolysis and electroplating are provided. The roles of microbes in important redox processes like photosynthesis, aerobic respiration, nitrogen fixation, nitrification, denitrification, sulfate reduction, and methane formation are summarized.
ACID_BASE.pptx
This document provides information on acid-base reactions and oxidation-reduction (redox) reactions. It defines acids and bases, and explains that in acid-base reactions, acids donate protons to bases. Neutralization reactions between acids and bases produce water and a salt. The document also discusses how to determine oxidation states of elements in compounds and identify the oxidized and reduced substances in redox reactions. It provides steps for balancing redox equations, including dividing the reaction into partial equations and adding electrons to balance charges. Examples of assigning oxidation states and balancing redox reactions are included.
Balancing Equations
The document discusses four types of chemical reactions: synthesis, decomposition, single replacement, and double replacement. It provides examples of each type of reaction and describes how to balance chemical equations by ensuring the same number and type of atoms are on both sides of the equation. The key is only changing the coefficients in front of the formulas and not the subscripts inside the formulas.
1. redox reactions summary presentation.ppt
Redox reactions involve the transfer of electrons between reactants. In an electrochemical cell, a spontaneous redox reaction occurs between two half-cells separated by a salt bridge. In the oxidation half-reaction, electrons are lost at the anode. In the reduction half-reaction, electrons are gained at the cathode. The standard electrode potential (E0) of a half-reaction indicates its tendency to be reduced or oxidized relative to the standard hydrogen electrode. The cell potential (Ecell) is equal to the cathode potential minus the anode potential and determines if the cell reaction is spontaneous.
Similar to The compounds of interest are Na2S and H2SO4. Th.pdf (20)
generally, compounds with Na+ or NO3- are soluble in water and will .pdf
generally, compounds with Na+ or NO3- are soluble in water and will .pdf
10) Ionic compounds are a combination of metals a.pdf
10) Ionic compounds are a combination of metals a.pdf
1. Copper is above silver in the activity series. Thus Cu metal will.pdf
1. Copper is above silver in the activity series. Thus Cu metal will.pdf
Oxidation reduction reactions By MUHAMMAD FAHAD ANSARI 12 IEEM 14
Oxidation reduction reactions By MUHAMMAD FAHAD ANSARI 12 IEEM 14
More from deepua8
Balancing ANY chemical equation is done exactly t.pdf
Balancing ANY chemical equation is done exactly the same. Easiest way is to start
with a primary element such as Carbon, then just work your way back and forth between
products and reactants until everything is balanced. Be sure to check your equation once you\'re
done to make sure everything is balanced completely. Types of Chemical reactions: 1.)
Combustion: A combustion reaction is when oxygen combines with another compound to form
water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An
example of this kind of reaction is the burning of propane: C3H8 + 5O2 --> 3CO2 + 4H2O
*Hint: Combustion reactions always have O2 in the reactants and CO2 and H2O somewhere in
the products. 2.) Synthesis: A synthesis reaction is when two or more simple compounds
combine to form a more complicated one. These reactions come in the general form of: A + B --
-> AB One example of a synthesis reaction is the combination of iron and sulfur to form iron
(II) sulfide: 8 Fe + S8 ---> 8 FeS 3.) Decomposition: A decomposition reaction is the opposite
of a synthesis reaction - a complex molecule breaks down to make simpler ones. These reactions
come in the general form: AB ---> A + B One example of a decomposition reaction is the
electrolysis of water to make oxygen and hydrogen gas: 2 H2O ---> 2 H2 + O2 4.) Single
displacement: This is when one element trades places with another element in a compound.
These reactions come in the general form of: A + BC ---> AC + B One example of a single
displacement reaction is when magnesium replaces hydrogen in water to make magnesium
hydroxide and hydrogen gas: Mg + 2 H2O ---> Mg(OH)2 + H2 5) Double displacement: This is
when the anions and cations of two different molecules switch places, forming two entirely
different compounds. These reactions are in the general form: AB + CD ---> AD + CB One
example of a double displacement reaction is the reaction of lead (II) nitrate with potassium
iodide to form lead (II) iodide and potassium nitrate: Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3 6)
Acid-base: This is a special kind of double displacement reaction that takes place when an acid
and base react with each other. The H+ ion in the acid reacts with the OH- ion in the base,
causing the formation of water. Generally, the product of this reaction is some ionic salt and
water: HA + BOH ---> H2O + BA One example of an acid-base reaction is the reaction of
hydrobromic acid (HBr) with sodium hydroxide: HBr + NaOH ---> NaBr + H2O *Hint: An acid
plus a base always produces a salt and H2O as a product.
