Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
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Ka=-log(Ka)=-log(5.6*10^-10)=9.25 According to Henderson Hasselbalch Equation pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6) pH=8.77 (2) NH3 + HCl =======> NH4Cl I 0.2 0.10 0.6 C 0.2-0.1 -0.1 +0.1 E 0.1 0 0.7 pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7) pH=8.40 (3) NH4Cl + NaOH =============> NH3 + NaCl + H2O I 0.6 0.20 0.2 C -0.20 -0.20 +0.20 E 0.4 0 0.4 pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25 Solution Ka=-log(Ka)=-log(5.6*10^-10)=9.25 According to Henderson Hasselbalch Equation pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6) pH=8.77 (2) NH3 + HCl =======> NH4Cl I 0.2 0.10 0.6 C 0.2-0.1 -0.1 +0.1 E 0.1 0 0.7 pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7) pH=8.40 (3) NH4Cl + NaOH =============> NH3 + NaCl + H2O I 0.6 0.20 0.2 C -0.20 -0.20 +0.20 E 0.4 0 0.4 pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25.
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![Ka=-log(Ka)=-log(5.6*10^-10)=9.25
According to Henderson Hasselbalch Equation
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6)
pH=8.77
(2)
NH3 + HCl =======> NH4Cl
I 0.2 0.10 0.6
C 0.2-0.1 -0.1 +0.1
E 0.1 0 0.7
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7)
pH=8.40
(3)
NH4Cl + NaOH =============> NH3 + NaCl + H2O
I 0.6 0.20 0.2
C -0.20 -0.20 +0.20
E 0.4 0 0.4
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25
Solution](https://image.slidesharecdn.com/ka-logka-log5-230409005115-4f420b0d/85/Ka-log-Ka-log-5-610-10-9-25-According-to-Henderson-Hasselb-pdf-1-320.jpg)
![Ka=-log(Ka)=-log(5.6*10^-10)=9.25
According to Henderson Hasselbalch Equation
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6)
pH=8.77
(2)
NH3 + HCl =======> NH4Cl
I 0.2 0.10 0.6
C 0.2-0.1 -0.1 +0.1
E 0.1 0 0.7
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7)
pH=8.40
(3)
NH4Cl + NaOH =============> NH3 + NaCl + H2O
I 0.6 0.20 0.2
C -0.20 -0.20 +0.20
E 0.4 0 0.4
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
TOPIC 18 : ACIDS AND BASES
Dissociation of water
Understanding 18.1 Lewis acids and bases
Understanding 18.2 Calculations involving acids and bases
pKb = 4.7 pOH = 4.7 + log 0.20 0.25 =4.6 pH = 1.pdf
pKb = 4.7 pOH = 4.7 + log 0.20 / 0.25 =4.6 pH = 14 - 4.6 =9.4 moles NH3 = 0.100
L x 0.25 = 0.025 moles NH4+ = 0.100 L x 0.20 = 0.020 moles HCl = 0.0001 L x 12 M =0.0012
NH3 + H+ >> NH4+ moles NH3 = 0.025 - 0.0012 =0.024 moles NH4+ = 0.020 + 0.0012 =0.021
total volume =0.1001 L concentration NH3 = 0.024 / 0.1001 =0.24 M concentration NH4+ =
0.021 / 0.1001 =0.21 M pOH = 4.7 + log 0.21 / 0.24 =4.6 pH = 9.4
Solution
pKb = 4.7 pOH = 4.7 + log 0.20 / 0.25 =4.6 pH = 14 - 4.6 =9.4 moles NH3 = 0.100
L x 0.25 = 0.025 moles NH4+ = 0.100 L x 0.20 = 0.020 moles HCl = 0.0001 L x 12 M =0.0012
NH3 + H+ >> NH4+ moles NH3 = 0.025 - 0.0012 =0.024 moles NH4+ = 0.020 + 0.0012 =0.021
total volume =0.1001 L concentration NH3 = 0.024 / 0.1001 =0.24 M concentration NH4+ =
0.021 / 0.1001 =0.21 M pOH = 4.7 + log 0.21 / 0.24 =4.6 pH = 9.4.
