16. a). Cellwall and cellulose17. e). a , b & c of the above18. .pdf
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16. a). Cellwall and cellulose 17. e). a , b & c of the above 18. c). In botany, a fruit is the seed-bearing structure in flowering plants (also known as angiosperms) formed from the ovary after flowering. 19. A lichen is an organism that results from a mutualistic relationship between a fungus and a photosynthetic organism. The other organism is usually a cyanobacterium or green alga. The fungus grows around the bacterial or algal cells. The fungus benefits from the constant supply of food produced by the photosynthesizer. 20. Mycorrhizae are symbiotic relationships that form between fungi and plants. The fungi colonize the root system of a host plant, providing increased water and nutrient absorption capabilities while the plant provides the fungus with carbohydrates formed from photosynthesis. Solution 16. a). Cellwall and cellulose 17. e). a , b & c of the above 18. c). In botany, a fruit is the seed-bearing structure in flowering plants (also known as angiosperms) formed from the ovary after flowering. 19. A lichen is an organism that results from a mutualistic relationship between a fungus and a photosynthetic organism. The other organism is usually a cyanobacterium or green alga. The fungus grows around the bacterial or algal cells. The fungus benefits from the constant supply of food produced by the photosynthesizer. 20. Mycorrhizae are symbiotic relationships that form between fungi and plants. The fungi colonize the root system of a host plant, providing increased water and nutrient absorption capabilities while the plant provides the fungus with carbohydrates formed from photosynthesis..
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17 Lecture Ppt
This document summarizes key concepts about protists from Chapter 17:
- Protists are a diverse group of eukaryotic organisms that may represent the oldest eukaryotes. They vary greatly in size, nutrition, and habitat.
- Protists include protozoans like amoebas, flagellates, and ciliates that move using pseudopods, flagella, or cilia. They also include algae, slime molds, and parasitic sporozoans.
- Algae like diatoms, dinoflagellates, red algae, brown algae, and green algae are important primary producers in aquatic ecosystems and sources of food and oxygen. Some are
1. It is widely agreed that the plant kingdom arose from A)Eumycota..pdf
1. It is widely agreed that the plant kingdom arose from A)Eumycota. B)Chrysophyta.
C)Phaeophyta. D)Rhodophyta. E)Chlorophyta.
2. A universal feature of the life cycle of plants is A)morphologically identical haploid and
diploid stages. B)genetically identical haploid and diploid stages. C)alteration of generations
between heteromorphic haploid gametophytes and diploid sporophytes. D)All of the above
E)None of the above
3. Several evolutionary adaptations to land are shared by all plants. These shared adaptations do
not include A)waxy protective coverings. B)support against gravity. C)means of taking up water
from the soil. D)protective structures for the new sporophyte. E)water transport by xylem.
4. Ferns are in which phylum? A) Lycophyta B) Anthocerophyta C) Hepatophyta D) Pterophyta
E) Bryophyta
5. The bryophytes are dependent on water for reproduction because A) sperm are passively
transported to eggs by water. B) gametogenesis only occurs when the plants are moist. C) eggs
and sperm are released into water and then unite. D) sperm must swim through water to reach
and fertilize eggs. E) None of the above
6. All plants produce __________ by mitosis and __________ by meiosis. A)spores, gametes
B)gametes, gametes C)gametes, spores D)spores, spores E)spores, gametes and spores
7. In a heterosporous life cycle, the microspore develops into the _____ gametophyte, while the
megaspore develops into the ______ gametophyte. A)female, male B)male, female C)diploid,
haploid D)haploid, diploid E)None of the above
8. Asexual reproduction in liverworts is accomplished by A)gametophytes. B)spores. C)gemmae.
D)tracheids.
