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Answer to 1 the A N of H?

Updated: 11/3/2022
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1 the Atomic Number of Hydrogen

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What is the difference between multiplication and convolution?

multiplication is point to point and convolustion is point to multi-point ex multiplication-- s[n]=x[n].h[n] s[0]=[x[0].h[0] s[1]=[x[1].h[1] s[2]=[x[2].h[2] . . . .. s[n-1]=[x[n-1].h[n-1] convollustion s[n]=x[n]*h[n] s[0]=[x[0].h[0]+x[0].h[1]+x[0].h[2]+.......+x[0].h[n-1] s[1]=[x[1].h[0]+x[1].h[1]+x[1].h[2]+.......+x[1].h[n-1] s[2]=[x[2].h[2]+x[2].h[1]+x[2].h[2]+.......+x[2].h[n-1] . . . s[n-1]=[x[n-1].h[0]+x[n-1].h[1]+x[n-1].h[2]+.......+x[n-1].h[n-1].


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The Lewis structure for HNNN can be represented as N≡N-H. Each nitrogen atom is bonded to the adjacent nitrogen atom through a triple bond (denoted by ≡) and one nitrogen atom is bonded to a hydrogen atom.


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The geometric-harmonic mean of grouped data can be formed as a sequence defined as g(n+1) = square root(g(n)*h(n)) and h(n+1) = (2/((1/g(n)) + (1/h(n)))). Essentially, this means both sequences will converge to the mean, which is the geometric harmonic mean.


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n n n n n n n n n n n n n n n o o o o o o o o o o o o o t t t t t t t t t t t h h h h h h h h h h h h h h i i i i i i i i i i n n n n n n n n n n g g g g g g gg g g g g g g g gg gg g


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1 the Atomic Number of Hydrogen


What does niu mean niu mean?

n n n n n n n n n n n n n n n o o o o o o o o o o o o o t t t t t t t t t t t h h h h h h h h h h h h h h i i i i i i i i i i n n n n n n n n n n g g g g g g gg g g g g g g g gg gg g


2 1 H is a n of hydrogen?

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The A N of H is one?

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HNNH has 6 sigma bonds. Each nitrogen atom forms two sigma bonds with hydrogen atoms, and the two nitrogen atoms are connected by a single sigma bond.


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#include<stdio.h> typedef struct hole { int id,size,inf,ef,filled,pid,psize; }hole; typedef struct process { int id,size,comp; }process; hole h[50]; process p[50]; int m,n,intfrag,exfrag; void output() { int i; intfrag=exfrag=0; printf("\n\n\nHole Process IF EF\n"); for(i=0;i<m;i++) { printf("\n%d %d %d %d",h[i].size,h[i].psize,h[i].inf,h[i].ef); intfrag=intfrag+h[i].inf; exfrag=exfrag+h[i].ef; } } void final() { int i; for(i=0;i<n;i++) { if(p[i].comp==0) { printf("There is no memory for the process of size %d\n",p[i].size); } } } void input() { int i,max,flag; max=flag=0; printf("\nEnter total no of holes:\n"); scanf("%d",&m); for(i=0;i<m;i++) { h[i].id=i; printf("\nEnter size of hole %d:\n",i+1); scanf("%d",&h[i].size); h[i].inf=0; h[i].ef=h[i].size; h[i].filled=0; h[i].pid=0; h[i].psize=0; } for(i=0;i<m;i++) { if(h[i].size<0) { printf("\nInvalid input as hole size is negative"); flag=1; } if(h[i].size>max) { max=h[i].size; } } printf("\nEnter total no of processes:\n"); scanf("%d",&n); for(i=0;i<n;i++) { p[i].id=i; printf("\nEnter size of process %d:\n",i+1); scanf("%d",&p[i].size); p[i].comp=0; } for(i=0;i<n;i++) { if(p[i].size<0) { printf("\nInvalid input as process size is negative"); flag=1; } if(p[i].size>max) { printf("\nInvalid input as process size is greater than max size of hole\n"); flag=1; } } if(flag==0) { output(); } } void first() { int i,j; for(i=0;i<n;i++) { for(j=0;j<m;j++) { if(p[i].size<=h[j].size && p[i].comp==0 && h[j].filled==0) { p[i].comp=1; h[j].filled=1; h[j].psize=p[i].size; h[j].id=p[i].id; h[j].inf=h[j].size-p[i].size; h[j].ef=0; output(); } } } printf("\n\n\nInternal Fragmentation:%d",intfrag); printf("\n\n\nExternal Fragmentation:%d\n\n",exfrag); final(); } void best() { int min,i,j,sub,var; for(i=0;i<n;i++) { min=999; sub=0; for(j=0;j<m;j++) { if(p[i].size<=h[j].size && p[i].comp==0 && h[j].filled==0) { sub=h[j].size-p[i].size; if(min>sub) { min=sub; var=j; } } } p[i].comp=1; h[var].filled=1; h[var].psize=p[i].size; h[var].id=p[i].id; h[var].inf=h[var].size-p[i].size; h[var].ef=0; output(); } printf("\n\n\nInternal Fragmentation:%d",intfrag); printf("\n\n\nExternal Fragmentation:%d\n\n",exfrag); final(); } void next() { int i,j,a; a=j=0; for(i=0;i<n;i++) { for(j=a;j<m;j++) { if(p[i].size<=h[j].size && p[i].comp==0 && h[j].filled==0) { p[i].comp=1; h[j].filled=1; h[j].psize=p[i].size; h[j].id=p[i].id; h[j].inf=h[j].size-p[i].size; h[j].ef=0; output(); a=(j+1)%m; } } } printf("\n\n\nInternal Fragmentation:%d",intfrag); printf("\n\n\nExternal Fragmentation:%d\n\n",exfrag); final(); } void main() { int ch; printf("\n\n\n**********MEMORY ALLOCATION**********\n\n"); printf("\n1.First Fit Algorithm\n2.Best Fit Algorithm\n3.Next Fit Algorithm"); printf("\nEnter your choice:\n"); scanf("%d",&ch); switch(ch) { case 1: input(); first(); break; case 2: input(); best(); break; case 3: input(); next(); break; } } Written by: Fabianski Benjamin


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In N2H4, the oxidation number of nitrogen (N) is -2. Each hydrogen (H) has an oxidation number of +1.