B yellow to blue as the amount of hydronium ion .pdf

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B yellow to blue as the amount of hydronium ion increase Solution B yellow to blue as the amount of hydronium ion increase.

1.KClO(3) ---- Potassium chlorate 2.H(2)SO(2) ------ sulphurous acid 3.SrF(2) ----- strontium fluoride 5.Hg(NO(3))(2) ----- Mercuric nitrate 6.SnCl(4) ---- Tinchloride 7.Se(3)Cl(2) ---- tri selenium dichloride 8.Hg(2)O ---- mercury (I) oxide 9.NCl(3) ---- Nitrogen trichloride Solution 1.KClO(3) ---- Potassium chlorate 2.H(2)SO(2) ------ sulphurous acid 3.SrF(2) ----- strontium fluoride 5.Hg(NO(3))(2) ----- Mercuric nitrate 6.SnCl(4) ---- Tinchloride 7.Se(3)Cl(2) ---- tri selenium dichloride 8.Hg(2)O ---- mercury (I) oxide 9.NCl(3) ---- Nitrogen trichloride.

Seven Requirements of CTC 1 Age 2 Relationship .pdf

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Seven Requirements of CTC: 1 Age 2 Relationship 3 Support 4 Dependent Status 5 Citizenship 6 Length of Residency 7 Family Income 1. Age test To qualify, a child must have been under age 17 (i.e., 16 years old or younger) at the end of the tax year for which you claim the credit. 2. Relationship test The child must be your own child, a stepchild, or a foster child placed with you by a court or authorized agency. An adopted child is always treated as your own child. (\"An adopted child\" includes a child lawfully placed with you for legal adoption, even if that adoption is not final by the end of the tax year.)You can also claim your brother or sister, stepbrother, stepsister. And you can claim descendants of any of these qualifying people—such as your nieces, nephews and grandchildren—if they meet all the other tests. 3. Support test To qualify, the child cannot have provided more than half of his or her own financial support during the tax year. 4. Dependent test You must claim the child as a dependent on your tax return.Bear in mind that in order for you to claim a child as a dependent, he or she must: 1) be your child (or adoptive or foster child), sibling, niece, nephew or grandchild; 2) be under age 19, or under age 24 and a fulltime student for at least five months of the year; or be permanently disabled, regardless of age; 3) have lived with you for more than half the year; and 4) have provided no more than half his or her own support for the year. 5. Citizenship test The child must be a U.S. citizen, a U.S. national or a U.S. resident alien. (For tax purposes, the term \"U.S. national\" refers to individuals who were born in American Samoa or in the Commonwealth of the Northern Mariana Islands.) 6. Residence test The child must have lived with you for more than half of the tax year for which you claim the credit. There are important exceptions, however: A child who was born (or died) during the tax year is considered to have lived with you for the entire year.Temporary absences by you or the child for special circumstances, such as school, vacation, business, medical care, military services or detention in a juvenile facility, are counted as time the child lived with you. (There are also some exceptions to the residency test for children of divorced or separated parents. For details, see the instructions for Form 1040, lines 51 and 6c, or Form 1040A, lines 33 and 6c.) 7. Family income test The child tax credit is reduced if your modified adjusted gross income (MAGI) is above certain amounts, which are determined by your tax-filing status. In 2017, the phase out threshold is $55,000 for married couples filing separately; $75,000 for single, head of household, and qualifying widow or widower filers; and $110,000 for married couples filing jointly. For each $1,000 of income above the threshold, your available child tax credit is reduced by $50. Once all this requirements are met than further this question shall arise Tax Professional shou.

the solvents that are used to combine together to.pdf

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the solvents that are used to combine together to make the solvent pair should be miscible. Hexane is nonpolar solvent, while water is a polar solvent. Thus, they are not miscible and will stay as two layers after they are combined together. Thus, they cannot work together as a solvent pair in recrystallization Solution the solvents that are used to combine together to make the solvent pair should be miscible. Hexane is nonpolar solvent, while water is a polar solvent. Thus, they are not miscible and will stay as two layers after they are combined together. Thus, they cannot work together as a solvent pair in recrystallization.