9. You are walking along a roadside and find a plant with the following characteristics: very thin
waxy cuticle, stomata, simple leaves in whorls around a central stem, independent sporophyte
and gametophyte, sporangia in strobili. This plant is most likely a member of which of the
following phyla? A)Bryophyta B)Sphenophyta C)Pterophyta D)Lycophyta
10. Plants differ from algae in that only plants A)are photosynthetic. B)are multicellular.
C)possess chlorophyll. D)have multicellular embryos protected by the parent. E)are eukaryotic.
11. Which statement about the alternation of generations in plants is not true? A) The plant life
cycle is characterized by diploid and haploid forms. B)Meiosis occurs in sporangia. C)Gametes
are always produced by meiosis. D)The zygote is the first cell of the sporophyte generation.
E)The gametophyte and sporophyte differ genetically.
12. The most abundant gymnosperm phylum today is A) Cycadophyta. B) Ginkgophyta. C)
Gnetophyta. D) Coniferophyta. E) None of the above
13. Coniferous gymnosperms, such as pines, depend primarily on __________ for pollination;
thus, the plants produce large quantities of pollen that disperse over large areas during the spring.
A) insects B) birds C) water D) wind E) mammals
14. An evolutionary trend that runs throughout the plant kingdom is that the sporophyte
generation ___ and .
18 Lecture Ppt
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Recommended
17 Lecture Ppt
This document summarizes key concepts about protists from Chapter 17:
- Protists are a diverse group of eukaryotic organisms that may represent the oldest eukaryotes. They vary greatly in size, nutrition, and habitat.
- Protists include protozoans like amoebas, flagellates, and ciliates that move using pseudopods, flagella, or cilia. They also include algae, slime molds, and parasitic sporozoans.
- Algae like diatoms, dinoflagellates, red algae, brown algae, and green algae are important primary producers in aquatic ecosystems and sources of food and oxygen. Some are
1. It is widely agreed that the plant kingdom arose from A)Eumycota..pdf
1. It is widely agreed that the plant kingdom arose from A)Eumycota. B)Chrysophyta.
C)Phaeophyta. D)Rhodophyta. E)Chlorophyta.
2. A universal feature of the life cycle of plants is A)morphologically identical haploid and
diploid stages. B)genetically identical haploid and diploid stages. C)alteration of generations
between heteromorphic haploid gametophytes and diploid sporophytes. D)All of the above
E)None of the above
3. Several evolutionary adaptations to land are shared by all plants. These shared adaptations do
not include A)waxy protective coverings. B)support against gravity. C)means of taking up water
from the soil. D)protective structures for the new sporophyte. E)water transport by xylem.
4. Ferns are in which phylum? A) Lycophyta B) Anthocerophyta C) Hepatophyta D) Pterophyta
E) Bryophyta
5. The bryophytes are dependent on water for reproduction because A) sperm are passively
transported to eggs by water. B) gametogenesis only occurs when the plants are moist. C) eggs
and sperm are released into water and then unite. D) sperm must swim through water to reach
and fertilize eggs. E) None of the above
6. All plants produce __________ by mitosis and __________ by meiosis. A)spores, gametes
B)gametes, gametes C)gametes, spores D)spores, spores E)spores, gametes and spores
7. In a heterosporous life cycle, the microspore develops into the _____ gametophyte, while the
megaspore develops into the ______ gametophyte. A)female, male B)male, female C)diploid,
haploid D)haploid, diploid E)None of the above
8. Asexual reproduction in liverworts is accomplished by A)gametophytes. B)spores. C)gemmae.
D)tracheids.
9. You are walking along a roadside and find a plant with the following characteristics: very thin
waxy cuticle, stomata, simple leaves in whorls around a central stem, independent sporophyte
and gametophyte, sporangia in strobili. This plant is most likely a member of which of the
following phyla? A)Bryophyta B)Sphenophyta C)Pterophyta D)Lycophyta
10. Plants differ from algae in that only plants A)are photosynthetic. B)are multicellular.
C)possess chlorophyll. D)have multicellular embryos protected by the parent. E)are eukaryotic.
