c ne
carbon atom, monomer, macromolecule, and polymer.
c. hydrogen
The element with the largest number of moles would be Neon (Ne) since all substances contain the same mass (10g), and Neon has the lowest molar mass among those elements.
The Lewis dot diagram of CH2CHCN should be simple to draw from this structural formula. It starts with a C atom bonded to two H atoms and double bonded to another C atom. This C atom is single bonded to one H atom and single bonded to another C atom. The final C atom is triple bonded to a nitrogen atom.
To determine which substance contains the largest number of moles, you need to divide the given mass of each substance by its molar mass to find the number of moles. The substance with the largest number of moles will have the smallest molar mass. In this case, Neon (Ne) has the smallest molar mass, so it contains the largest number of moles.
largest of a, b, c :a > b ? a > c ? a : c : b > c ? b : c
// return the larger of any 2 numbers int larger (int a, int b) { return a>b?a:b; } // return the largest of any 3 numbers int largest (int a, int b, int c) { return larger (larger (a, b), c)); } // return the middle value of any 3 numbers int middle (int a, int b, int c) { if (a>b) a^=b^=a^=b; // swap a and b (b is now the larger of the two) if (b>c) b^=c^=b^=c; // swap b and c (c is now the largest of all three) return larger (a, b); // return the larger of a and b }
void main() { int a,b,c; clrscr(); printf("Enter the value of a:"); scanf("%d",&a); printf("\nEnter the value of b:"); scanf("%d",&b); printf("\nEnter the value of c:"); scanf("%d",&c); if(a>b) { if(a>c) { if(b>c) { printf("c is smallest\n"); printf("b is middle\n"); printf("a is largest\n"); } else { printf("b is smallest\n"); printf("c is middle\n"); printf("a is largest\n"); } } else { printf("b is smallest\n"); printf("a is middle\n"); printf("c is largest\n"); } } else if(b>c) { if(a>c) { printf("c is smallest\n"); printf("a is middle\n"); printf("b is largest\n"); } else { printf("a is smallest\n"); printf("c is middle\n"); printf("b is largest\n"); } } else { printf("a is smallest\n"); printf("b is middle\n"); printf("c is largest\n"); } getch(); }
Let the numbers be A,B and C and assume that A ≠ B ≠ C 1) Subtract B from A. If the result is a positive number then A > B otherwise B > A. 2) Subtract the greater of A and B from C. If the result is a positive number then C is the largest number otherwise the subtrahend (either A or B) is the largest number.
Find the largest of two, then find the largest of that value and the third value. int* max (int* a, int* b) { return (a*) > (b*) ? a : b; } int* max_of_three (int* a, int* b, int* c) { return max (max (a, b), c); }
for the largest number: #include<stdio.h> void main() { int a,b,c,number,largestnumber; a=99; b=9; c=77; if(a>b) { number=a; } else if(b>c) { number=b; } else { number=c; } largestnumber=number; printf("%d",largestnumber); }
#define max2(a,b) (b>a?b:a) #define max3(a,b,c) (max2(a,max(b,c)))
#define max (a, b) ((a) >= (b)) ? (a) : (b)
#include <iostream.h> #include <conio.h> void main() { clrscr(); int largest(int,int,int); cout<<"Enter 3 Integer Numbers\n"; int a,b,c; cin>>a>>b>>c; int result; result=largest(a,b,c); cout<<"\n\nLargest Value of Inputed is "<<result; getch(); } inline largest(int a,int b,int c) { int z; z=(a>b)?((a>c)?a:c):((b>c)?b:c); return(z); }
A plus b plus c equals d. A is the largest answer b is the smallest answer and d is less than 6?''
Atom economy or atom utilization is simply the Mr or molecular weight of desired product formed divided by the Mr of all the reactants. Say we have a reaction A + B -----> C C + D ------> E AU = (Mr of E/(Mr of A + B + C)) x 100
carbon atom, monomer, macromolecule, and polymer.