Solution
Balancing ANY chemical equation is done exactly the same. Easiest way is to start
with a primary element such as Carbon, then just work your way back and forth between
products and reactants until everything is balanced. Be sure to check your equation once you\'re
done to make sure everything is balanced completely. Types of Chemical reactions: 1.)
Combustion: A combustion reaction is when oxygen combines with an.
36.Kovacs reagent is used in Indole test. Kovacs reagent is 4 (p)-.pdf
Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and
therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular
weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =-
114.1/(2*1.14225) =40 kJ
Solution
Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and
therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular
weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =-
114.1/(2*1.14225) =40 kJ.
10.Real number1.Baseb. Consists of a set and rule for combining2.Bin.pdf
pH = pKa + log ( [ HEPES] / [HEPES protonated] 7.8 = 7.47 + log ( [HEPES] /
[HEPES protonated]) 7.8-7.47 = log ( [HEPES] / [HEPES protonated]) [HEPES] / [HEPES
protonated] = 10 ^ 0.33 = 2.138 [HEPES] = 2.138 * [HEPES protonated] Given [HEPES] =
0.025M [HEPES protonated] = 0.025/2.138 =0.0117M For preparation of solution amount of
HEPES = 0.0117+0.025 = 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required
0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required has to be used
Solution
pH = pKa + log ( [ HEPES] / [HEPES protonated] 7.8 = 7.47 + log ( [HEPES] /
[HEPES protonated]) 7.8-7.47 = log ( [HEPES] / [HEPES protonated]) [HEPES] / [HEPES
protonated] = 10 ^ 0.33 = 2.138 [HEPES] = 2.138 * [HEPES protonated] Given [HEPES] =
0.025M [HEPES protonated] = 0.025/2.138 =0.0117M For preparation of solution amount of
HEPES = 0.0117+0.025 = 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required
0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required has to be used.
1. Yeasts grow by budding. The cell buds and separates into 2 cells..pdf
O-N=O is a resonance structure. In a resonance structure, the atoms themselves do
not move, but the bonds move. In O-O, the elements cannot change to a different element. In
N=C, another atom cannot be added. And in O=N-F, the atoms cannot be moved around.
Solution
O-N=O is a resonance structure. In a resonance structure, the atoms themselves do
not move, but the bonds move. In O-O, the elements cannot change to a different element. In
N=C, another atom cannot be added. And in O=N-F, the atoms cannot be moved around..
1. 252.125Solution1. 252.125.pdf
Most carbon compounds are poor conductors of electricity as we have seen . From
the data on the boiling and melting points of the above compounds, we can conclude that the
forces of attraction between these molecules are not very strong. Since these compounds are
largely nonconductors of electricity, we can conclude that the bonding in these compounds does
not give rise to any ions.
Solution
Most carbon compounds are poor conductors of electricity as we have seen . From
the data on the boiling and melting points of the above compounds, we can conclude that the
forces of attraction between these molecules are not very strong. Since these compounds are
largely nonconductors of electricity, we can conclude that the bonding in these compounds does
not give rise to any ions..
(B) 0.815Solution(B) 0.815.pdf
It\'s due to the reverse of the formation reaction of CdS CdCl2 + H2S(g) <--> CdS
+ 2HCl(aq) if we make the HCl strong enough we can force the equilibrium the other way, and
CdCl2 is soluble.
Solution
It\'s due to the reverse of the formation reaction of CdS CdCl2 + H2S(g) <--> CdS
+ 2HCl(aq) if we make the HCl strong enough we can force the equilibrium the other way, and
CdCl2 is soluble..
Suppose AFnSolution Suppose AFn.pdf
Has Bond in the product that are weaker than the reactants
Solution
Has Bond in the product that are weaker than the reactants.