11. Which statement about the alternation of generations in plants is not true? A) The plant life
cycle is characterized by diploid and haploid forms. B)Meiosis occurs in sporangia. C)Gametes
are always produced by meiosis. D)The zygote is the first cell of the sporophyte generation.
E)The gametophyte and sporophyte differ genetically.
12. The most abundant gymnosperm phylum today is A) Cycadophyta. B) Ginkgophyta. C)
Gnetophyta. D) Coniferophyta. E) None of the above
13. Coniferous gymnosperms, such as pines, depend primarily on __________ for pollination;
thus, the plants produce large quantities of pollen that disperse over large areas during the spring.
A) insects B) birds C) water D) wind E) mammals
14. An evolutionary trend that runs throughout the plant kingdom is that the sporophyte
generation ___ and .
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Classifying the Symbiotic Relationships
Do you know that a tiny organism living inside you helps keep you alive? That organism is E. coli, a bacterium that lives in your intestines. E. coli provides you with vitamins; you provide E. coli with food. This relationship is called
mutualism
, because both you and E. coli benefit.
Directions:
Read the following descriptions of each relationship. Then complete columns two and three of the table. In column two, state whether
each
organism benefits, suffers, or is not affected. In column three, classify the relationships as mutualism, commensalism, or parasitism.
A. Orchid vines grow up the trunks of trees. The areas at the tops of trees have more light than the areas underneath trees. The trees are not affected by the orchids.
B. A type of bacteria that lives on the roots of bean and pea plants converts nitrogen gas in the air into nitrogen compounds that the plants need for growth.
C. A lamprey eel fastens itself to a host fish, such as lake perch, and feeds on it. When the fish dies, the lamprey finds another host.
D. A remora attaches itself to sharks by a rough, flattened fin on the top of its head. It feeds on shreds of the sharks’ prey.
E. Fungal hyphae live on the roots of plants. The fungal hyphae improve the nutrient uptake of the roots, protect the plants from pathogens, and produce plant growth hormones. The fungi get carbohydrates from the plant.
F. Hookworms attached to the intestinal walls of humans get nutrients from the host’s blood and tissue juices. Infected humans experience digestive disorders and anemia.
G.
Polyergus rufescens
ants steal the eggs of
Formica fusca
ants. When the eggs hatch, the Formica ants become slave workers for the Polyergus ants.
H. Species of small fishes and shrimp remove parasites from larger fishes. The small fishes get nutrients from the parasites and the large fishes are cleaned of parasites.
I. Yucca moths lay their eggs in the seed pods of yucca plants. Larvae that hatch from the yucca moth’s eggs feed on some, but not all, the seeds. The plant is pollinated by the moth.
J. Brittle stars make their homes in tropical sponges.
Scroll down for the chart…
Organism
Effect of Relationship
(benefits, suffers, unaffected)
Type of Symbiosis
A
orchid
tree
B
bacteria
bean/pea roots
C
lamprey eel
fish
D
remora
shark
E
fungal hyphae
roots
F
hookworm
human
G
Polyergus ant
Formica ant
H
small fish
large fish
I
yucca moth
yucca plant
J
brittle star
tropical sponge
Taken from Addison-Wesley Publishing Company
.
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Plants are predominantly photosynthetic eukaryotes that obtain their energy from sunlight via photosynthesis. They are characterized by sexual reproduction and alternation of generations, though asexual reproduction is also common. There are about 320,000 known plant species, the majority of which reproduce via seeds. Plants provide oxygen, form the basis of ecosystems, and have been domesticated as food, medicines, and other uses throughout human history.
