A25-ml sample of river water was titrated with 0.0010 m k2cr2o7 and required 8.3 ml to reach the end point. what is the chemical oxygen demand, in milligrams of o2 per liter, of the sample?

First let us calculate the moles of K2Cr2O7 that was supplied.

moles K2Cr2O7 = 0.0010 M * 0.0083 L = 8.3x10^-6 mol

From the chemical formula itself, we see that there are 7 O for every mole of K2Cr2O7 or 3.5 O2. Therefore:

moles O2 = 8.3x10^-6 mol K2Cr2O7 * (3.5 mol O2 / 1 mol K2Cr2O7)

moles O2 = 2.905x10^-5 mol O2

Calculating for the mass of O2 in mg:

mass O2 = 2.905x10^-5 mol O2 * (32 g / mol) * (1000 mg / g)

mass O2 = 0.9296 mg

Therefore the chemical oxygen demand (COD) is:

COD = 0.9296 mg / (0.025 L)

COD = 37.184 mg/L

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

explanation: no but i do not usually comment