What does one J equal
A Joule can equal a (Pascal * Meter^3)
Pacals are force units described as Force/Area
A Liter is a 3 Dimensional unit
Using Boyle's Law (P1V1 = P2V2), we can calculate the new volume needed to contain the gas at a pressure of 50 kPa. Given that P1=100 kPa, V1=5.0 L, and P2=50 kPa, we can solve for V2: (100 kPa * 5.0 L) / 50 kPa = V2. The new volume, V2, required to hold the gas at 50 kPa would be 10.0 L.
how many moles are contained in 4.67 L sample of gas at 33 degrees celcius and 199 kpa
Helium has a density of 0.1785 kg/m3 = 0.1785 g/L = 0.0001785 kg/dm3 = 0.0001785 kg/L = 0.0001785 g/cm3 = 0.0001785 g/mL.Density at stp (0 °C and 101.325 kPa): 0.1786 g/L
The mass, 5.00 g, must be converted to moles. The molar mass of Cl2 is 71 g/mole. Also convert 750 ml to 0.75 L 5.00 g x (1mole / 71 g) = 0.0704 mol (PV / nT)a = (PV / nT)b ; where a and b represent different conditions (120 kpa)(0.75 L) / [(0.0704 mole)(T)] = (101.325 kpa)(22.4 L) / [(1 mole)(273 K)] Solve for T, the unknown temperature. T = (1 mole)(273 K) / [(101.325 kpa)(22.4 L)] x [(120 kpa)(0.75 L)] / (0.0704 mol) T = 153.7 K round to 154 K
1 l is equal to 1000 ml. 1.8l is equal to 1800 ml.
3.8 kPa
Using Boyle's Law (P1V1 = P2V2), we can calculate the new volume needed to contain the gas at a pressure of 50 kPa. Given that P1=100 kPa, V1=5.0 L, and P2=50 kPa, we can solve for V2: (100 kPa * 5.0 L) / 50 kPa = V2. The new volume, V2, required to hold the gas at 50 kPa would be 10.0 L.
#include<stdio.h> #include<string.h> int max(int a,int b) { return a>b?a:b; }//end max() int main() { char a[]="xyxxzxyzxy"; char b[]="zxzyyzxxyxxz"; int n = strlen(a); int m = strlen(b); int i,j; for(i=n;i>=1;i--) a[i] = a[i-1]; for(i=m;i>=1;i--) b[i] = b[i-1]; int l[n+1][m+1]; printf("\n\t"); for(i=0;i<=n;i++) { for(j=0;j<=m;j++) { if(i==0 j==0) l[i][j]=0; else if(a[i] == b[j] ) l[i][j] = l[i-1][j-1] + 1; else l[i][j] = max(l[i][j-1],l[i-1][j]); printf("%d |",l[i][j]); } printf("\n\t"); } printf("Length of Longest Common Subsequence = %d\n",l[n][m]); return 0; }
8.31 L-kPa/mol-K
8.31 L-kPa/mol-K
#include<stdio.h> main() { printf("Enter Order Of The Matrix\n"); int l,m; int i,j,k,t,x; scanf("%d%d",&l,&m); int a[l][m],temp[l*m]; printf("Enter The Elements Row Wise...\n"); for(i=0;i<l;i++) { for(j=0;j<m;j++) { scanf("%d",&a[i][j]); } } printf("Matrix A \n "); for(i=0;i<l;i++) { printf("\n"); for(j=0;j<m;j++) { printf("%d\t",a[i][j]); } } /*----------------------------------------------------------------------*/ x=0; for(i=0;i<l;i++) { for(j=0;j<m;j++) { temp[x]=a[i][j]; x++; } } for(i=0;i<l*m;i++) { for(j=0;j<(l*m)-1;j++) { if(temp[j]>temp[j+1]) { t=temp[j]; temp[j]=temp[j+1]; temp[j+1]=t; } } } x=0; for(i=0;i<l;i++) { for(j=0;j<m;j++) { a[i][j]=temp[x]; x++; } } printf("\nMatrix A After Sort \n "); for(i=0;i<l;i++) { printf("\n"); for(j=0;j<m;j++) { printf("%d\t",a[i][j]); } } }
More pressure means less volume. Calculate the ratio of pressure, then divide the 4.2 liters by that ratio.This assumes: * That the temperature doesn't change. * That the gas behaves like an ideal gas.
The Life and Rhymes of--- - 2005 L-L- Cool J 1-3 was released on: USA: February 2006
1 mL = 0.001 L.
#include<stdio.h> #include<conio.h> void main() { int i,j,k=0,l; clrscr(); for(i=1;i<=5;i++) { for(j=1;j<=i;j++) { k++; printf("%d ",k%2); } for(l=i;l<=4;l++) { k++; } printf("\n"); } getch(); }
#include<stdio.h> #include<conio.h> void main() { clrscr(); int n,i,j,temp,a[50]; printf("Enter how many elements="); scanf("%d",&n); printf("Enter %d elements",n); for(i=1;i<=n;i++) { scanf("%d",&a[i]); } for(i=1;i<=n-1;i++) { for(j=1;j<=n-1;j++) { if(a[j]>a[j+1]) { temp=a[j]; a[j]=a[j+1]; a[j+1]=temp; } } } printf("\nSorted Array:\n"); for(i=1;i<=n;i++) { printf("\n%d",a[i]); } getch(); }
Use for loop declare string array str[] and string variable l= string length of string array j=l for i=0 to i=l/2 then temp=str[i] str[i]=str[j-1] str[j-1]=temp j=j-1 now print str array it will be reversed