A stream leaving a sewage pond (containing 80 mg/L of sewage) moves as a plug with a velocity of 40 m/hr. A concentration of 50 mg/L is measured 5,000 m downstream. What is the 1st order decay rate constant in the stream?

Decay rate constant,k  = 0.00376/hr

Explanation:

IsT Order  Rate of reaction is given as

In At/ Ao = -Kt

where [A]t is the final concentration at time  t  and  [A]o  is the inital concentration at time 0, and  k  is the first-order rate constant.

Initial concentration = 80 mg/L

Final concentration = 50 mg/L

Velocity = 40 m/hr

Distance= 5000 m

Time taken = Distance / Time

              5000m / 40m/hr = 125 hr

In At/ Ao = -Kt

In 50/80 = -Kt

-0.47 = -kt

- K= -0.47 / 125

k = 0.00376

Decay rate constant,k  = 0.00376/hr

answer:

// here is code in C++.

#include <bits/stdc++.h>

using namespace std;

//main function

int main()

{

  // variable

 int n;

 cout<<"enter a number:";

 // read the input

 cin>>n;

 // find the square root of the absolute value

 float square_root=sqrt(abs(n));

 // print the square_root

 cout<<"square root of "<<n<<" is: "<<square_root<<endl;

  return 0;

}

Explanation:

Read a number from user and assign it to "n".Then find its absolute value with abs() function.Then calculate the square root of the absolute value with sqrt() function.Print the square root of the number.

Output:

enter a number:-17

square root of -17 is: 4.12311

                                                         This is easy

                                                     sinserly Natalie 7 grader (this is not that hard)