The most simple property to calculate this is Ka.pdf

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The most simple property to calculate this is Ka*Kb=e-14 -->Ka=(e-14)/Kb Ka for NH4+: Kb of NH3=(1.8e-5) --> Ka=(e-14)/(1.8e-5)=5.56e-10 Ka for CH3NH3+: Kb for CH3NH2=4.4e-4 --> Ka=(e-14)/(4.4e-4)=2.27e-11 Kb for F-: Ka for HF=6.8e-4 --> Kb=(e- 14)/(6.8e-4)=1.47e-11 Solution The most simple property to calculate this is Ka*Kb=e-14 -->Ka=(e-14)/Kb Ka for NH4+: Kb of NH3=(1.8e-5) --> Ka=(e-14)/(1.8e-5)=5.56e-10 Ka for CH3NH3+: Kb for CH3NH2=4.4e-4 --> Ka=(e-14)/(4.4e-4)=2.27e-11 Kb for F-: Ka for HF=6.8e-4 --> Kb=(e- 14)/(6.8e-4)=1.47e-11.

The images of the compounds are not visible. Kin.pdf

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Step1 Moles of water produced =18 Step2 The init.pdf

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Sorry. SnO2 is amphoteric. This one is a tricky o.pdf

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SO2 Solu.pdf

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This short document appears to be discussing a solution to a problem or issue numbered as I and II. It briefly mentions I and II at the beginning and end but does not provide any other context or details about the topic being discussed. The middle section indicates there is a solution presented but without more information it is difficult to understand what problem is being addressed or what the proposed solution entails.

Group 11 has three elements namely, Cu, Ag and Au.pdf

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Cu, (H2), Zn, Mg Activity series basedon hyd.pdf

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answewrSolutionanswewr.pdf

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cis- So.pdf

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When a compound is completely oxidized with oxygen, the products wil.pdf

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Calcium metal reacts with liquid water to produce.pdf

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two;Solutiontwo;.pdf

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The mineralogical and structural adjustment of solid rocks to physic.pdf

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The mineralogical and structural adjustment of solid rocks to physical and chemical conditions that have been imposed at depths below the near surface zones of weathering and diagenesis and which differ from conditions under which the rocks in question originated. The word \"Metamorphism\" comes from the Greek: meta = change, morph = form, so metamorphism means to change form. In geology this refers to the changes in mineral assemblage and texture that result from subjecting a rock to conditions such pressures, temperatures, and chemical environments different from those under which the rock originally formed. There are two major kinds of metamorphism: regional and contact. Regional metamorphism. Most metamorphic rocks are the result of regional metamorphism (also called dynamothermal metamorphism). These rocks were typically exposed to tectonic forces and associated high pressures and temperatures. They are usually foliated and deformed and thought to be remnants of ancient mountain ranges. Contact metamorphism. Contact metamorphism (also called thermal metamorphism) is the process by which the country rock that surrounds a hot magma intrusion is metamorphosed by the high heat flow coming from the intrusion. The zone of metamorphism that surrounds the intrusion is called the halo (or aureole) and rarely extends more than 100 meters into the country rock. Geostatic pressure is usually a minor factor, since contact metamorphism generally takes place less than 10 kilometers from the surface.Metamorphism is defined as follows: The mineralogical and structural adjustment of solid rocks to physical and chemical conditions that have been imposed at depths below the near surface zones of weathering and diagenesis and which differ from conditions under which the rocks in question originated. The word \"Metamorphism\" comes from the Greek: meta = change, morph = form, so metamorphism means to change form. In geology this refers to the changes in mineral assemblage and texture that result from subjecting a rock to conditions such pressures, temperatures, and chemical environments different from those under which the rock originally formed. There are two major kinds of metamorphism: regional and contact. Regional metamorphism. Most metamorphic rocks are the result of regional metamorphism (also called dynamothermal metamorphism). These rocks were typically exposed to tectonic forces and associated high pressures and temperatures. They are usually foliated and deformed and thought to be remnants of ancient mountain ranges. Contact metamorphism. Contact metamorphism (also called thermal metamorphism) is the process by which the country rock that surrounds a hot magma intrusion is metamorphosed by the high heat flow coming from the intrusion. The zone of metamorphism that surrounds the intrusion is called the halo (or aureole) and rarely extends more than 100 meters into the country rock. Geostatic pressure is usually a minor factor, since contact metamorp.

