In this context, 1 J (Joule) is equal to 1 kPa L (kiloPascal liter) because they both represent the same amount of energy. 1 Joule is the standard unit of energy in the International System of Units (SI), while 1 kiloPascal liter is a non-SI unit that can also be used to measure energy but is more commonly used in certain fields like engineering and fluid mechanics.
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
The relative density of helium is less than 1, which means it is lighter than air. This is why helium is commonly used to make balloons float.
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
#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; }
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.
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]); } } }
The Life and Rhymes of--- - 2005 L-L- Cool J 1-3 was released on: USA: February 2006
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.
#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(); }
1 mL = 0.001 L.
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