PriorityQueue.cs Jim Mischel using System; using Sy.pdf
// PriorityQueue.cs
//
// Jim Mischel
using System;
using System.Collections;
using System.Collections.Generic;
using System.Runtime.InteropServices;
namespace Mischel.Collections
{
[Serializable]
[ComVisible(false)]
public struct PriorityQueueItem
{
private TValue _value;
public TValue Value
{
get { return _value; }
}
private TPriority _priority;
public TPriority Priority
{
get { return _priority; }
}
internal PriorityQueueItem(TValue val, TPriority pri)
{
this._value = val;
this._priority = pri;
}
}
[Serializable]
[ComVisible(false)]
public class PriorityQueue : ICollection,
IEnumerable>
{
private PriorityQueueItem[] items;
private const Int32 DefaultCapacity = 16;
private Int32 capacity;
private Int32 numItems;
private Comparison compareFunc;
///
/// Initializes a new instance of the PriorityQueue class that is empty,
/// has the default initial capacity, and uses the default IComparer.
///
public PriorityQueue()
: this(DefaultCapacity, Comparer.Default)
{
}
public PriorityQueue(Int32 initialCapacity)
: this(initialCapacity, Comparer.Default)
{
}
public PriorityQueue(IComparer comparer)
: this(DefaultCapacity, comparer)
{
}
public PriorityQueue(int initialCapacity, IComparer comparer)
{
Init(initialCapacity, new Comparison(comparer.Compare));
}
public PriorityQueue(Comparison comparison)
: this(DefaultCapacity, comparison)
{
}
public PriorityQueue(int initialCapacity, Comparison comparison)
{
Init(initialCapacity, comparison);
}
private void Init(int initialCapacity, Comparison comparison)
{
numItems = 0;
compareFunc = comparison;
SetCapacity(initialCapacity);
}
public int Count
{
get { return numItems; }
}
public int Capacity
{
get { return items.Length; }
set { SetCapacity(value); }
}
private void SetCapacity(int newCapacity)
{
int newCap = newCapacity;
if (newCap < DefaultCapacity)
newCap = DefaultCapacity;
// throw exception if newCapacity < NumItems
if (newCap < numItems)
throw new ArgumentOutOfRangeException(\"newCapacity\", \"New capacity is less
than Count\");
this.capacity = newCap;
if (items == null)
{
items = new PriorityQueueItem[newCap];
return;
}
// Resize the array.
Array.Resize>(ref items, newCap);
}
public void Enqueue(PriorityQueueItem newItem)
{
if (numItems == capacity)
{
// need to increase capacity
// grow by 50 percent
SetCapacity((3 * Capacity) / 2);
}
int i = numItems;
++numItems;
while ((i > 0) && (compareFunc(items[(i - 1) / 2].Priority, newItem.Priority) < 0))
{
items[i] = items[(i - 1) / 2];
i = (i - 1) / 2;
}
items[i] = newItem;
//if (!VerifyQueue())
//{
// Console.WriteLine(\"ERROR: Queue out of order!\");
//}
}
public void Enqueue(TValue value, TPriority priority)
{
Enqueue(new PriorityQueueItem(value, priority));
}
private PriorityQueueItem RemoveAt(Int32 index)
{
PriorityQueueItem o = items[index];
--numItems;
// move the last item to fill the hole
PriorityQueueItem tmp = items[numItems];
// If you forget to clear this, you have a potential memory leak.
items[numItems] = default(PriorityQueueItem);
if (num.
the balanced equations are H3PO4(aq) + 3[NaOH](.pdf
the balanced equations are: H3PO4(aq) + 3[NaOH](aq) = Na3PO4(aq) + 3[H2O](l)
2[HI](aq) + LiO(aq) = LiI2(aq) + H2O(l) 2[HNO3](aq) + Ca(OH)2(s) = Ca(NO3)2(aq) +
2[H2O](l)
Solution
the balanced equations are: H3PO4(aq) + 3[NaOH](aq) = Na3PO4(aq) + 3[H2O](l)
2[HI](aq) + LiO(aq) = LiI2(aq) + H2O(l) 2[HNO3](aq) + Ca(OH)2(s) = Ca(NO3)2(aq) +
2[H2O](l).