Cálculos teóricos de la valoración de acido fuerte mas base fuerte
valoracion acido base fuerte
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5. A water sample with a total alkalinity of 2.00 x 10 molesL has .pdf
5. A water sample with a total alkalinity of 2.00 x 10 moles/L has a pH of 7.00. Compute the
equilibrium concentrations of CO2, HCOs , co,, and OH Note that water and carbon dioxide can
replace carbonic acid in the Kai reaction without changing the value of Kal H2O + CO2 H+ +
HCO, Ka1 = 4.5 x 10-7
Solution
The equilibrim equations involved are:
H2O +CO2 <---->H+ +HCO3- ka1=4.5*10^-7
HCO3- <--->CO32- +H+ ka2=4.8*10^-11
Also given, pH=-log [H+]=7.0
[H+]=10^-7 M ...........(a)
[OH-]=kw/[H+]=10^-14/[H+]=10^-14.............(b)
pH=pka +log [base]/[acid] [henderson-hasselbach equation]
pka2=-log ka2=-log(4.8*10^-11)=10.3
7.0=10.3 +log[CO32-]/[HCO3-]
or, [CO32-]/[HCO3-]=5.012e-4.
or,[CO32-]=(5.012e-4.)[HCO3-]..................(1)
Also, H2O +CO2 <---->H+ +HCO3- ka1=4.5*10^-7=[H+][HCO3-]/[CO2]
for x=[H+]=[HCO3-] (at equilibrium)
[CO2]=[CO2]o-x , [CO2]o=initial concentration,but x is very less for very less dissociation ( ka1
is very small)
[CO2]eq=[CO2]o
4.5*10^-7=[H+][HCO3-]/[CO2]=x^2/[CO2]
or,[CO2]eq=x^2/(4.5*10^-7)=[HCO3-]eq/(4.5*10^-7)
[CO2]eq=[HCO3-]eq^2/(4.5*10^-7) ...................(2)
total alkalnity=[HCO3-] +2[CO32-] +[OH-] -[H+]=2.0*10^-3 mol/L
plugging in eqn (a),(b),(1) & (2) in the equation for total alkalinity,
total alkalnity=[HCO3-] +2(5.012e-4.)[HCO3-]. =2.0*10^-3 mol/L
or, [HCO3-]=1.997*10^-3 mol/L
[CO32-]eq=(5.012e-4.)[HCO3-].=(5.012e-4.)(2.0*10^-3 mol/L)=1.002*10^-6 mol/l
[H+]eq=[OH-]eq=10^-7 mol/L
[CO2]=[HCO3-]eq^2/(4.5*10^-7) =(1.997*10^-3 mol/L)^2/(4.5*10^-7)=8.862 mol/L
[CO2]eq=8.862*10^3 mol/L.
a. 0.010 M NaCl pH will be approximately 7 because the solution.pdf
a. 0.010 M NaCl
pH will be approximately 7 because the solution will be neutral
b. 0.010 M NaOH
0.01 M NaOH
pOH= -log(0.01) =2
pH +pOH =14
pH = 12
c. 0.010 M NH3
NH3 + H2O NH4(+) + OH-
Kb = 1.8*10^-5 = [X]2/(1 - X)
the x in denominator can be ignored as the value is much smaller than 1
1.8*10^-5 = [X]2
[OH]=[X]=4.243*10^-3
pOH = -log[OH]=2.37
pH +pOH =14
pH = 11.63
d. 0.010 M malonic acid
pka1=2.85,pka2=5.70
concentration doesnot matter for diprotic acids
pH= (pka1 + pka2) /2
pH=(2.85+5.70)/2=4.275
Solution
a. 0.010 M NaCl
pH will be approximately 7 because the solution will be neutral
b. 0.010 M NaOH
0.01 M NaOH
pOH= -log(0.01) =2
pH +pOH =14
pH = 12
c. 0.010 M NH3
NH3 + H2O NH4(+) + OH-
Kb = 1.8*10^-5 = [X]2/(1 - X)
the x in denominator can be ignored as the value is much smaller than 1
1.8*10^-5 = [X]2
[OH]=[X]=4.243*10^-3
pOH = -log[OH]=2.37
pH +pOH =14
pH = 11.63
d. 0.010 M malonic acid
pka1=2.85,pka2=5.70
concentration doesnot matter for diprotic acids
pH= (pka1 + pka2) /2
pH=(2.85+5.70)/2=4.275.