The general idea is to find a root of the polynomial and then apply .pdf

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The general idea is to find a root of the polynomial and then apply Horner\'s method to remove the corresponding factor according to the Ruffini rule. This iterative scheme is numerically unstable; the approximation errors accumulate during the successive factorizations, so that the last roots are determined with a polynomial that deviates widely from a factor of the original polynomial. To reduce this error, it is advisable to find the roots in increasing order of magnitude. Wilkinson\'s polynomial illustrates that high precision may be necessary when computing the roots of a polynomial given its coefficients: the problem of finding the roots from the coefficients is in general ill-conditioned. The most simple method to find a single root fast is Newton\'s method. One can use Horner\'s method twice to efficiently evaluate the value of the polynomial function and its first derivative; this combination is called Birge–Vieta\'s method. This method provides quadratic convergence for simple roots at the cost of two polynomial evaluations per step. Closely related to Newton\'s method are Halley\'s method and Laguerre\'s method. Using one additional Horner evaluation, the value of the second derivative is used to obtain methods of cubic convergence order for simple roots. If one starts from a point x close to a root and uses the same complexity of 6 function evaluations, these methods perform two steps with a residual of O(|f(x)|9), compared to three steps of Newton\'s method with a reduction O(|f(x)|8), giving a slight advantage to these methods. When applying these methods to polynomials with real coefficients and real starting points, Newton\'s and Halley\'s method stay inside the real number line. One has to choose complex starting points to find complex roots. In contrast, the Laguerre method with a square root in its evaluation will leave the real axis of its own accord. Another class of methods is based on translating the problem of finding polynomial roots to the problem of finding eigenvalues of the companion matrix of the polynomial. In principle, one can use any eigenvalue algorithm to find the roots of the polynomial. However, for efficiency reasons one prefers methods that employ the structure of the matrix, that is, can be implemented in matrix-free form. Among these methods are the power method, whose application to the transpose of the companion matrix is the classical Bernoulli\'s method to find the root of greatest modulus. The inverse power method with shifts, which finds some smallest root first, is what drives the complex (cpoly) variant of the Jenkins–Traub method and gives it its numerical stability. Additionally, it is insensitive to multiple roots and has fast convergence with order 1 + 2.6 {\\displaystyle 1+\\Phi \\approx 2.6} even in the presence of clustered roots. This fast convergence comes with a cost of three Horner evaluations per step, resulting in a residual of O(|f(x)|2+3), which is slower than three steps of New.

The answer is d. XThe most active nonmetal has the highest electro.pdf

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The correct order is116124310139Rest of the statem.pdf

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Simply note that SO(n) is the kernel of the group homomorphism det.pdf

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All liquids in containers have a concave meniscus.pdf

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All liquids in containers have a concave meniscus. This is because surface molecules being attracted to each other and also to the sides of the containers. Lakes also have a concave meniscus, but they are so large we don\'t notice it. Solution All liquids in containers have a concave meniscus. This is because surface molecules being attracted to each other and also to the sides of the containers. Lakes also have a concave meniscus, but they are so large we don\'t notice it..

Mutants of the Drosophila dunce and rutabaga genes, which encode a c.pdf

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Mutants of the Drosophila dunce and rutabaga genes, which encode a cAMP-specific phosphodiesterase and a calcium/calmodulin-responsive adenylyl cyclase, respectively, are deficient in short-term memory. Altered synaptic plasticity has been demonstrated at neuromuscular junctions in these mutants, but little is known about how their central neurons are affected. This problem was examined by using the \"giant\" neuron culture, which offers a unique opportunity to analyze mutational effects on neuronal activity and the underlying ionic currents in Drosophila. On the basis of instantaneous frequency and first latency of spikes evoked by current steps, four categories of firing patterns (tonic, adaptive, delayed, and interrupted) were identified in wild-type neurons, revealing interesting parallels to those commonly observed in vertebrate CNS neurons. The distinct firing patterns are correlated with expression of different ratios of 4-aminopyridine- and tetra ethylammonium-sensitive K+ currents. Subsets of dnc and rut neurons display abnormal spontaneous spikes and altered firing patterns. Altered frequency coding in mutant neurons was demonstrated further by using stimulation protocols involving conditioning with previous activity. Abnormal spike activity and reduced K+ current remain in double-mutant neurons, suggesting that the opposite effects on cAMP metabolism by dnc and rut do not counterbalance the mutual functional defects. The aberrant spontaneous activity and altered frequency coding in different stimulus paradigms may present problems in the stability and reliability of neural circuits for information processing during certain behavioral tasks, raising the possibility of modulation in neuronal excitability as a cellular mechanism underlying learning and memory (Zhao, 1997). Not all memory genes first identified in other contexts, however, play a significant role in place memory. The DopEcR gene has been implicated in several behaviors, including a 30 min memory after courtship conditioning. This G-protein linked receptor is responsive to both dopamine and the steroid hormone ecdysone. Remarkably, DopEcR has been shown to interact with the cAMP cascade through double mutant and pharmacological tests. Using conditions that induce a robust and lasting place memory, the DopEcR mutant flies do not show a defect in memory directly after training or at 1 h post-training. This is despite the fact that the rut and cAMP-phosphodiesterase genes (dunce) are critical for place memory. It may be that DopEcR is not required for this type of learning and would be consistent with the independence of place memory from dopamine signaling. Alternatively, other redundant pathways may compensate for the reduction in DopEcR function caused by the DopEcRPB1 allele. One might further speculate that other types of behavioral plasticity, such as reversal learning or memory enhancement after unpredicted high temperature exposures in the heat-box might be more sen.

