O2- in oxides, O1- in peroxides
2 Zn + O2 = 2 ZnO (zinc oxide, a white powder)
Because the sodium metal will react with the oxygen in the alcohol. CH3CH2OH + Na =
Atoms do have the same properties as long as stable form of its corresponding element is not a molecule. For example H1 (Hydrogen) which is a single atom has the same properties that of Hydrogen as it doesn't require to be in a molecular form to be stable. On the other hand O1 (Oxygen) doesn't have the same properties as Oxygen that we breathe as that Oxygen is actually O2 which is actually necessary for it to remain stable.
The proportion by number of atoms in carbon dioxide is 1:2. The atomic weight of carbon is 12.011 and the atomic weight of oxygen is 15.9994. Therefore, the proportion by mass of carbon to oxygen in carbon dioxide is 12.011/(2 X 15.9994) = 0.37536 or 37.536 %, to the justified number of significant digits.
.o1 uF
bulbasaur!i like charizard better!
O2- in oxides, O1- in peroxides
#include<stdio.h> #include<conio.h> #include<string.h> #include<stdlib.h> #include<process.h> void main() { clrscr(); char *s1,*s2,*o1,*o2,temp1,temp2; printf("Enter first statement:"); gets(s1); printf("Enter second statement:"); gets(s2); if(s1[0]!=s2[0]) { printf("Sorry"); getch(); exit(0); } o1[0]=s1[0]; o1[1]='-'; o1[2]='>'; for(int i=3;s1[i]==s2[i];i++) o1[i]=s1[i]; temp1=i; temp2=i; o1[i++]='Z'; o1[i++]='\0'; o2[0]='Z'; o2[1]='-'; o2[2]='>'; int p=3; for(int j=temp1;j<strlen(s1);j++) { o2[p]=s1[j]; p++; } o2[p++]='/'; for(j=temp2;j<strlen(s2);j++) { o2[p]=s2[j]; p++; } o2[p++]='\0'; puts(o1); puts(o2); getch(); }
0.1 litre = 10 centilitres
O1 visa
Usually two oxegen atoms (O2) though can be one (O1).
Usually two oxegen atoms (O2) though can be one (O1).
they catch the ball. but there is to o1 on each team catches against the separate team
700,000,000,000,000.O1
12 teenagers can eat 12 pizzas in one-and-a-half days so they can eat 48 in 6 days I always had difficulty with these problems. It is computed as follows: If P1 people can eat O1 objects in T1 days then P2 people can eat O2 objects in T2 days P1T1/O1 = P2T2/O2 In the example P1 = 1.5 T1 = 1.5 O1 = 1.5 P2 = 12 T2 = 6 solve for O2: O2 = P2 O1 T2/ P1 T1 = 48
6 inches in one half foot.