Recommended
TOPIC 18 : ACIDS AND BASES
Dissociation of water
Understanding 18.1 Lewis acids and bases
Understanding 18.2 Calculations involving acids and bases
pKb = 4.7 pOH = 4.7 + log 0.20 0.25 =4.6 pH = 1.pdf
pKb = 4.7 pOH = 4.7 + log 0.20 / 0.25 =4.6 pH = 14 - 4.6 =9.4 moles NH3 = 0.100
L x 0.25 = 0.025 moles NH4+ = 0.100 L x 0.20 = 0.020 moles HCl = 0.0001 L x 12 M =0.0012
NH3 + H+ >> NH4+ moles NH3 = 0.025 - 0.0012 =0.024 moles NH4+ = 0.020 + 0.0012 =0.021
total volume =0.1001 L concentration NH3 = 0.024 / 0.1001 =0.24 M concentration NH4+ =
0.021 / 0.1001 =0.21 M pOH = 4.7 + log 0.21 / 0.24 =4.6 pH = 9.4
Solution
pKb = 4.7 pOH = 4.7 + log 0.20 / 0.25 =4.6 pH = 14 - 4.6 =9.4 moles NH3 = 0.100
L x 0.25 = 0.025 moles NH4+ = 0.100 L x 0.20 = 0.020 moles HCl = 0.0001 L x 12 M =0.0012
NH3 + H+ >> NH4+ moles NH3 = 0.025 - 0.0012 =0.024 moles NH4+ = 0.020 + 0.0012 =0.021
total volume =0.1001 L concentration NH3 = 0.024 / 0.1001 =0.24 M concentration NH4+ =
0.021 / 0.1001 =0.21 M pOH = 4.7 + log 0.21 / 0.24 =4.6 pH = 9.4.
Cálculos teóricos de la valoración de acido fuerte mas base fuerte
valoracion acido base fuerte
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5. A water sample with a total alkalinity of 2.00 x 10 molesL has .pdf
5. A water sample with a total alkalinity of 2.00 x 10 moles/L has a pH of 7.00. Compute the
equilibrium concentrations of CO2, HCOs , co,, and OH Note that water and carbon dioxide can
replace carbonic acid in the Kai reaction without changing the value of Kal H2O + CO2 H+ +
HCO, Ka1 = 4.5 x 10-7
Solution
The equilibrim equations involved are:
H2O +CO2 <---->H+ +HCO3- ka1=4.5*10^-7
HCO3- <--->CO32- +H+ ka2=4.8*10^-11
Also given, pH=-log [H+]=7.0
[H+]=10^-7 M ...........(a)
[OH-]=kw/[H+]=10^-14/[H+]=10^-14.............(b)
pH=pka +log [base]/[acid] [henderson-hasselbach equation]
pka2=-log ka2=-log(4.8*10^-11)=10.3
7.0=10.3 +log[CO32-]/[HCO3-]
or, [CO32-]/[HCO3-]=5.012e-4.
or,[CO32-]=(5.012e-4.)[HCO3-]..................(1)
Also, H2O +CO2 <---->H+ +HCO3- ka1=4.5*10^-7=[H+][HCO3-]/[CO2]
for x=[H+]=[HCO3-] (at equilibrium)
[CO2]=[CO2]o-x , [CO2]o=initial concentration,but x is very less for very less dissociation ( ka1
is very small)
[CO2]eq=[CO2]o
4.5*10^-7=[H+][HCO3-]/[CO2]=x^2/[CO2]
or,[CO2]eq=x^2/(4.5*10^-7)=[HCO3-]eq/(4.5*10^-7)
[CO2]eq=[HCO3-]eq^2/(4.5*10^-7) ...................(2)
total alkalnity=[HCO3-] +2[CO32-] +[OH-] -[H+]=2.0*10^-3 mol/L
plugging in eqn (a),(b),(1) & (2) in the equation for total alkalinity,
total alkalnity=[HCO3-] +2(5.012e-4.)[HCO3-]. =2.0*10^-3 mol/L
or, [HCO3-]=1.997*10^-3 mol/L
[CO32-]eq=(5.012e-4.)[HCO3-].=(5.012e-4.)(2.0*10^-3 mol/L)=1.002*10^-6 mol/l
[H+]eq=[OH-]eq=10^-7 mol/L
[CO2]=[HCO3-]eq^2/(4.5*10^-7) =(1.997*10^-3 mol/L)^2/(4.5*10^-7)=8.862 mol/L
[CO2]eq=8.862*10^3 mol/L.
