In a neutral atom, yes.
This is not an atom, but a non-existing Carbon anion (-1, negatively charged)
The element with an atomic number of 8 is Oxygen. It has 2 electrons in the first shell and 6 in the second, leaving a valence of 2. Oxygen, therefore, can form covalent bonds with only two Hydrogen atoms. This is also known as water.
Nitrogen's atomic number is 7, which means it's got two orbitals. The inner one has two electrons in it, like every atom except hydrogen has. The outer orbital has five electrons. An orbital can have as many as eight electrons. It can get them either from itself or by sharing with other atoms. So...it has five electrons of its own, shares three with three hydrogen atoms, and makes a stable compound.
Mol-ib-dee-num
Two ions are said to be isoelectronic with each other if they have the same number of valence electrons and the same structure, regardless of the nature of elements involved. some examples are: cations like K+, Ca2+,Sc3+ CO, N2, valence electrons of C- 4, valence electrons of O -6, total num-10 valence electron of N -5 , THEREFORE TOTAL ELECTRONS-5+5=10 .Hence isoelectronic
This is not an atom, but a non-existing Carbon anion (-1, negatively charged)
int RevNum( int num ) { const int base = 10; int result = 0; do { result *= base; result += num % base; } while( num /= base); return( result ); }
int RevNum( int num ) { const int base = 10; int result = 0; do { result *= base; result += num % base; } while( num /= base); return( result ); }
function complement(num, base){ var result = 0, column = 0; if(base < 2) return null; while(num > 0){ digit = num % base;comp = base - 1 - digit;result += comp * Math.pow(base, column);column++;num -= digit;num /= base;} return result; }
int RevNum( int num ) { const int base = 10; int result = 0; int remain = 0; do { remain = num % base; result *= base; result += remain; } while( num /= base); return( result ); }
(defun prime (num) (if (< 2 num) (do ((dividend 2 (1 + dividend)) (chk-to (sqrt num))) ((equal (rem num dividend) 0)) (when (<= chk-to dividend) (return t))) t))
/*The coding style used in this source code is for convenience. * It is widely used style of coding. */ #include <stdio.h> void main() { int number, modulus, reverse; reverse = 0; printf("Enter a number \n"); scanf("%d", &number); while(number != 0) { modulus = number % 10; reverse = (reverse * 10) + modulus; number =number / 10; } printf("The reversed number is %d", reverse); getch(); }
unsigned binary_to_gray (unsigned num) { return num ^ (num >> 1); } unsigned gray_to_binary (unsigned num) { /* note: assumes num is no more than 32-bits in length */ num ^= (num >> 16); num ^= (num >> 8); num ^= (num >> 4); num ^= (num >> 2); num ^= (num >> 1); return num ; }
#include<stdio.h> #include<conio.h> void main() { int i=0,,res=0; int num,ar[5]; printf("\nEnter the number.\n"); scanf("%d",&num); while(num>0) { ar[i]=num%10; num/=10; i++; } i--; while(i>=0) res=res+ar[i]; printf("Result is %d",res); getch(); }
num num num
#include<iostream> #include<iomanip> #include<math.h> // for floor, log and pow bool is_fib(unsigned num) { // Algorithm: num is Fibonacci if 5*num*num+4 or 5*num*num-4 is a perfect square. // Note: // Since 5*num*num could result in a huge number which could easily overflow, // we'll use a modified algorithm that utilises numbers smaller than num. double root5 = std::sqrt(5.0); double phi = (1 + root5) / 2; long idx = (long)floor( log(num*root5) / log(phi) + 0.5 ); long n = (long)floor( pow(phi, idx)/root5 + 0.5); return (n == num); } int main() { // Print non-Fibonacci numbers in range: 0-50 for (unsigned num=0; num<=50; ++num) if (!is_fib(num)) std::cout <<std::setw(6) <<num <<" is not a Fibonacci number" <<std::endl; }
Nikhil Aggarwalstrong_number: The sum of factorials of digits of a number is equal to the original number.void strong_number(){int num,i,p,r,sum=0,save_num;printf("\nEnter a number");scanf("%d",&num);save_num=num;while(num){i=1,p=1;r=num%10;while(i