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This document summarizes research on endophytic microbes and their potential applications in crop management. Key points include:
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Fungi
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Classifying the Symbiotic RelationshipsDo you know that a tiny o.docx
Classifying the Symbiotic Relationships
Do you know that a tiny organism living inside you helps keep you alive? That organism is E. coli, a bacterium that lives in your intestines. E. coli provides you with vitamins; you provide E. coli with food. This relationship is called
mutualism
, because both you and E. coli benefit.
Directions:
Read the following descriptions of each relationship. Then complete columns two and three of the table. In column two, state whether
each
organism benefits, suffers, or is not affected. In column three, classify the relationships as mutualism, commensalism, or parasitism.
A. Orchid vines grow up the trunks of trees. The areas at the tops of trees have more light than the areas underneath trees. The trees are not affected by the orchids.
B. A type of bacteria that lives on the roots of bean and pea plants converts nitrogen gas in the air into nitrogen compounds that the plants need for growth.
C. A lamprey eel fastens itself to a host fish, such as lake perch, and feeds on it. When the fish dies, the lamprey finds another host.
D. A remora attaches itself to sharks by a rough, flattened fin on the top of its head. It feeds on shreds of the sharks’ prey.
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F. Hookworms attached to the intestinal walls of humans get nutrients from the host’s blood and tissue juices. Infected humans experience digestive disorders and anemia.
G.
Polyergus rufescens
ants steal the eggs of
Formica fusca
ants. When the eggs hatch, the Formica ants become slave workers for the Polyergus ants.
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J. Brittle stars make their homes in tropical sponges.
Scroll down for the chart…
Organism
Effect of Relationship
(benefits, suffers, unaffected)
Type of Symbiosis
A
orchid
tree
B
bacteria
bean/pea roots
C
lamprey eel
fish
D
remora
shark
E
fungal hyphae
roots
F
hookworm
human
G
Polyergus ant
Formica ant
H
small fish
large fish
I
yucca moth
yucca plant
J
brittle star
tropical sponge
Taken from Addison-Wesley Publishing Company
.
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Plants are predominantly photosynthetic eukaryotes that obtain their energy from sunlight via photosynthesis. They are characterized by sexual reproduction and alternation of generations, though asexual reproduction is also common. There are about 320,000 known plant species, the majority of which reproduce via seeds. Plants provide oxygen, form the basis of ecosystems, and have been domesticated as food, medicines, and other uses throughout human history.
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PriorityQueue.cs Jim Mischel using System; using Sy.pdf
// PriorityQueue.cs
//
// Jim Mischel
using System;
using System.Collections;
using System.Collections.Generic;
using System.Runtime.InteropServices;
namespace Mischel.Collections
{
[Serializable]
[ComVisible(false)]
public struct PriorityQueueItem
{
private TValue _value;
public TValue Value
{
get { return _value; }
}
private TPriority _priority;
public TPriority Priority
{
get { return _priority; }
}
internal PriorityQueueItem(TValue val, TPriority pri)
{
this._value = val;
this._priority = pri;
}
}
[Serializable]
[ComVisible(false)]
public class PriorityQueue : ICollection,
IEnumerable>
{
private PriorityQueueItem[] items;
private const Int32 DefaultCapacity = 16;
private Int32 capacity;
private Int32 numItems;
private Comparison compareFunc;
///
/// Initializes a new instance of the PriorityQueue class that is empty,
/// has the default initial capacity, and uses the default IComparer.