a. 0.010 M NaCl pH will be approximately 7 because the solution.pdf
a. 0.010 M NaCl
pH will be approximately 7 because the solution will be neutral
b. 0.010 M NaOH
0.01 M NaOH
pOH= -log(0.01) =2
pH +pOH =14
pH = 12
c. 0.010 M NH3
NH3 + H2O NH4(+) + OH-
Kb = 1.8*10^-5 = [X]2/(1 - X)
the x in denominator can be ignored as the value is much smaller than 1
1.8*10^-5 = [X]2
[OH]=[X]=4.243*10^-3
pOH = -log[OH]=2.37
pH +pOH =14
pH = 11.63
d. 0.010 M malonic acid
pka1=2.85,pka2=5.70
concentration doesnot matter for diprotic acids
pH= (pka1 + pka2) /2
pH=(2.85+5.70)/2=4.275
Solution
a. 0.010 M NaCl
pH will be approximately 7 because the solution will be neutral
b. 0.010 M NaOH
0.01 M NaOH
pOH= -log(0.01) =2
pH +pOH =14
pH = 12
c. 0.010 M NH3
NH3 + H2O NH4(+) + OH-
Kb = 1.8*10^-5 = [X]2/(1 - X)
the x in denominator can be ignored as the value is much smaller than 1
1.8*10^-5 = [X]2
[OH]=[X]=4.243*10^-3
pOH = -log[OH]=2.37
pH +pOH =14
pH = 11.63
d. 0.010 M malonic acid
pka1=2.85,pka2=5.70
concentration doesnot matter for diprotic acids
pH= (pka1 + pka2) /2
pH=(2.85+5.70)/2=4.275.
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Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
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B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
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Anatomy:
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B. Stem:
1. Primary structure: Transverse structure shows ridges and grooves. cortex clorenchyma
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2. Secondary structure: Parenchymalous cells present between the vascular bundles to form a
ring of cambium, ray initials and fusiform initial are present.
C. Leaves: Transverse structure of leaves shows epidermis, cuticle, xylem, phloem, air space,
spongy parenchyma tissues. The vascular bundles are collateral closed and primary phloem and
primary xylem are also present.
Adaptations:
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branched scrubs, commonly climbing over vegetation many species spread by rhizomes Ephedra
Trifura. Under favorable condition it grows tree like habitat.
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In pharmaceutical industries it works as to access Ephedra drug materials.
Traditionally it is used by indigenous people for medicinal treatments of asthma, hay fever and
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Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
A.
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public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate, double thrust)
{
currentFuelFlowRate=0.0;
verticalSpeed=0.0;
altitude=al;
amountOfFuel=fuel;
massOfTheLander=mass;
maxFuelConsumptionRate=rate;
maxThrust=thrust;
}
/* functions to get the values of instance variables*/
public double getCurrentFuelFlowRate()
{
return currentFuelFlowRate;
}
public double getVerticalSpeed()
{
return verticalSpeed;
}
public double getAltitude()
{
return altitude;
}
public double getAmountOfFuel()
{
return amountOfFuel;
}
public double getMassOfTheLander()
{
return massOfTheLander;
}
public double getMaxFuelConsumptionRate()
{
return maxFuelConsumptionRate;
}
public double getMaxThrust()
{
return maxThrust;
}
/* updates the value of currentFuelFlowRate*/
public void setCurrentFuelFlowRate(double rate)
{
currentFuelFlowRate=rate;
}
/* simulating the passage for a small amount of time t*/
public void setPassage(double t)
{
if(currentFuelFlowRate>0)
{
if(amountOfFuel==0)
{
currentFuelFlowRate=0;
}
}
/* velocity is the verticalSpeed*/
double f=maxThrust*currentFuelFlowRate;
double m=massOfTheLander;
verticalSpeed=t*((f/m)-1.62);
//v is verticalSpeed
altitude=t*verticalSpeed;
//ship is landed
if(altitude<0)
{
altitude=0;
verticalSpeed=0;
}
//r is currentFuelFlowRate
//c is maxFuelConsumptionRate
double changeInRemainFuel=t*currentFuelFlowRate*maxFuelConsumptionRate;
amountOfFuel=amountOfFuel-changeInRemainFuel;
if(amountOfFuel<0)
{
amountOfFuel=0;
}
}
}
////////////////////////////////////////////////////////
public class LunarLanderDemo {
public static void main(String[] args) {
LunarLander L=new LunarLander(1000,1700,900,10,5000);
/*give values between 0.0 , 1.0*/
L.setCurrentFuelFlowRate(0.6);
L.setPassage(0.1);
System.out.println(\"Simulation of Lunar Lander Passage at 0.1 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
L.setCurrentFuelFlowRate(0.3);
L.setPassage(0.09);
System.out.println(\"Simulation of Lunar Lander Passage at 0.09 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
}
}
Solution
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate.
