To determine [H3O+] in 0.00250 M HNO2(aq)(Ka=4.5.×10−4), the first equation we would obtain in solving for x=[H3O+] isKa=[H3O+][NO2−][HNO2]=x20.00250−x Normally when x<5% Minitial, we solve this equation using the quadratic formula. This tutorial will lead you through an alternative method called successive approximations.

Now you will solve the same problem as above, but using the quadratic formula instead of iterations, to show that the same value of x is obtained either way. Using the quadratic equation to calculate [H3O+] in 0.00250 M HNO2, what are the values of a, b, c and x , where a, b, and c are the coefficients in the quadratic equation ax2+bx+c=0, and x is [H3O+]? Recall that Ka=4.5×10−4 .

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b

answerspecific gravity;