C code on linked list #include stdio.h #include stdlib.h.pdf
The C code defines functions to create and manipulate a linked list. It takes 5 user-input values, inserts them into an empty linked list, prints the list, takes another input value and inserts it, then prints the final list.
Definition of Log-Normal DistributionA statistical distr.pdf
e =mc^2. m is the defected mass, or mass that isn\'t there and is instead used as
binding energy. For Mg, you have 25 nucleons: 12 protons and 13 neutrons. So your expected
mass would be 1.007825x12 + 1.008665x13 = 25.206545. The defected mass is the supposed
mass minus the actual mass. so its 25.206545-24.985839= .220706 amu. Then you have to
convert this into kilograms. so, divide 220706 amu by 1 kg = 6.022 x 10^26 and you get:
3.664996*10^-28
Solution
e =mc^2. m is the defected mass, or mass that isn\'t there and is instead used as
binding energy. For Mg, you have 25 nucleons: 12 protons and 13 neutrons. So your expected
mass would be 1.007825x12 + 1.008665x13 = 25.206545. The defected mass is the supposed
mass minus the actual mass. so its 25.206545-24.985839= .220706 amu. Then you have to
convert this into kilograms. so, divide 220706 amu by 1 kg = 6.022 x 10^26 and you get:
3.664996*10^-28.
Step1 NaOH (aq) ----- Na(+)(aq) + Cl(-)(aq) Ste.pdf
calculate the mol of NaOH so mol= concentration * volume n=C x V n= 0.175 x
0.0206 = 0.0036 then you get the ratio of acid to base from the equation above which is 1:1 so
the number of mol of base is equal to the number of mol of acid n(acid)=0.0036 the
concentration of that which is C=n/V C= 0.0036/0.025 C=0.144M
Solution
calculate the mol of NaOH so mol= concentration * volume n=C x V n= 0.175 x
0.0206 = 0.0036 then you get the ratio of acid to base from the equation above which is 1:1 so
the number of mol of base is equal to the number of mol of acid n(acid)=0.0036 the
concentration of that which is C=n/V C= 0.0036/0.025 C=0.144M.
The compounds with low oxidation states (O.S.) be.pdf
The oxidation states of p-block main group elements like groups 13, 14, 15 and 16 become more stable at lower oxidation states further down the column. For group 14 elements C, Si, and Ge, the +4 oxidation state is most stable while the +2 state requires kinetic stabilization. For Sn, both +2 and +4 states are important but +4 is more common, and for Pb, the +2 state is most stable due to the inert pair effect caused by relativistic contraction of the s orbital. For transition metals, higher oxidation states also become less favorable down the column as ionization energies increase.
A) ionic compounds generally formed between the c.pdf
A) ionic compounds generally formed between the combination of 1st and 2nd
group elements with 7th group elements . ex: NaCl, MgF, CaCl2 and so on B) Molecular
compounds means vanderwall forces acting btw the molecule these may be crystals or liguids .ex
Iodine(I2) is molecular crystal ,Quartz, phosphorous(P4), sulphur(S8) C) hydrocarbons in
general contain carbon and hydrogen in some cases hetero atoms may present in the compound .
ex CH4 , C2H6, C2H2, C6H5N02, C6H12O6 (glucose) etc D) in general 5th, 6th, 7th groups
oxides , oxy acids behaves as acid . ex: N20, N205, F20,HNO3, HClO4, HCl, HF, H2SO4,
H2S208 etc along with these metal with lower oxidation state can also act as acidic compound
ex: MnO MnO2 MnO3 in the above three oxides oxidation states of metals in orderly +2,+4,+6.
so first compound can act as an acid last one is a base
Solution
A) ionic compounds generally formed between the combination of 1st and 2nd
group elements with 7th group elements . ex: NaCl, MgF, CaCl2 and so on B) Molecular
compounds means vanderwall forces acting btw the molecule these may be crystals or liguids .ex
Iodine(I2) is molecular crystal ,Quartz, phosphorous(P4), sulphur(S8) C) hydrocarbons in
general contain carbon and hydrogen in some cases hetero atoms may present in the compound .
ex CH4 , C2H6, C2H2, C6H5N02, C6H12O6 (glucose) etc D) in general 5th, 6th, 7th groups
oxides , oxy acids behaves as acid . ex: N20, N205, F20,HNO3, HClO4, HCl, HF, H2SO4,
H2S208 etc along with these metal with lower oxidation state can also act as acidic compound
ex: MnO MnO2 MnO3 in the above three oxides oxidation states of metals in orderly +2,+4,+6.
so first compound can act as an acid last one is a base.