Mass of gold = 0.017100 x 1000= 0.17 kg = 170 gSolutionMass.pdf

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#include stdio.h#include stdlib.h#include string.hstruct.pdf

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#include #include #include struct product { int id; char name; int price; int qty; }; struct Bill { int pid; char pname; int pprice; }; char mygetch(); int getid(); int billfileno(); void manageproduct(); void purchaseproduct(); void generatebill(); void addproduct(); void displayallproducts(); struct product findproduct(int id); void updateproduct(int id, int qty); char fproduct[]={\"product.txt\"}; char fbill[]={\"bill.txt\"}; int total=0; int main() { File *fp; int ch; while(1) { printf(\"welcome product management system\ \"); printf(\"manage product\ \"); printf(\"\\purchase product\ \"); printf(\"generate bill\ \"); printf(\"exit\ \"); printf(\"enter your choice\"); scanf(\"%d\",&ch); switch(ch) { case1:manageproduct(); break; case2:purchaseproduct(); break; case3:generate bill(); break; case4:exit(); } mygetch(); } return 0; int getid() { file *fp; int value=0; fp=fopen(\"product.txt\",\"r\"); if(fp==null) { fp=fopen(\"product.txt\",\"w\"); fprint(fp,\"%d\",0); fclose(fp); fp=fopen(\"product.txt\",\"r\"); } fscanf(fp,\"%d\",&value); fclose(fp); fp=fopen(\"product.txt\",\"w\"); fprintf(\"fp,\"%d\",value+1); fclose(fp);return value+1; void manageproduct() { int ch init back=0; while(1) { printf(\"welcome product management system\ \"); printf(\"1.add new product\ \"); printf(\"2.display all products\ \"); printf(\"back\ \"); printf(\"enter your choice\ \"); scanf(\"%d\",&ch); switch(ch) { case1:addproduct(); break; case2:displayallproduct(); break; case3:(back==1) break; else { mygetch(); } } } void addproduct() { file *fp; struct product t1; fp=fopen(fproduct,\"ab\"); t1.id=getid(); printf(\"enter product name\"); scanf(\"%s\",&t1.name); printf(\"enter product price\"); scanf(\"%d\",&t1.price); printf(\"enter product qty\"); scanf(\"%d\",&t1.qty); fwrite(&t1,sizeof(t1),1,fp); fclose(fp); void dis[playallproducts() { file *fp; struct product t; int id,found=0; fp=fopen(fproduct,\"rb\"); printf(\"product details\ \"); printf(\"id\\t name \\tqty\\tprice\ \ \"); while(1) { fread(&t,sizeof(t),1,fp); if(feof(fp)) { break; } printf(\"%d\\t\",t.id); printf(\"%s\\t\",t.name); printf(\"%d\\t\",t.qty); printf(\"%d\\t\",t.price); fclose(fp); } void purchase product() char ch1,ch2; int id; while(1) { displayallproducts(); _fpurge(stdin); printf(\"do u want to purchase y/n\"); scanf(\"%c\",&ch1); if(ch1==\'y\') { file *fp; struct bill t1; struct product t2; fp=fopen(fbill,\"ab\"); printf(\"enter the product id to purchase\"); scanf(\"%d\",&id); t2=findproduct(id); t1.pid=t2.id; strcpy(t1.pname,t2.name); t1.pprice=t2.price; fwrite(&t1,sizeof(t1),1,fp); fclose(fp); } _fpurge(stdin); printf(\"do u want to continue y/n\"); scanf(\"%c\",&ch2); if(ch2==\'n\') { break; } mygetch(); } } struct product findproduct(int id) { file *fp; struct product t; fp=fopen(fproduct,\"rb\"); if(feof(fp)) { break; } if(t.id==id) { updateproduct(id,t.qty-1); break; } } fclose(fp); return t; } void updateproduct(int id,int qty) { file *fp,*fp1; struct product t,t1; int found=.

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