///
public PriorityQueue()
: this(DefaultCapacity, Comparer.Default)
{
}
public PriorityQueue(Int32 initialCapacity)
: this(initialCapacity, Comparer.Default)
{
}
public PriorityQueue(IComparer comparer)
: this(DefaultCapacity, comparer)
{
}
public PriorityQueue(int initialCapacity, IComparer comparer)
{
Init(initialCapacity, new Comparison(comparer.Compare));
}
public PriorityQueue(Comparison comparison)
: this(DefaultCapacity, comparison)
{
}
public PriorityQueue(int initialCapacity, Comparison comparison)
{
Init(initialCapacity, comparison);
}
private void Init(int initialCapacity, Comparison comparison)
{
numItems = 0;
compareFunc = comparison;
SetCapacity(initialCapacity);
}
public int Count
{
get { return numItems; }
}
public int Capacity
{
get { return items.Length; }
set { SetCapacity(value); }
}
private void SetCapacity(int newCapacity)
{
int newCap = newCapacity;
if (newCap < DefaultCapacity)
newCap = DefaultCapacity;
// throw exception if newCapacity < NumItems
if (newCap < numItems)
throw new ArgumentOutOfRangeException(\"newCapacity\", \"New capacity is less
than Count\");
this.capacity = newCap;
if (items == null)
{
items = new PriorityQueueItem[newCap];
return;
}
// Resize the array.
Array.Resize>(ref items, newCap);
}
public void Enqueue(PriorityQueueItem newItem)
{
if (numItems == capacity)
{
// need to increase capacity
// grow by 50 percent
SetCapacity((3 * Capacity) / 2);
}
int i = numItems;
++numItems;
while ((i > 0) && (compareFunc(items[(i - 1) / 2].Priority, newItem.Priority) < 0))
{
items[i] = items[(i - 1) / 2];
i = (i - 1) / 2;
}
items[i] = newItem;
//if (!VerifyQueue())
//{
// Console.WriteLine(\"ERROR: Queue out of order!\");
//}
}
public void Enqueue(TValue value, TPriority priority)
{
Enqueue(new PriorityQueueItem(value, priority));
}
private PriorityQueueItem RemoveAt(Int32 index)
{
PriorityQueueItem o = items[index];
--numItems;
// move the last item to fill the hole
PriorityQueueItem tmp = items[numItems];
// If you forget to clear this, you have a potential memory leak.
items[numItems] = default(PriorityQueueItem);
if (num.
the balanced equations are H3PO4(aq) + 3[NaOH](.pdf
the balanced equations are: H3PO4(aq) + 3[NaOH](aq) = Na3PO4(aq) + 3[H2O](l)
2[HI](aq) + LiO(aq) = LiI2(aq) + H2O(l) 2[HNO3](aq) + Ca(OH)2(s) = Ca(NO3)2(aq) +
2[H2O](l)
Solution
the balanced equations are: H3PO4(aq) + 3[NaOH](aq) = Na3PO4(aq) + 3[H2O](l)
2[HI](aq) + LiO(aq) = LiI2(aq) + H2O(l) 2[HNO3](aq) + Ca(OH)2(s) = Ca(NO3)2(aq) +
2[H2O](l).
Molecular wt. of NaOH = 40 No. of moles of NaOH =.pdf
Molecular wt. of NaOH = 40 No. of moles of NaOH = 0.3/40 = 0.0075 moles
Molarity = (No. of moles of NaOH)/(Volume of
Solution
) = 0.0075/0.17 = 0.0441 moles/lt.
images are not visible. post the question again .pdf
images are not visible. post the question again
Solution
images are not visible. post the question again.
VSEPR is easy if you follow the right steps. 1. .pdf
VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer)
electrons around the central atom (remember the charge) 2. Work out how many of these are
involved in bonding (remember you may have single or double bonds) 3. Remember the lone
pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the
number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a
double bond, so there are 2 of Se\'s electrons in a double bond There are two Se=O bonds, so 4
electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So
overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is
trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule.
For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4
P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons.
Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are
single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved
in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electron groups around the central
ion, so the shape is trigonal bipyramid if you draw in the lone pairs (trigonal plane with 2 groups
above and below the plane. You can determine that these two groups must be the lone pairs since
they repel more than the I-Cl bonds). Removing the lone pairs gives you a trigonal planar
molecule. The number of groups around the central ion determines the shape. 2 is linear 3 is
trigonal planar
Solution
VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer)
electrons around the central atom (remember the charge) 2. Work out how many of these are
involved in bonding (remember you may have single or double bonds) 3. Remember the lone
pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the
number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a
double bond, so there are 2 of Se\'s electrons in a double bond There are two Se=O bonds, so 4
electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So
overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is
trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule.