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Cooling. The reaction is giving off energy(heat) so you need to control this in some
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Similar to Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
Tabla de Ka y pKa
Tabla de Ka y pKa nos muestra acidos con sus respectivas constantes de acides y pka's. En la tabla se ve el nombre del acido, su formula quimica, su acido conjugado su base conjugada, y la fuerza de estos.
IB Chemistry on Titration Curves between Acids and Bases
IB Chemistry on Titration Curves between Acids and Bases
IB Chemistry on Titration Curves between Acids and Bases
IB Chemistry on Titration Curves between Acids and Bases
Cálculos teóricos de la valoración de acido debil mas base fuerte
Cálculos teóricos de la valoración de acido debil mas base fuerte
Similar to Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf (10)
IB Chemistry on Titration Curves between Acids and Bases
IB Chemistry on Titration Curves between Acids and Bases
IB Chemistry on Titration Curves between Acids and Bases
IB Chemistry on Titration Curves between Acids and Bases
Cálculos teóricos de la valoración de acido debil mas base fuerte
Cálculos teóricos de la valoración de acido debil mas base fuerte
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O = O bond is shortest Oxygen electronegativity d.pdf
O = O bond is shortest Oxygen electronegativity difference is higher and also it is
double bond
Solution
O = O bond is shortest Oxygen electronegativity difference is higher and also it is
double bond.
NaCl Na is the symbol for Sodium which bonds to .pdf
NaCl Na is the symbol for Sodium which bonds to Cl which is the symbol for
Chlorine making Sodium Chloride.
Solution
NaCl Na is the symbol for Sodium which bonds to Cl which is the symbol for
Chlorine making Sodium Chloride..
NiCl2 because it not dissociate .pdf
NiCl2 because it not dissociate
Solution
NiCl2 because it not dissociate.
iodide, I-, has the -1 oxidation state. iodine (I.pdf
iodide, I-, has the -1 oxidation state. iodine (I) has 0 oxidation state. I- - e ---> I
Thus, this change is an oxidation reaction.
Solution
iodide, I-, has the -1 oxidation state. iodine (I) has 0 oxidation state. I- - e ---> I
Thus, this change is an oxidation reaction..
Lithium is in group 1 on the peridic table. This .pdf
Lithium is in group 1 on the peridic table. This means that it has only 1 valence
electron. Lithium will tend to lose that electron when it ionizes and become an ion. Remeber, an
ion is any atom, or sometimes molecule, with a charge. When Li loses the electron, it loses one
of its negative charges so the atom becomes an ion with a +1 charge because it now has 3
positively charged protons and only 2 negatively charged electrons.
Solution
Lithium is in group 1 on the peridic table. This means that it has only 1 valence
electron. Lithium will tend to lose that electron when it ionizes and become an ion. Remeber, an
ion is any atom, or sometimes molecule, with a charge. When Li loses the electron, it loses one
of its negative charges so the atom becomes an ion with a +1 charge because it now has 3
positively charged protons and only 2 negatively charged electrons..
Solution1.cpp#include iostreamheader for input output function.pdf
Solution
1.cpp
#include //header for input output function
using namespace std;//it tells the compiler to link std name space
int main()
{//main function
int n=1,m=-1;
if(n<-m)//if else statement
{
cout<.
Systematic positionKingdom PlantaeSubkingdom Tracheobionta (Vas.pdf
Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
Anatomy:
A. Root : It shows usually Gymnosperm structure. Cortex, endoderm, inner periderm, vascular
camblum, primary and secondary xylem was present.
B. Stem:
1. Primary structure: Transverse structure shows ridges and grooves. cortex clorenchyma
elongated, endoderm is present in single layer.
