Elements prefer to live in their natural forms.N2 and O2 are more stable than NO.Added:Air is a mixture of N2 and O2 molecules, where NO is a compound, chemically formed by reaction: N2 + O2 --> 2 NO. This reaction does not take place at normal conditions as in the air around us.
n2 has more ionization energy than no.
hydried is more stable than H-atom
Ammonia is less stable than water.
1 mole N2 = 28.0134g 1 mole N2 = 6.022 x 1023 molecules N2 28.0134g N2 = 6.022 x 1023 molecules N2 (4.00 x 1023 molecules N2) x (28.0134g/6.022 x 1023 molecules) = 18.6g N2
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
#include<stdio.h> #include<stdlib.h> void display(float **,int); float** add(float **,float **,int,int,int); int main() { float **p1,**p2,**p3,**p4; int i,j,n1,n2,k=0,x; printf("Enter no of terms of a pollynomial:\n"); scanf("%d",&n1); printf("Enter no of terms of another pollynomial:\n"); scanf("%d",&n2); p1=(float **) malloc(n1*sizeof(float *)); p2=(float **) malloc(n2*sizeof(float *)); for(i=0;i<n1;i++) p1[i]=(float *) malloc(2*sizeof(float)); for(i=0;i<n2;i++) p2[i]=(float *) malloc(2*sizeof(float)); printf("Enter the first pollynomial:\n"); for(i=0;i<n1;i++) { printf("\nEnter value and exponent:"); scanf("%f %f",&p1[i][0],&p1[i][1]); } printf("Enter the second pollynomial:\n"); for(i=0;i<n2;i++) { printf("\nEnter value and exponent:"); scanf("%f %f",&p2[i][0],&p2[i][1]); } printf("\nFirst pollynomial:\n"); display(p1,n1); printf("\nSecond pollynomial:\n"); display(p2,n2); for(i=0;i<n1;i++) for(j=0;j<n2;j++) if(p1[i][1]==p2[j][1]) k++; x=n1+n2-k; p3=add(p1,p2,n1,n2,x); printf("\nAdded polynomial:\n"); display(p3,x); return 0; } void display(float **p,int n) { int i; printf("%fx^%d",p[0][0],(int)p[0][1]); for(i=1;i<n;i++) printf("+%fx^%d",p[i][0],(int)p[i][1]); } float** add(float **p1,float **p2,int n1,int n2,int n) { int i,j,k; float **p3; p3=(float **)malloc(n*sizeof(float*)); for(i=0;i<n;i++) p3[i]=(float *)malloc(2*sizeof(float)); i=0; j=0; k=0; while(i<n1 && j<n2) { if(p1[i][1]==p2[j][1]) { p3[k][0]=p1[i][0]+p2[j][0]; p3[k][1]=p1[i][1]; k++; i++; j++; } else if(p1[i][1]<p2[j][1]) { p3[k][0]=p1[i][0]; p3[k][1]=p1[i][1]; k++; i++; } else { p3[k][0]=p2[j][0]; p3[k][1]=p2[j][1]; k++; j++; } } while(i<n1) { p3[k][0]=p1[i][0]; p3[k][1]=p1[i][1]; k++; i++; } while(j<n2) { p3[k][0]=p2[j][0]; p3[k][1]=p2[j][1]; k++; j++; } return p3; }
Three pairs of electrons.
Elements prefer to live in their natural forms.N2 and O2 are more stable than NO.Added:Air is a mixture of N2 and O2 molecules, where NO is a compound, chemically formed by reaction: N2 + O2 --> 2 NO. This reaction does not take place at normal conditions as in the air around us.
n2 has more ionization energy than no.
O2, because it have more electrons than N2
N2 is a very stable, nonreactive , non-polar gas so is almost insoluble in water.
P2 is a grading indicator used by gemologists certified by other than the Gemological Institute of America (GIA). P2 corresponds to the GIA grade I2: at least two (visible) Inclusions. You can read more about diamond grading by different gemologists, below.
Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
The composition of Buffer P2 is:200 mM NaOH1% SDS (w/v)Buffer P2 is the lysis buffer
Es has an atomic number of 99 and not a natural element. It is highly radioactive and probably belongs to the metals family. All metals are basically toxic; the same for radioactive substances. Anyway, I try to avoid any pure element other than O2 or N2, in stable form and at room temperature. Even O2 or N2 will be toxic in large amounts, so be careful.
No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k. Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.