A ethers ethers contain R-O-R linkage not carbony.pdf
A ethers ethers contain R-O-R linkage not carbonyl
Solution
A ethers ethers contain R-O-R linkage not carbonyl.
The fact that a molecule vibrates does not in its.pdf
The fact that a molecule vibrates does not in itself insure that the molecule will
exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the
vibrational motion associated with that mode must produce a change in the dipole moment of the
molecule. HCl, for example, with a center of positive charge at the H atom and a center of
negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment
changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the
vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further,
stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared
inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many
molecules that, although possessing no permanent dipole moment, still undergo vibrations that
cause changes in the value of the dipole moment from 0 to some non-zero value. An example is
CO2, which has no permanent dipole moment because the individual bond dipoles exactly
cancel. However, when CO2 undergoes a bending vibration, its dipole moment changes from
zero to some non-zero value. This vibration produces a change in dipole moment and is IR
active. These vibrational modes are responsible for the \"greenhouse\" effect in which heat
radiated from the earth is absorbed (trapped) by CO2 molecules in the atmosphere. The arrows
indicate the directions of motion. Vibrations labeled A and B represent the stretching of the
chemical bonds, one in a symmetric (A) fashion, in which both C=O bonds lengthen and contract
together (in-phase), and the other in an asymmetric (B) fashion, in which one bond shortens
while the other lengthens. The asymmetric stretch (B) is infrared active because there is a change
in the molecular dipole moment during this vibration.
Solution
The fact that a molecule vibrates does not in itself insure that the molecule will
exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the
vibrational motion associated with that mode must produce a change in the dipole moment of the
molecule. HCl, for example, with a center of positive charge at the H atom and a center of
negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment
changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the
vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further,
stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared
inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many
molecules that, although possessing no permanent dipole moment, still undergo vibrations that
cause changes in the value of the dipole moment from 0 to some non-zero value. An example is
CO2, which has no permanent dipole moment because the individual bond dipo.
sucrose has a formula of C12H22O11while ammonia h.pdf
sucrose has a formula of C12H22O11while ammonia has a formula of NH3. NH3 is
polar and has anelectronegative N which can conduct charge.
Solution
sucrose has a formula of C12H22O11while ammonia has a formula of NH3. NH3 is
polar and has anelectronegative N which can conduct charge..
reduction strength Fe Pb As .pdf
The document states that iron has a greater reduction strength than lead, which has a greater reduction strength than arsenic. The reduction strengths from highest to lowest are: Fe > Pb > As.
sp3d since it has 1s 3p and 1d orbitals available.pdf
sp3d since it has 1s 3p and 1d orbitals available for hybridization
Solution
sp3d since it has 1s 3p and 1d orbitals available for hybridization.
No standard potential data given. .pdf
This document does not provide any standard potential data. It simply states that no standard potential data is given in two brief sentences without further context or details.
x2 + 4xSolutionx2 + 4x.pdf
This short document provides the equation x2 + 4x without showing any work or solution steps. It appears to present a quadratic equation but does not give any context or solve the equation.
moles of Al = 1227 Since chlorine is in excess ,.pdf
moles of Al = 12/27 Since chlorine is in excess , Moles of Alcl3 = 12/27 * 1 = 4/9
Solution
moles of Al = 12/27 Since chlorine is in excess , Moles of Alcl3 = 12/27 * 1 = 4/9.
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Step1 NaOH (aq) ----- Na(+)(aq) + Cl(-)(aq) Ste.pdf
The compounds with low oxidation states (O.S.) be.pdf
The compounds with low oxidation states (O.S.) be.pdf
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A) ionic compounds generally formed between the c.pdf
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