2. Secondary structure: Parenchymalous cells present between the vascular bundles to form a
ring of cambium, ray initials and fusiform initial are present.
C. Leaves: Transverse structure of leaves shows epidermis, cuticle, xylem, phloem, air space,
spongy parenchyma tissues. The vascular bundles are collateral closed and primary phloem and
primary xylem are also present.
Adaptations:
Mostly xerophytes scrubs or woody climbers Ephedra Pedunculata. Most species of ephedra are
branched scrubs, commonly climbing over vegetation many species spread by rhizomes Ephedra
Trifura. Under favorable condition it grows tree like habitat.
Ephedra Foliata - scrambling scrub, Ephedra Compacta - Branching type, Ephedra
Camphylophoda - Spendulus branches, Ephedra Intermedia - Shoot, Ephedra Saxatilis - scrub.
Values and functions:
In pharmaceutical industries it works as to access Ephedra drug materials.
Traditionally it is used by indigenous people for medicinal treatments of asthma, hay fever and
cold.
Alkaloids obtained from the species of Ephedra is used in herbal medicine.
Solution
Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
A.
Query to list out all the languages in the table LANGUAGES.SELECT .pdf
Query to list out all the languages in the table LANGUAGES.
SELECT LANGUAGE FROM LANGUAGES
Solution
Query to list out all the languages in the table LANGUAGES.
SELECT LANGUAGE FROM LANGUAGES.
public class LunarLander { double currentFuelFlowRate; d.pdf
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate, double thrust)
{
currentFuelFlowRate=0.0;
verticalSpeed=0.0;
altitude=al;
amountOfFuel=fuel;
massOfTheLander=mass;
maxFuelConsumptionRate=rate;
maxThrust=thrust;
}
/* functions to get the values of instance variables*/
public double getCurrentFuelFlowRate()
{
return currentFuelFlowRate;
}
public double getVerticalSpeed()
{
return verticalSpeed;
}
public double getAltitude()
{
return altitude;
}
public double getAmountOfFuel()
{
return amountOfFuel;
}
public double getMassOfTheLander()
{
return massOfTheLander;
}
public double getMaxFuelConsumptionRate()
{
return maxFuelConsumptionRate;
}
public double getMaxThrust()
{
return maxThrust;
}
/* updates the value of currentFuelFlowRate*/
public void setCurrentFuelFlowRate(double rate)
{
currentFuelFlowRate=rate;
}
/* simulating the passage for a small amount of time t*/
public void setPassage(double t)
{
if(currentFuelFlowRate>0)
{
if(amountOfFuel==0)
{
currentFuelFlowRate=0;
}
}
/* velocity is the verticalSpeed*/
double f=maxThrust*currentFuelFlowRate;
double m=massOfTheLander;
verticalSpeed=t*((f/m)-1.62);
//v is verticalSpeed
altitude=t*verticalSpeed;
//ship is landed
if(altitude<0)
{
altitude=0;
verticalSpeed=0;
}
//r is currentFuelFlowRate
//c is maxFuelConsumptionRate
double changeInRemainFuel=t*currentFuelFlowRate*maxFuelConsumptionRate;
amountOfFuel=amountOfFuel-changeInRemainFuel;
if(amountOfFuel<0)
{
amountOfFuel=0;
}
}
}
////////////////////////////////////////////////////////
public class LunarLanderDemo {
public static void main(String[] args) {
LunarLander L=new LunarLander(1000,1700,900,10,5000);
/*give values between 0.0 , 1.0*/
L.setCurrentFuelFlowRate(0.6);
L.setPassage(0.1);
System.out.println(\"Simulation of Lunar Lander Passage at 0.1 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
L.setCurrentFuelFlowRate(0.3);
L.setPassage(0.09);
System.out.println(\"Simulation of Lunar Lander Passage at 0.09 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
}
}
Solution
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate.
