N2 is the chemical formula for Nitrogen.
0 in N2
0 in N2
3N2H4 --> 4NH3 + N2 is the correctly balanced equation.
3Mn + N2 ----> Mn3N2 its actually 3mg+N2(arrow) Mg3N2
N2+ and N2- I just did it on mastering chem and it worked I'm pretty sure its because when you count the valence electrons in N2+ and N2- you get 9 and 11 respectively because these are odd there has to be an unpaired electron in each
Nitrogen
n2-1 and n2-4 are trivial cases because of n2-m2=(n-m)(n+m). So the only prime of the form n2-1 is 3 and of the form n2-4 is 5.
Well, nitrogen has no formula, futher more, nitrogen GAS has a formula known as 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
0 in N2
0 in N2
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
3N2H4 --> 4NH3 + N2 is the correctly balanced equation.
3Mn + N2 ----> Mn3N2 its actually 3mg+N2(arrow) Mg3N2
(N2) + 3(H2) = 2(NH3)
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2-1) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
void main() { int i; float n1,n2; abc: printf("Enter two nos "); scanf("%f%f",&n1,&n2); printf("\n %f + %f = %f " ,n1,n2,n1+n2); printf("\n %f - %f = %f " ,n1,n2,n1-n2); printf("\n %f x %f = %f " ,n1,n2,n1*n2); printf("\n %f / %f = %f " ,n1,n2,n1/n2); printf("\npress 5 to make another calculation"); scanf("%d",&i); if (i==5) goto abc; }