Consider the titration of50.0 mL of 0.217 M hydrazoic acid (HN3, Ka=2.6 x 10^-5) with 0.183M NaOH. The pH of the soluion before any NaOH is added is____ The answer is 2.62 I cant figure out what to do after you get the moles Solution Concentration of acid = 0.217M [H+] = sqrt (Ka* C) [H+] = 2.37 * 10 ^ -3 so pH = -log([H+]) = 2.624.

1. Create a function mean comp that compares the mean value of the odd numbered elements with
the mean value of the even numbered elements and returns: if they are equal, if the mean of the
odd numbered elements is greater, if the mean of the even numbered elements is greater.
Solution
A sample Java Program methods
public static void main(String []args){
float[] list = {1.0f, 7.0f, 8.0f, 1.0f};
System.out.println(meancomp(list));
}
static int meancomp(float[] list) {
float evenSum = 0.0f;
float oddSum = 0.0f;
for (int i = 0; i < list.length; i++) {
if (i % 2 == 1)
evenSum = evenSum + list[i];
else
oddSum = oddSum + list[i];
}
int noOfEvenElements = list.length / 2;
int noOfOddElements = list.length / 2 + list.length % 2;
float evenMean = evenSum / (float) noOfEvenElements;
float oddMean = oddSum / (float) noOfOddElements;
int result = Float.compare(evenMean, oddMean);
if (result == 0)
return 0;
else if (result < 0)
return 1;
else
return 2;
}