For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4
P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons.
Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are
single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved
in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electro.
This is quite a broad question. I will try to acc.pdf
This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends on
\"ligands\" (complexing molecule/ion). e.g. Co2+ in water is pink [Co(H2O)6]2+, while Co2+ in
NH3 is blue [Co(NH3)4]2+. See \"spectrochemical series\" for more detail about this. 6. To find
\"exact\" color of metal ions, you need a supercomputer to do so. (I\'m not sure whether today
computer can calculate it). So, it\'s better to memorize the colors. Personally, I think you\'ll
understand about metal\'s color when you study transition metals, like Crystal Field Theory, and
Ligand Field Theory.
Solution
This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends .
}public staatic boolean isValidElement(String token){public vo.pdf
}
public staatic boolean isValidElement(String token)
{
public void setParts(String IP)
{
IP=\"192.168.20.40\";
}
public void setParts(int[], data)
{
data[0]=toString(IP);
data[1]=toString(IP);
data[2]=toString(IP);
data[3]=toString(IP);
}
Solution
}
public staatic boolean isValidElement(String token)
{
public void setParts(String IP)
{
IP=\"192.168.20.40\";
}
public void setParts(int[], data)
{
data[0]=toString(IP);
data[1]=toString(IP);
data[2]=toString(IP);
data[3]=toString(IP);
}.
y=41.2xSolutiony=41.2x.pdf
This short document presents an equation, y=4*1.2x, and labels it as a solution. No other context or explanation is provided about what problem this solution pertains to or what the variables represent.
What is an example of big data either from your personal experience .pdf
What is an example of big data either from your personal experience or outside research?
Answer:
Big Data: Big data means really a big data, it is a collection of large datasets that cannot be
processed using traditional computing techniques. Big data is not merely a data, rather it has
become a complete subject, which involves various tools, technqiues and frameworks.
Big data involves the data produced by different devices and applications. Given below are some
of the fields that come under the Big Data.
Black Box Data : It is a component of helicopter, airplanes, and jets, etc. It captures voices of the
flight crew, recordings of microphones and earphones, and the performance information of the
aircraft.
Social Media Data : Social media such as Facebook and Twitter hold information and the views
posted by millions of people across the globe.
Stock Exchange Data : The stock exchange data holds information about the ‘buy’ and ‘sell’
decisions made on a share of different companies made by the customers.
Power Grid Data : The power grid data holds information consumed by a particular node with
respect to a base station.
Transport Data : Transport data includes model, capacity, distance and availability of a vehicle.
Search Engine Data : Search engines retrieve lots of data from different databases.
How is it similar or different to the example in the case? In response to your peers’ posts, discuss
how big data in the examples provided impact an organization in terms of the advantages and
disadvantages?
Big data is really critical to our life and its emerging as one of the most important technologies in
modern world. Follow are just few benefits which are very much known to all of us:
Using the information kept in the social network like Facebook, the marketing agencies are
learning about the response for their campaigns, promotions, and other advertising mediums.
Using the information in the social media like preferences and product perception of their
consumers, product companies and retail organizations are planning their production.
Disadvantages of Big Data:
- Unknown population representation
- Issues of data quality
- Typically not very multivariate (at the person level)
- Privacy and confidentiality issues
- Difficult to assess accuracy and uncertainty
Solution
What is an example of big data either from your personal experience or outside research?
Answer:
Big Data: Big data means really a big data, it is a collection of large datasets that cannot be
processed using traditional computing techniques. Big data is not merely a data, rather it has
become a complete subject, which involves various tools, technqiues and frameworks.
