Boron has the largest empirical atomic radius: 85 pm.
5th element - boron 6th element - carbon 7th element - nitrogen 8th element - oxygen 9th element - fluorine 10th element - neon
boron
There are 14. In alphabetical order they are: Boron (B) Carbon (C) Fluorine (F) Hydrogen (H) Iodine (I) Nitrogen (N) Oxygen (O) Phosphorus (P) Potassium (K) Sulphur (S), Tungsten (W), Uranium (U), Vanadium (V) Yttirium (Y)
Fluorine (F). It has the highest electron affinity which essentially means it is the most willing to gain an electron. Fluorine is also the MOST reactive element on the periodic table. Key to remember, as you go from left to right across the periodic table, the activity increases. As you go from top to bottom the activity decreases.
The most electronegative elements are found in the top right corner of the Periodic Table (excluding inert gases). So they are: Boron - B Carbon - C Nitrogen - N Oxygen - O Fluorine - F Fluorine is the most electronegative element.
F (fluorine) is the largest and heaviest (atomic number 9). Boron, nitrogen, and oxygen are numbers 5, 7, and 8.
#include<stdio.h> #include<conio.h> int main(void) { float a,b,c=0, d=0, e=0,f=0; printf("Please enter two numbers:\n"); scanf("%f %f", &a, &b); c=a+b; d=a-b; e=a*b; f=a/b; printf("The sum of %f and %f is :%f\n", a,b,c); printf("The subtraction of %f and %f is :%f\n", a,b,d); printf("The multiplication of %f and %f is :%f\n", a,b,e); printf("The division of %f by %f is :%f\n",a,b,f); getch(); }
void main() { int a,b,c; clrscr(); printf("Enter the value of a:"); scanf("%d",&a); printf("\nEnter the value of b:"); scanf("%d",&b); printf("\nEnter the value of c:"); scanf("%d",&c); if(a>b) { if(a>c) { if(b>c) { printf("c is smallest\n"); printf("b is middle\n"); printf("a is largest\n"); } else { printf("b is smallest\n"); printf("c is middle\n"); printf("a is largest\n"); } } else { printf("b is smallest\n"); printf("a is middle\n"); printf("c is largest\n"); } } else if(b>c) { if(a>c) { printf("c is smallest\n"); printf("a is middle\n"); printf("b is largest\n"); } else { printf("a is smallest\n"); printf("c is middle\n"); printf("b is largest\n"); } } else { printf("a is smallest\n"); printf("b is middle\n"); printf("c is largest\n"); } getch(); }
The most active element is fluorine (F) because it has the highest electronegativity of the options listed, meaning it has a strong tendency to attract and bond with other atoms.
ITS EASY...TRY THIS OUT..TRAPEZOIDAL METHOD#include#include#includefloat valcal(float x){return (x*x*x);}int main(){float a,b,h,c,I;int n,i;printf("THE TRAPEZOIDAL RULE:\n");printf("---------------------");printf("\n\n\nEnter the two limits and the no. of divisions:\n");scanf("%f %f %d",&a, &b, &n);h=(b-a)/n;//printf("\nVALUE of h: %f\n", h);c=a;I=valcal(a)+valcal(b);//printf("\nVALUE FOR a: %f\n", valcal(a));//printf("\nVALUE FOR b: %f\n", valcal(b));for(i=1;i=b){printf("\n\nc>b\n\n");break;}//printf("\nVALUE FOR %f: is %f\n",c, valcal(c));I=I+(2*valcal(c));//printf("\nI right now is %f", I);}printf("\n\n\nThe integration of x*x*x is: %f",(h*I)/2);printf("\n\n\n");system("pause");}SIMPSON'S 1/3RD METHOD#include#include#includefloat valcal(float x){return (1/(1+x*x));}int main(){float a,b,h,c,I;int n,i;printf("THE SIMPSON'S ONE-THIRD RULE:\n");printf("------------------------------");printf("\n\n\nEnter the two limits and the no. of divisions:\n");scanf("%f %f %d",&a, &b, &n);h=(b-a)/n;//printf("\nVALUE of h: %f\n", h);c=a;I=valcal(a)+valcal(b);//printf("\nVALUE FOR a: %f\n", valcal(a));//printf("\nVALUE FOR b: %f\n", valcal(b));for(i=1;ib){printf("\n\nc>b\n\n");break;}//printf("\nVALUE FOR %f: is %f\n",c, valcal(c));if(i%2==0)I=I+(2*valcal(c));elseI=I+4*valcal(c);//printf("\nI right now is %f", I);}printf("\n\n\nThe integration of x*x*x is: %f",(h*I)/3);printf("\n\n\n");system("pause");}NEED MORE HELP...MAIL ME YOUR PROB... SEE YA
#include<stdio.h> #include<conio.h> main() { int a[100]; int n,largest,index,position; printf("enter the number of elements in the array"); scanf("%d",&n); printf("enter %d elements",n); for(index=0;index<n;index++) scanf("%d",&a[index]); largest=a[0]; position=0; for(index=1;index<n;index++) if(a[index]>largest) { largest=a[index]; position=index; } printf("largest element in the array is %d\n",largest); printf("largets element's position in the array is %d\n",position+1); getch(); }
∫ f'(x)(af(x) + b)n dx = (af(x) + b)n + 1/[a(n + 1)] + C C is the constant of integration.
what element C or N has the highest ionization energy
S b n s s f b f s f n f 7 4 1 7 7 5 3 2 6 1
T| Capitals on the notes signify sharps (the black keys) A| B| Also see my Alicia Keys tab of the Empire State of Mind 2 for the chords! N| PEACE! A| ~Steph~ B| B| E| R5|F-F-F----- R| R4|F-F-F----- .| L3|F-F-F-F--- C| L3|----C----- O| L2|b-b---F--- M| -| R6|F-F-F---F---F---F-F-F--- T| R6|C-C-C---C---C---C-C-C--- A| R5|A-A-A---A---A---A-A-A--- B| N| L3|F-f----- A| L3|C-C----- B| L2|----b--- B| L2|----F--- E| L1|----b--- R| .| R6|F-F-F---F---F---F-F-F--- C| R6|C-C-C---C---C---C-C-C--- O| R5|A-A-A---A---A---A-A-A--- M| -| L3|F-F-F--- T| L3|--C----- A| L2|b---F--- B| N| R6|F-F-F---F---F---F-F-F--- A| R6|C-C-C---C---C---C-C-C--- B| R5|A-A-A---A---A---A-A-A--- B| E| L3|F-f----- R| L3|C-C----- .| L2|----b--- C| L2|----F--- O| L1|----b---
The element with the longest covalent radius among Li, B, N, and F is Li. This is because as you move down a group in the periodic table, the atomic radius increases due to the addition of new electron shells. The covalent radius generally follows this trend.
Factor 2n2 - 44n + 242 = f(n) the quick way: f(n) = 2 (n2 - 22n + 121) notice by sight that 22 is 2 * 11 and 121 is 11 squared. Set b = 11 f(n) = 2 (n2 - 2bn + b2 ) Now, factor the expression in brackets. The + and - signs show that both factors will have minus signs, so expect the form (a - b)(c -d). Check that both n and b occur in the middle term, so (n-b)(n-b) and therefore, f(n) = 2 (n - 11)2 Which is the answer.