Cooling. The reaction is giving off energy(heat) .pdf
Cooling. The reaction is giving off energy(heat) so you need to control this in some
cases or it can be problematic
Solution
Cooling. The reaction is giving off energy(heat) so you need to control this in some
cases or it can be problematic.
php global $bsize,$playerToken,$myToken,$gameOver,$winArr,$rowAr.pdf
Tic Tac Toe in PHP\';
for($i=0; $i<$size; $i++) {
for($j=0; $j<$size; $j++) {
$count++;
$retVal.=\'\';
}
$retVal.=\'
\';
}
$retVal.=\'\';
echo $retVal;
}
function genBox2($size,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$count=0;
$retVal=\'\';
for($i=0; $i<$size; $i++) {
for($j=0; $j<$size; $j++) {
$count++;
$retVal.=\'\';
}
$retVal.=\'
\';
}
$retVal.=\'\';
echo $retVal;
}
function isEmpty($who) {
if($who==\"\")
return 1;
else
return 0;
}
function move($bsize,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$count=0;
$maxCount=0;
$pos=0;
$retVal=0;
# Build Row Array
for($i=0; $i<$bsize; $i++) {
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$count++;
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
}
$rowArr[$i]=$maxCount;
if($fullCounter==$bsize)
$rowArr[$i]=-1;
}
# Building Column Array
for($i=0; $i<$bsize; $i++) {
$count=$i+1;
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
$count+=$bsize;
}
$colArr[$i]=$maxCount;
if($fullCounter==$bsize)
$colArr[$i]=-1;
}
# Building Diagonal Array
for($i=0; $i<2; $i++) {
if($i==0)
$count=$i+1;
else
$count=$bsize;
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
if($i==0)
$count+=$bsize+1;
else
$count+=$bsize-1;
}
$digArr[$i]=$maxCount;
if($fullCounter==$bsize)
$digArr[$i]=-1;
}
# Finding Max Values
$maxRow=myMax(0,$bsize,\"row\",$rowArr);
$maxCol=myMax(0,$bsize,\"col\",$colArr);
$maxDig=myMax(0,$bsize,\"dig\",$digArr);
$maxArrs[0]=myMax(1,$bsize,\"row\",$rowArr);
$maxArrs[1]=myMax(1,$bsize,\"col\",$colArr);
$maxArrs[2]=myMax(1,$bsize,\"dig\",$digArr);
//$maxArrs=array(max(1,$bsize,\"row\",$rowArr),max(1,$bsize,\"col\",$colArr),max(1,$bsize,\"
dig\",$digArr));
if(myMax(0,$bsize,\"x\",$maxArrs)==0)
$pos=$bsize*($maxRow+1)-$bsize;
if(myMax(0,$bsize,\"x\",$maxArrs)==1)
$pos=$maxCol;
if(myMax(0,$bsize,\"x\",$maxArrs)==2)
if($maxDig==0)
$pos=$maxDig;
else
$pos=$bsize-1;
$retFlag=0;
for($y=0; $y<$bsize; $y++) {
if(!$retFlag) {
if($arr[$pos]==\"\") {
$retVal=$pos;
$retFlag=1;
}
if(myMax(0,$bsize,\"x\",$maxArrs)==0)
$pos++;
if(myMax(0,$bsize,\"x\",$maxArrs)==1)
$pos+=$bsize;
if(myMax(0,$bsize,\"x\",$maxArrs)==2)
if($maxDig==0)
$pos+=$bsize+1;
else
$pos+=$bsize-1;
}
}
return $retVal;
}
function myMax($what,$bsize,$type,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$max=-1;
$maxIndex=-1;
if($type!=\"dig\") {
for($i=0; $i<$bsize; $i++) {
if($arr[$i]>$max) {
$max=$arr[$i];
$maxIndex=$i;
}
}
}
if($type==\"dig\") {
for($i=0; $i<2; $i++) {
if($arr[$i]>$max) {
$ma.
c. cytochrome C It is the enzyme involved in the .pdf
c. cytochrome C It is the enzyme involved in the reaction, so cannot exchange net
electrons
Solution
c. cytochrome C It is the enzyme involved in the reaction, so cannot exchange net
electrons.
C is wrong 14 7N contains 7 protons and 7 neutron.pdf
C is wrong 14 7N contains 7 protons and 7 neutrons , N/P = 1 which lies in belt of
stability
Solution
C is wrong 14 7N contains 7 protons and 7 neutrons , N/P = 1 which lies in belt of
stability.
1.The following are the main types of service providers to a mutua.pdf
1.
The following are the main types of service providers to a mutual fund:
Service provider refers to the sponsor of funds. From the public who are ready to invest, the fund
sponsor raises funds. The fund sponsor is one among the following:
The duty of the sponsor is to invest the funds of the investor in stocks, bonds, and shares
according to the preference of the investors.