Big data involves the data produced by different devices and applications. Given below are some
of the fields that come under the Big Data.
Black Box Data : It is a component of helicopter, airplanes, and jets, etc. It captures voices of the
flight crew, recordings of microphones and earphones, and the performance.
S-Se=S with 3 lone pairs on the first S, 1 lone.pdf
S-Se=S with 3 lone pairs on the first S, 1 lone pair on the Se, and 2 lone pairs on
the second S
Solution
S-Se=S with 3 lone pairs on the first S, 1 lone pair on the Se, and 2 lone pairs on
the second S.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
The formula for calculating Alveolar ventilation is as followsAlv.pdf
The formula for calculating Alveolar ventilation is as follows:
Alveolar ventilation = (tidal volume - dead space) * respiratory rate
Alveolar ventilation = ( 500 – 100) * 12
Alveolar ventilation = 4800
Solution
The formula for calculating Alveolar ventilation is as follows:
Alveolar ventilation = (tidal volume - dead space) * respiratory rate
Alveolar ventilation = ( 500 – 100) * 12
Alveolar ventilation = 4800.
D. This element can only have a -2 oxidation stat.pdf
D. This element can only have a -2 oxidation state in covalent compounds. is false.
note: such as S can be +4 in SO2
Solution
D. This element can only have a -2 oxidation state in covalent compounds. is false.
note: such as S can be +4 in SO2.
solAn object will allocated statically when that object is needed.pdf
sol:
An object will allocated statically when that object is needed for entire life time of that program
1)if object needs recusrion and storage of object conservly then it is stored in stack
2)f object needs dynamic storage then it should be allocated in heap
Solution
sol:
An object will allocated statically when that object is needed for entire life time of that program
1)if object needs recusrion and storage of object conservly then it is stored in stack
2)f object needs dynamic storage then it should be allocated in heap.
P = 12SolutionP = 12.pdf
This very short document states that P equals one-half. No other information or context is provided.
carbon moves by A. interstitial diffusion So.pdf
Carbon moves through materials by interstitial diffusion. Interstitial diffusion is a process where atoms diffuse through the spaces in a material's crystal lattice rather than displacing other atoms from their lattice sites. Carbon atoms are small enough to move through the spaces in many materials' crystal lattices, allowing carbon to diffuse interstitially.
NH4Cl = acid NaOH = base They react 11 Leftover = 0.26-0..pdf
NH4Cl = acid
NaOH = base
They react 1:1
Leftover = 0.26-0.192 = 0.068M NH4Cl
pH = -log(0.068) = 1.1675
pH + pOH = 14
pOH = 14 - 1.1675 = 12.8325
[OH-] = 10^-12.8325 = 1.471 * 10^-13 M
Solution
NH4Cl = acid
NaOH = base
They react 1:1
Leftover = 0.26-0.192 = 0.068M NH4Cl
pH = -log(0.068) = 1.1675
pH + pOH = 14
pOH = 14 - 1.1675 = 12.8325
[OH-] = 10^-12.8325 = 1.471 * 10^-13 M.
D is the transition represent!! Solution .pdf
This very short document appears to be discussing a transition and representing the letter D. It states "D is the transition represent!!" but does not provide any other context or information.
CO2 is non polar because C has no lone pairs, so .pdf
CO2 is non polar because C has no lone pairs, so the structure is linear and hence
dipole cancell each other SO2 has lone pair, so the structure is bent due to the repulsion and
hence the structure has a net dipole moment, so polar.
Solution
CO2 is non polar because C has no lone pairs, so the structure is linear and hence
dipole cancell each other SO2 has lone pair, so the structure is bent due to the repulsion and
hence the structure has a net dipole moment, so polar..
As more halide ions are added, it loses its color.pdf
As more halide ions are added, it loses its color and becomes more white due to the
ion concentration.
Solution
As more halide ions are added, it loses its color and becomes more white due to the
ion concentration..
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