If the investors demand for 100% safety of the invested funds, the fund sponsor invests in such
shares or bonds. If the investor is ready to meet risks even to 100%, then the fund sponsor will
invest accordingly. For this service, the investor has to pay the fund sponsor up to a certain limit.
2.
Determining the fees of mutual fund advisor:
There are certain restrictions that are imposed by SEBI. An advisor has to follow the same in
order to claim his fees. Mutual fund distributors are generally exempted from being registered as
investment advisors.
For a no-load large-cap fund, the percentage of fees is 0.95 whereas for small cap funds it is
1.20%. Therefore, if the advisor charges more than the specified percentage, the shareholders can
challenge that the fees is excessive.
3.
Differences between mutual funds and closed-end funds:
One of the major differences between the mutual funds and the closed-end funds is in the date of
maturity. A mutual fund can be withdrawn whenever the funds are required. However, it is not
the case in closed-end funds. In closed-end funds, the funds can be taken only at its date of
maturity. Some funds can be taken before the date of maturity. However, the rate of interest will
either be very little or even nil.
The closed-end funds have declined in popularity because of the returns and the maturity date.
Investing in this leads to overtrading and poor returns.
4.
Differences between ETFs and UITs from index mutual funds:
ETF is similar to closed-end funds. They trade closer to NAVs and therefore can be created and
retired whenever required. Mostly ETFs are passively managed with the help of indexes.
However, there are actively managed investments too. They are allowed to sell short and buy on
margin. They are even allowed to purchase a single share.
UITs are sold by investment advisors rather than purchasing from a secondary market. This does
not have a board of directors and is registered with Securities and Exchange Board. This
portfolio has different types of securities too.
Index mutual fund is a type of mutual fund in which a sale and buy index is set. When the rate
falls down to the set index, the funds are sold/bought on its own.
Solution
1.
The following are the main types of service providers to a mutual fund:
Service provider refers to the sponsor of funds. From the public who are ready to invest, the fund
sponsor raises funds. The fund sponsor is one among the following:
The duty of the sponsor is to invest the funds of the investor in stocks, bonds, and shares
according to the preference of the investors.
If the investors demand.
D. planned and directed means to improve the functioning of the clie.pdf
D. planned and directed means to improve the functioning of the client system
Solution
D. planned and directed means to improve the functioning of the client system.
domain - -x-1 0 = x+10 =x-1i.e. (-inf,-1)range-(-in.pdf
domain -
-x-1> 0
=> x+1<0
=>x<-1
i.e. (-inf,-1)
range-
(-inf,+inf)
Solution
domain -
-x-1> 0
=> x+1<0
=>x<-1
i.e. (-inf,-1)
range-
(-inf,+inf).
C. Keystroke loggers are a form of malware that records every key th.pdf
C. Keystroke loggers are a form of malware that records every key the computer user presses to
harvest usernames, passwords, social security numbers, and other confidential information and
then send that data to the person who caused the malware installation.
Solution
C. Keystroke loggers are a form of malware that records every key the computer user presses to
harvest usernames, passwords, social security numbers, and other confidential information and
then send that data to the person who caused the malware installation..
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Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
- 1. Ka=-log(Ka)=-log(5.6*10^-10)=9.25 According to Henderson Hasselbalch Equation pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6) pH=8.77 (2) NH3 + HCl =======> NH4Cl I 0.2 0.10 0.6 C 0.2-0.1 -0.1 +0.1 E 0.1 0 0.7 pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7) pH=8.40 (3) NH4Cl + NaOH =============> NH3 + NaCl + H2O I 0.6 0.20 0.2 C -0.20 -0.20 +0.20 E 0.4 0 0.4 pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25 Solution
- 2. Ka=-log(Ka)=-log(5.6*10^-10)=9.25 According to Henderson Hasselbalch Equation pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6) pH=8.77 (2) NH3 + HCl =======> NH4Cl I 0.2 0.10 0.6 C 0.2-0.1 -0.1 +0.1 E 0.1 0 0.7 pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7) pH=8.40 (3) NH4Cl + NaOH =============> NH3 + NaCl + H2O I 0.6 0.20 0.2 C -0.20 -0.20 +0.20 E 0.4 0 0.4 pH = pka + log[salt]/[acid] pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25